Math in Focus

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 9 Lesson 9.2 Stem-and-Leaf Plots detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 9 Lesson 9.2 Answer Key Stem-and-Leaf Plots

Math in Focus Grade 7 Chapter 9 Lesson 9.2 Guided Practice Answer Key

Display data in a stem-and-leaf plot.

Question 1.
The data show the math quiz scores for 15 students. Display the data in a stem-and-leaf plot.

STEP 1: Identify the least and greatest values, and identify the stems.
STEP 2: Place a leaf on the plot for each item of data.
STEP 3: Arrange the leaves in ascending order for each number on the stem.
STEP 4: Write a title and a key for the stem-and-leaf plot.
Answer:
Step 1: Least value:8; greatest value:44; stems:0, 1, 2, 3, 4
Step 2:  Constructing leaf for item of data

Step 3: Now arrange in ascending order

Step 4: Write the title, Remember, a Stem and Leaf plot can have multiple sets of leaves.
0|8 represents 8
1|0 represents 10
1|1 represents 11
1|1 represents 11
1|2 represents 12
1|3 represents 13
1|5 represents 15
1|8 represents 18 and so on… likewise, we represent upto
4|4 represents 44.

Solve.

Question 2.
The data below show the attendance, in the number of days, for some students during the first month of school.

a) Display the data in a stem-and-leaf plot.
Answer:
The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties.
Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

1|5 represents 15 days
1|8 represents 18 days
1|9 represents 19 days
1|9 represents 19 days
2|0 represents 20 days
2|0 represents 20 days
Likewise, we have to write all the representations.

b) How many students are represented by the data?
Answer: 25
Count the number of leaves present in the Stem-and-leaf plot
5, 8, 9, 9, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3.
1|5 represents 15 days
1|8 represents 18 days
1|9 represents 19 days
1|9 represents 19 days
2|0 represents 20 days
2|0 represents 20 days
2|0 represents 20 days
2|0 represents 20 days
2|0 represents 20 days
2|1 represents 21 days
2|1 represents 21 days
2|1 represents 21 days
2|2 represents 22 days
2|2 represents 22 days and so on… up to
2|3 represents 23 days.
Total 25.

c) What is the most common attendance, or mode?
Answer: 23
Mode definition: A mode is defined as the value that has a higher frequency in a given set of values. It is the value that appears the most number of times.
according to the definition:
23 has the highest frequency. It occurs more time.

d) What is the mean attendance?
Answer:
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
The given data:
15, 18, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 23, 23, 23.
Mean=15+18+19+19+20+20+20+20+20+21+21+21+22+22+22+22+22+23+23+23+23+23+23+23+23/25
Mean=528/25
Mean=21.12 days

e) What is the median attendance?
Answer:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data:
15, 18, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 23, 23, 23.
The middlemost number is 22, so the median is 22.

Solve. Round your answer to the nearest tenth if necessary.

Question 3.
The stem-and-leaf plot shows the heights, in inches, of the players on two football teams.

a) How many players are on each team?
Answer:
Observe the above-given stem-and-leaf plot.
There are two teams: Team A and Team B
In Team A, count the number of leaves:
The given data in the team A is 67, 68, 68, 69, 70, 70, 70, 71, 71, 72, 73.
There are 11 players in Team A.
Now coming to Team B:
The given data is 68, 68, 68, 70, 70, 71, 71, 71, 72, 72, 72
There are 11 players in Team B.

b) What is the median height for each team?
Answer:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data in the team A is 67, 68, 68, 69, 70, 70, 70, 71, 71, 72, 73.
Median for Team A:
The middlemost number is 70
Median for Team B:
The given data is 68, 68, 68, 70, 70, 71, 71, 71, 72, 72, 72
The middlemost number is 71

c) What is the mean height for each team?
Answer:
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
Finding Mean for Team A:
The given data in the team A is 67, 68, 68, 69, 70, 70, 70, 71, 71, 72, 73.
Mean for Team A= 67+68+68+69+70+70+70+71+71+72+73/11
Mean for Team A= 769/11
Mean for Team A=69.9
Finding mean for Team B:
The given data is 68, 68, 68, 70, 70, 71, 71, 71, 72, 72, 72
Mean for Team B=68+68+68+70+70+71+71+71+72+72+72/11
Mean for Team B=773/11
Mean for Team B=70.3

d) On average, which team has taller players?
Answer:
Median of Team A:70
Mean of Team A:69.9
Median of Team B:71
Mean of Team B:70.3
According to the above observations, we can write which team has taller players.
On average Team B has taller players because both it’s mean and median are greater. But the means and the medians for both teams are very close, so the players on Team B are not much taller on average.

Solve.

Question 4.
The weights, in pounds, of some newborn babies at a hospital are shown in the stem-and-leaf plot.

a) Find the median weight.
Answer:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data:
3.1, 6.0, 6.1, 6.2, 6.5, 6.5, 6.5, 6.6, 6.8, 7.
Median=6.5+6.5/2
Median=13/2
Median=6.5lb

b) Find the mean weight.
Answer:
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
The given data:
3.1, 6.0, 6.1, 6.2, 6.5, 6.5, 6.5, 6.6, 6.8, 7.
Mean=3.1+6.0+6.1+6.2+6.5+6.5+6.5+6.6+6.8+7/10
Mean=61.3/10
Mean=6.13lb

c) Compare the median and mean. Explain the differences.
Answer:
The median is 6.5lb while the mean is 6.13lb. The value of the median is much greater due to the presence of an outlier with the value of 3.1lb

d) What is the range of the data?
Answer:
In statistics, the range is the spread of your data from the lowest to the highest value in the distribution. It is a commonly used measure of variability.
– The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
R=H-L
– R = range
– H = highest value
– L = lowest value
The range is the easiest measure of variability to calculate. To find the range, follow these steps:
1. Order all values in your data set from low to high.
2. Subtract the lowest value from the highest value.
The given data:
3.1, 6.0, 6.1, 6.2, 6.5, 6.5, 6.5, 6.6, 6.8, 7.
Range=7-3.1
Range=3.9
Range=4 (rounded to the nearest 10)

e) Explain how the outlier 3.1 pounds affects the range of the data. How does the outlier affect the mean of the data?
Answer:
The range is the difference between the least and the greatest values in the data set. So, the range is greater because of the outlier. The mean is also smaller because of the outlier.

Math in Focus Course 2B Practice 9.2 Answer Key

Display the data using a stem-and-leaf plot. Indicate the range and median of each data set.

Question 1.
The masses, in grams, of russet potatoes, are 89, 110, 81, 92, 100, 96, 101, 109, 105, and 112.
Answer:
The given data: 89, 110, 81, 92, 100, 96, 101, 109, 105, and 112.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

8|1 represents 81 grams.
Range: The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
The given data: 81, 89, 92, 96, 100, 101, 105, 109, 110, 112.
R=H-L
– R = range
– H = highest value
– L = lowest value
The range is the easiest measure of variability to calculate. To find the range, follow these steps:
1. Order all values in your data set from low to high.
2. Subtract the lowest value from the highest value.
Highest value:112
Lowest value:81
Range=112-81
Range=31 grams.
Median:
The given data: 81, 89, 92, 96, 100, 101, 105, 109, 110, 112.
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median=100+101/2
Median=201/2
Median=100.5grams.

Question 2.
The lengths of salmon, in inches, are 28, 31, 32, 29, 26, 33, 29, 28, 30, and 31.

Answer:
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

2|6 represents 26 inches.
Range: The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
The given data: 26, 28, 28, 29, 29, 30, 31, 31, 32, 33
R=H-L
– R = range
– H = highest value
– L = lowest value
The range is the easiest measure of variability to calculate. To find the range, follow these steps:
1. Order all values in your data set from low to high.
2. Subtract the lowest value from the highest value.
Highest value:33
Lowest value:26
Range=33-26
Range=7 inches
Median:
The given data: 26, 28, 28, 29, 29, 30, 31, 31, 32, 33
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median=29+30/2
Median=59/2
Median=29.5 inches

Question 3.
The average hourly rate, in dollars, of workers in 8 professions are 26, 13, 10, 17, 12, 15, 14, and 15.
Answer:
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

1|0 represents 10$
the given data: 10, 12, 13, 14, 15, 15, 17, 26
range=maximum value-minimum value
maximum value=26
minimum value=10
range=26-10
range=16$
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
the given data: 10, 12, 13, 14, 15, 15, 17, 26
Median=14+15/2
Median=29/2
Median=14.5$
Mean = (Sum of all the observations/Total number of observations)
The given data: 26, 13, 10, 17, 12, 15, 14, and 15
Mean=26+13+10+17+12+15+14+15/8
Mean=122/8
Mean=15.25

Question 4.
The scores of a basketball competition are 70, 56, 93, 92, 72, 74, 76, 72, 78, and 59.
Answer:
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

5|6 represents 56 points
The given data: 56, 59, 70, 72, 72, 74, 76, 78, 92, 93
range=maximum value-minimum value
range=93-56
range=37 points
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data: 56, 59, 70, 72, 72, 74, 76, 78, 92, 93
Median=72+74/2
Median=146/2
Median=73 points.

Question 5.
The ages of passengers on board a bus are 33, 37, 5, 11, 63, 55, 56, 13, 15, and 19.
Answer:
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

0|5 represents 5 years
the given data: 5, 11, 13, 15, 19, 33, 37, 55, 56, 63
Range=maximim value-minimum value
Range=63-5
range=58 years.
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
the given data: 5, 11, 13, 15, 19, 33, 37, 55, 56, 63
Median=19+33/2
Median=52/2
Median=26 years.

Use the data in the table to answer the following.

The table shows the weights, in pounds, of 30 pumpkins.

Question 6.
Draw the stem-and-leaf plot of the data.
Answer:
The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties.
Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

0|8 represents 8 lb

Question 7.
What is the range of the data?
Answer:
The given data: 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15, 16, 16, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 21, 22
In statistics, the range is the spread of your data from the lowest to the highest value in the distribution. It is a commonly used measure of variability.
– The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
R=H-L
– R = range
– H = highest value
– L = lowest value
The range is the easiest measure of variability to calculate. To find the range, follow these steps:
1. Order all values in your data set from low to high.
2. Subtract the lowest value from the highest value.
Range=22-8
Range=14 lb

Question 8.
What is the mode?
Answer:
The given data: 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15, 16, 16, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 21, 22
Mode definition: A mode is defined as the value that has a higher frequency in a given set of values. It is the value that appears the most number of times.
according to the definition:
16 has the highest frequency. It occurs more time.

Question 9.
What is the median?
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data: 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15, 16, 16, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 21, 22
Median=15+16/2
Median=31/2
Median=15.5 lb
The median is nothing but Q2. Now we have to find Q1(lower quartile), Q3 (upper quartile)
Q1 data set: 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15
Q1=11
Q1 is the median of lower quartile
Q3 data set: 16, 16, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 21, 22
Q3=18
Q2 is the median of upper quartile.

Question 10.
What is the interquartile range?
Answer:
The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third Quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus lower quartile.
Formula:
The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q_­1
where Q₁ is the first quartile and Q₃ is the third quartile of the series.
IQR=Q3-Q1
IQR=18-11
IQR=7 lb

Use the data in the table to answer the following.

The table shows the attendance, in working days, of workers in a manufacturing plant over a period of one year.

Question 11.
Draw a stem-and-leaf plot.
Answer:
The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties.
Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves

2 2|0 represents 220 days

Question 12.
What is the mean number of working days in for this particular year?
Answer:
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
The given data: 220, 230, 231, 232, 232, 233, 234, 234, 235, 236, 237, 238, 239, 239, 239, 239, 239, 239, 239, 239, 239, 240, 240, 240, 240, 240, 240, 240, 240, 240.
Mean=7103/30
Mean=236.7
Mean=237 (rounded to near number)

Question 13.
What is the mode?
Answer:
Mode definition: A mode is defined as the value that has a higher frequency in a given set of values. It is the value that appears the most number of times.
according to the definition:
The given data: 220, 230, 231, 232, 232, 233, 234, 234, 235, 236, 237, 238, 239, 239, 239, 239, 239, 239, 239, 239, 239, 240, 240, 240, 240, 240, 240, 240, 240, 240.
239 days and 240 days have the highest frequency. It occurs more time.

Question 14.
What is the median?
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data: 220, 230, 231, 232, 232, 233, 234, 234, 235, 236, 237, 238, 239, 239, 239, 239, 239, 239, 239, 239, 239, 240, 240, 240, 240, 240, 240, 240, 240, 240.
Median=239+239/2
Median=478/2
Median=239 days.

Solve. Show your work.

Question 15.
The weights of 10 pumpkins, in pounds, harvested at a farm is shown in the stem-and-leaf plot.

Compare the median and mean. Explain the difference in the values.
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data: 2.1, 2.2, 2.6, 2.7, 3.1, 3.3, 3.6, 3.9, 4.1, 8.1
Median=3.1+3.3/2
Median=6.4/2
Median=3.2 lb
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
Mean=2.1+2.2+2.6+2.7+3.1+3.3+3.6+3.9+4.1+8.1/10
Mean=35.7/10
Mean=3.57 lb
The median is 3.2 pounds while the mean is 3.57 pounds; the value of the mean is greater due to the presence of the outlier with the value of 8.1 pounds.

Question 16.
The speeds, in miles per hour, tracked by a police patrol car along a certain highway, are shown in the stem-and-leaf plot.

Explain how the outliers 40, 43, and 45 miles per hour affect the range of the data. How do these outliers affect the mean of the data?
Answer:
The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
R=H-L
– R = range
– H = highest value
– L = lowest value
Range=maximum value- minimum value
maximum value=81
minimum value=40
Range=41
The range is the difference between the least and the greatest values in the data set. So, the range is greater because of the outlier.

Question 17.
Math Journal The waiting time, in minutes, at a certain bus interchange for two bus routes on a particular day are recorded in the plot.

With the data shown, discuss some advantages and disadvantages of the stem-and-leaf plot.
Answer:
Advantages:
– It is easy to identify the minimum and maximum values from the data set.
– The range, median, and mode can be identified easily from a stem-and-leaf plot.
– The individual data values are preserved, unlike in a histogram.
Disadvantages:
– When the range is too narrow, there would be too many leaves.
– When the range is wide, there would be some stems that have no leaf, especially when outliers cause the range to widen.
– Bus route B has a relatively small range, with data concentrated within a few stems. Bus route A has a range that is much wider. So, when both bus routes are plotted in a double stem-and-leaf plot, bus route A has many stems without any leaf and the stem is much longer.

Question 18.
Math Journal Describe a data set that could be represented by a stem-and-leaf plot. Explain how using a stem-and-leaf plot could help you analyze the data set.
Answer:
The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties.
Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.
Stem-and-leaf plot graphs are usually used when there are large amounts of numbers to analyze.
To compare two sets of data, you can use a back-to-back stem-and-leaf plot. For instance, if you want to compare the scores of two sports teams, you can use the stem-and-leaf plot.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 9 Lesson 9.1 Interpreting Quartiles and Interquartile Range detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 9 Lesson 9.1 Answer Key Interpreting Quartiles and Interquartile Range

Math in Focus Grade 7 Chapter 9 Lesson 9.1 Guided Practice Answer Key

Calculate.

Question 1.
The table shows the average monthly temperatures in degrees Fahrenheit in New York. Find the range of the these temperatures.

Answer:
Definition of range: In statistics, the range of a data set is a measure of the spread or the dispersion of the observations. It is the difference between the largest and the smallest observed values in a data set.
Formula:
Range=max-min
Where max=maximum value in a data set.
Min=Minimum value in a data set.
How to calculate range steps:
1. List the elements of your data set.
2. Identify the highest and lowest numbers in the set.
3. Subtract the smallest number in your data set from the largest number.
4. Label the range clearly.
The given temperature data set: 31.5, 33.6, 42.4, 52.5, 62.7, 71.6, 76.8, 75.5, 68.2, 57.5, 47.6, 36.6
Now arrange the data set from lowest to highest:
31.5, 33.6, 36.6, 42.4, 47.6, 52.5, 57.5, 62.7, 68.2, 71.6, 75.5, 76.8
Range= 76.8-31.5
Range=45.3

Complete.

Question 2.
The heights, in centimeters, of the tomato seedling plants in a greenhouse are listed below. Find the first, second, and third quartiles of the heights of the seedling plants.

First arrange the heights in ascending order:
Q₂ is the median of the data. So, Q₂ =
Q₁ is the median of the lower half of the data:
Q₁ =
Q₃ is the median of the upper half of the data:
Q₃ =
Answer:
– In statistics, we generally deal with a large amount of numerical data. We have various concepts and formulas in statistics to evaluate the large data. One of such best applications is quartiles.
– Quartiles are the values that divide the given data set into four parts, making three points. Thus, quartiles are the values that divide the given data into three quarters. The middle part of the quartiles tells the central point of distribution. The difference between the higher and lower quartiles so formed gives the interquartile range.
– The quartiles formula is used to divide a given set of numbers into quarters. There are three quartiles formed, dividing the given data into four quarters. We are arranging the data in ascending order. The first quartile lies between the first number and the median, the second quartile is the median, and the third quartile lies between the median and the last number.
When the set of numbers is arranged in ascending order, then the formula used to calculate the quartiles are given below:
1. First quartile(Q1)=(n+1/4)th term
2. Second quartile(Q2)=(n+1/2)th term.
3. Third quartile=(Q3)=[3(n+1)/4)]th term.
The given data set: 9.3, 4.3, 5.2, 3.9, 10, 7, 6, 6.4, 9.5, 7.7, 10.6, 4.8, 8, 3.2
Now arrange the set from lowest to highest:
3.2, 3.9, 4.3, 4.8, 5.2, 6, 6.4, 7, 7.7, 8, 9.3, 9.5, 10, 10.6
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=6.4+7/2
Q2=6.5
Q2 is the median of the data.
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Q1=3.2, 3.9, 4.3, 4.8, 5.2, 6, 6.4
Q1=4.8
Q₁ is the median of the lower half of the data
Q3=7, 7.7, 8, 9.3, 9.5, 10, 10.6
Q3=9.3
Q₃ is the median of the upper half of the data

Solve.

Question 3.
Out of 50 multiple choice questions, 19 students scored the following numbers of correct answers.

Find the interquartile range and interpret what it means.
Answer:
the given data set: 36, 49, 45, 30, 27, 50, 45, 44, 20, 18, 14, 19, 31, 50, 26, 33, 10, 42, 30
Now arrange the data into lowest to highest:
10, 14, 18, 19, 20, 26, 27, 30, 30, 31, 33, 36, 42, 44,  45, 45, 49, 50, 50
Definition of interquartile range: The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third
A quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus the lower quartile.
Interquartile range formula: The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below:
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q1
Where Q1 is the first quartile, Q3 is the third quartile.
How to calculate Interquartile range:
The procedure to calculate the interquartile range is given as follows:
– Arrange the given set of numbers into increasing or decreasing order.
– Then count the given values. If it is odd, then the centre value is median otherwise obtain the mean value for two centre values. This is known as the Q₂ value. If there are an even number of values, the median will be the average of the middle two values.
– Median equally cuts the given values into two equal parts. They are described as Q₁ and Q₃ parts.
– The median of data values below the median represents Q₁.
– The median of data values above the median value represents Q₃.
– Finally, we can subtract the median values of Q₁ and Q₃.
– The resulting value is the interquartile range.
Q2=31
Q2 is the median of the data.
n=19 (odd)
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
Q1=10, 14, 18, 19, 20, 26, 27, 30, 30
Q1=20
Q1 is the lower half of the data
Q3=33, 36, 42, 44,  45, 45, 49, 50, 50
Q3=45
Now we know Q1 and Q3
Interquartile range=Q3-Q1
Interquartile range=45-20
Interquartile range=25

Calculate.

Question 4.
A survey of 50 people asking the number of books they read in a year produced the following results.

a) Find the lower quartile, the upper quartile, and the interquartile range.
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Now calculate the cumulative frequency
Cumulative frequency is defined as a running total of frequencies. The frequency of an element in a set refers to how many of that element there are in the set. Cumulative frequency can also be defined as the sum of all previous frequencies up to the current point.

n=50
n/2=50/2=25
Median=1/2[(n/2)th term + (n/2+1)th term]
Median=1/2[ 25 + 26]
Median=1/2[51]
Median=25.5
Now check for the number, if the number was not there then check for the 25.5 above number and write the certain books
25.5 is not there, the next nearest number is 33 and the number of books they read is 5.
Median Q2=5
Q1 lower quartile=(n/4)th term
Q1=50/4
Q1=12.5
check for the next nearest number for 12.5 which is 13 and the corresponding number is 2
Therefore, Q1 is 2
Q3 upper quartile=(3n/4)th term
Q3=(3*50/4)
Q3=150/4
Q3=37.5
37.5 is not there, the next nearest number is 40 and the corresponding number is 6.
Q3=6
Interquartile range=Q3-Q1
Interquartile range=6-2
Interquartile range=4

b) Interpret what the interquartile range means.
Answer:
Definition of interquartile range: The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third
A quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus the lower quartile.
Interquartile range formula: The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below:
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q1
Where Q1 is the first quartile, Q3 is the third quartile.
How to calculate Interquartile range:
The procedure to calculate the interquartile range is given as follows:
– Arrange the given set of numbers into increasing or decreasing order.
– Then count the given values. If it is odd, then the centre value is median otherwise obtain the mean value for two centre values. This is known as the Q₂ value. If there are an even number of values, the median will be the average of the middle two values.
– Median equally cuts the given values into two equal parts. They are described as Q₁ and Q₃ parts.
– The median of data values below the median represents Q₁.
– The median of data values above the median value represents Q₃.
– Finally, we can subtract the median values of Q₁ and Q₃.
– The resulting value is the interquartile range.

Technology Activity

Materials
spreadsheet software

USE SPREADSHEET SOFTWARE TO FIND QUARTILES, INTERQUARTILE RANGE, AND RANGE

Work in pairs.

Step 1: Randomly pick 10 students. Count the number of letters in their first names and enter the values in a row of cells.

Step 2: Choose another 5 cells for Q₁, Q₂, Q₃, interquartile range, and range.

Step 3: Use the spreadsheet software’s function for finding quartiles, maximums, and minimums of data. In the 5 cells you choose in Step 2 generate statistics for the 10 values you entered in Step 1.

Step 4: What is the median of the data? What does this tell you about the first names in the data set? What does the interquartile range tell you about the first names in the data set?

Step 5: Randomly pick another 10 students and repeat the process. Compare the statistics for the two data sets. What differences do you notice in the interquartile range for the two data sets?

Math Journal Suppose you add a data value for a student whose first name has many more letters than anyone else’s. How would adding this data value affect the range and the interquartile range of the data set? Explain your thinking.
Answer:

Math in Focus Course 2B Practice 9.1 Answer Key

Find the range of each set of data values.

Question 1.
Weights of rabbits in pounds: 15, 9, 10, 22, 14.5, 12.5, 16, 8, and 11.
Answer:
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
The given data set: 15, 9, 10, 22, 14.5, 12.5, 16, 8, and 11.
Highest value=22
Lowest value=8
Range=highest value-lowest value
Range=22-8
Range=14

Question 2.
Number of pairs of shoes sold each day for a week at a retail shop: 67, 63, 66, 60, 58, 70, and 75.
Answer:
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
The given data set: 67, 63, 66, 60, 58, 70, and 75.
Highest value=75
Lowest value=58
Range=75-58
Range=17

Question 3.
Time taken in seconds to solve a puzzle by a group of contestants: 20, 25.6, 15.4, 20, 22.8, 18.7, 19, 24, 13.4, and 16.
Answer:
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
The given data set: 20, 25.6, 15.4, 20, 22.8, 18.7, 19, 24, 13.4, and 16
Highest value=25.6
Lowest value=13.4
Range=25.6-13.4
Range=12.2

Question 4.
Average monthly rainfall in centimeters in a city:
18.3, 17.8, 17.3, 8.9, 6.1, 2.0, 1.4, 2.2, 2.8, 5, 12.7, and 14.
Answer:
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
The given data set: 18.3, 17.8, 17.3, 8.9, 6.1, 2.0, 1.4, 2.2, 2.8, 5, 12.7, and 14.
Highest value=18.3
Lowest value=1.4
Range=18.3-1.4
Range=16.9

Find the first, second, and third quartiles of each set of data values.

Question 5.
Science test scores of 13 students:
78, 63, 56, 85, 62, 59, 78, 90, 83, 63, 84, 66, and 63.
Answer:
The given data set: 78, 63, 56, 85, 62, 59, 78, 90, 83, 63, 84, 66, and 63.
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
56,59,62,63,63,63,66,78,78,83,84,85,90
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=66
n=13
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
Q1=56,59,62,63,63,63
Q1=62+63/2
Q1=125/2
Q1=62.5
Thus, the first quartile Q1 is 62.5
Q3=78,78,83,84,85,90
Q3=83+84/2
Q3=167/2
Q3=83.5

Question 6.
Scores of a basketball team in a series of games:
82, 66, 70, 68, 54, 77, 80, 70, 82, and 65.
Answer:
The given data set: 82, 66, 70, 68, 54, 77, 80, 70, 82, and 65.
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
54, 65, 66, 68, 70, 70, 77, 80, 82, 82
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2= 70+70/2
Q2=140/2
Q2=70
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q₂.
n=10
Q1=54, 65, 66, 68, 70
Q1=66
Q₁ is the median of the lower half of the data
Q3=70, 77, 80, 82, 82
Q3=80
Q3 is the median of the upper half of the data

Question 7.
Ages of 20 children at a party:
6, 9, 5, 12, 7, 8, 9, 10, 7, 7, 7, 6, 9, 9, 12, 10, 8, 10, 11, and 9.
Answer:
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
5, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 11, 12, 12
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=9+9/2
Q2=18/2
Q2=9
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q₂.
n=20
Q1=5, 6, 6, 7, 7, 7, 7, 8, 8, 9
Q1=7+7/2
Q1=14/2
Q1=7
Q₁ is the median of the lower half of the data
Q3=9, 9, 9, 9, 10, 10, 10, 11, 12, 12
Q3=10+10/2
Q3=20/2
Q3=10
Q3 is the median of the upper half of the data

Question 8.
Heights of 15 tomato plants in inches:
36, 27, 18, 40, 29, 43, 24, 20, 15, 31, 39, 22, 27, 30, and 20.
Answer:
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
15, 18, 20, 20, 22, 24, 27, 27, 29, 30, 31, 36, 39, 40, 43
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=27
n=15
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
Q1=15, 18, 20, 20, 22, 24, 27
Q1=20
Q3=29, 30, 31, 36, 39, 40, 43
Q3=36

Use the information and table below to answer questions 9 to 12.

To create greater public awareness of the environment, 22 volunteers collected cans for recycling. The table shows the number of cans each volunteer collected.

Question 9.
Find the range of the number of cans collected.
Answer:
The given data: 25, 105, 100, 47, 61, 82, 75, 44, 61, 70, 53, 38, 50, 73, 54, 80, 36, 25, 28, 24, 110, 78
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
range=110-24
range=86

Question 10.
Calculate Q₁, Q₂, and Q₃.
Answer:
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
24, 25, 25, 28, 36, 38, 44, 47, 50, 53, 54, 61, 61, 70, 73, 75, 78, 80, 82, 100, 105, 110
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=54+61/2
Q2=115/2
Q2=57.5
n=22 (even number)
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q₂.
Q1=24, 25, 25, 28, 36, 38, 44, 47, 50, 53, 54
Q1=38
Q3=61, 61, 70, 73, 75, 78, 80, 82, 100, 105, 110
Q3=78

Question 11.
Find the interquartile range.
Answer:
Definition of interquartile range: The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third
A quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus the lower quartile.
Interquartile range formula: The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below:
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q1
Where Q1 is the first quartile, Q3 is the third quartile.
Q1=38; Q3=78
Interquartile range=Q3-Q1
Interquartile range=78-38
Interquartile range=40

Question 12.
Interpret Q₃ and the interquartile range.
Answer:
The third quartile (Q 3), also known as the upper quartile, splits the lowest 75% (or highest 25%) of data. It is the middle value of the upper half. The first quartile is also known as the 25th percentile, the second quartile as 50th percentile, and the third quartile as 75th percentile. The Interquartile range is from Q 1 to Q 3.

Use the information and table below to answer questions 13 to 16.

The table shows the results of a survey asking 50 people how many hours they spent using the Internet in a day.

Question 13.
Find the median, the lower quartile, and the upper quartile of the number of hours spent using the Internet.
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Now calculate the cumulative frequency
Cumulative frequency is defined as a running total of frequencies. The frequency of an element in a set refers to how many of that element there are in the set. Cumulative frequency can also be defined as the sum of all previous frequencies up to the current point.
Cumulative frequency for 1 hour=3
Cumulative frequency for 2 hour=3+0=3
Cumulative frequency for 3 hour=3+9=12
Cumulative frequency for 4 hour=5+12=17
Cumulative frequency for 5 hour=7+17=24
Cumulative frequency for 6 hour=11+24=35
Cumulative frequency for 7 hours=5+35=40
Cumulative frequency for 8 hours=1+40=41
Cumulative frequency for 9 hours=6+41=47
Cumulative frequency for 10 hours=47+3=50
n=50
therefore, n/2=50/2=25
24 is nearer to the 25
median=l+(n/2-cf)/f*c
l=lower limit of median class
n=total number of observations
cf=cumulative frequency of the class
preceding the median class
f=frequency of the median class
c=class length
Median=5+(25-17)/7*1
After calculating the above equation
Median=5+(18)/7*1
Meadian=5+(1.14)*1
Median=6.14
If we round the nearest number, then it will be 6 hours.
Lower quartile=(n/4)th term
Lower quartile=50/4
Lower quartile=12.5
We got 12.5 checks the number which is above the 12.5
We have 17: it is 4 hours
Therefore, Q1=4 hours
Q3=(3n/4)th term
Q3=(3*50/4)
Q3=37.5
Now check the number which is having above 37.5.
So the number is 40: 7 hours

 

Question 14.
Calculate the interquartile range.
Answer:
Interquartile range=upper quartile-lower quartile
Interquartile range=7-4
Interquartile range=3 hours.

Question 15.
Interpret the data values that are greater than the upper quartile.
Answer:
Upper quartile Q3=(3n/4)th term
Upper quartile Q3=(3*50/4)
Upper quartile Q3=37.5
Now check the number which is having above 37.5.
So the number is 40: 7 hours
25% of the people spent at least 7 hours on the internet.

Question 16.
Math journal A majority of the people surveyed spent 6 hours or less using the Internet.” Do you agree with this statement? Justify your argument by using the quartiles.
Answer:
The above quartiles we got are:
Q1= 4 hours
Q2=6 hours
Q3=7 hours.
yes, according to the quartiles Q1 and Q2 majority of the people surveyed spent 6 hours or less using the Internet.

Use the information and table below to answer questions 17 to 21.

The table shows the lifespans of 29 batteries that were tested.

Question 17.
Find the range of the lifespans.
Answer:
The given data set for lifespan: 2, 3.5, 4.5, 5, 6, 7.5, 8, 9
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
Maximum value=9
Minimum value=2
Range=maximum value-minimum value
Range=9-2
Range=7
Therefore, the range of batteries lifespan is 7 hours.

Question 18.
Math Journal Explain why the range is not a good statistic to use for evaluating the lifespans of these batteries.
Answer:
well, it depends on the type of battery and how you are using it. Some batteries have a limited lifespan and others can last for years. So we cannot evaluate the lifespans of these batteries by using the range.

Question 19.
Find the median, the lower quartile, and the upper quartile of the lifespans.
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Now calculate the cumulative frequency
Cumulative frequency is defined as a running total of frequencies. The frequency of an element in a set refers to how many of that element there are in the set. Cumulative frequency can also be defined as the sum of all previous frequencies up to the current point.

n=29 (given)
Therefore, n/2=29/2=14.5
Median=1/2[(n/2)th term + (n/2+1)th term]
Median=1/2[ 14.5 + 15.5]
Median=1/2[30]
Median=15
Now check for the number, if the number was not there then check for the 15 above number and write the certain battery lifespan.
15 is not there, the next nearest number is 18 and the battery lifespan is 6 hours.
Median Q2=6 hours.
Q1 lower quartile=(n/4)th term
Q1=29/4
Q1=7.25
Q1=3
the corresponding battery lifespan is 3.5 which is nothing but 4.
Therefore, Q1 is 4 hours.
Q3 upper quartile=(3n/4)th term
Q3=(3*29/4)
Q3=87/4
Q3=21.75
21.75 (22) is not there, the next nearest number is 24 and the battery lifespan is 8 hours.
Q3=8 hours.

Question 20.
Calculate the interquartile range.
Answer:
Upper quartile Q3=8
Lower quartile Q1=4
Interquartile range=upper quartile-lower quartile
Interquartile range=8-4
Interquartile range=4 hours.

Question 21.
Math Journal Compare the lifespans less than the lower quartile to the lifespans greater than the upper quartile. Based on this comparison, do you consider the difference in battery lifespans significant? Explain your reasoning.
Answer:
Q1 lower quartile=(n/4)th term
Q1=29/4
Q1=7.25
Q1=3
the corresponding battery lifespan is 3.5 which is nothing but 4.
Therefore, Q1 is 4 hours.
Q3 upper quartile=(3n/4)th term
Q3=(3*29/4)
Q3=87/4
Q3=21.75
21.75 (22) is not there, the next nearest number is 24 and the battery lifespan is 8 hours.
Q3=8 hours.
The difference is significant. The battery lifespans above the upper quartile are more than double the battery lifespans below the lower quartile.

Question 22.
Math Journal Another statistic used to analyze data is the midrange.

Is the midrange the same as the median of a set of data values? Explain your reasoning by using an example.
Answer:
Midrange in layman terms is the middle of any data set or simply the average, mean of the data. A midrange is a statistical tool that is also known as the measure of centre in statistics. Along with the existence of the midrange formula means, medium, average, mode, and range are also known as the measure of central tendency. The midrange of the data set is simply the value between the biggest value and the lowest value. In order to find the midrange of the data set the value is then divided by 2 after summing the lowest value present in the data set with the highest value present in the data set.
No, the midrange and median are not the same.
Example: The daily temperature recorded in the city of Colombia Bogata is 55, 65, 67, 69, 70, 80, 81, 87, 90. We need to calculate the mid-temperature in Bogata during this period.
According to the midrange formula:
midrange=greatest value-lowest value/2
midrange=90-55/2
midrange=35/2
midrange=17.5
midrange and median are not the same.

Solve.

Question 23.
The finishing times of 40 people in a 100-meter freestyle swimming event were collected. The quartiles for the data are shown:
Q₁ = 62.05 seconds, Q₂ = 69.16 seconds, and Q₃ = 71.43 seconds.

List the letters of all the following statements that are correct.
a) 50% of the swimmers managed to complete the 100-meter event in between 69.16 and 71.43 seconds.
Answer: wrong statement.
The finishing times have already been taken. But in this question the event is not completed so the statement is wrong.

b) 75% of the swimmers took longer than 62.05 seconds to complete the event.
Answer:
The event is completed so the statement is correct.
75%of swimmers mean:
75% of 40 people
75*40/100
=30
30 people took longer than 62.05 seconds to complete the event.

c) Swimmers with times greater than the upper quartile are fast swimmers.
Answer:
The statement is wrong.
The given time is 3 h 35 min 49 sec
The upper quartile time is 71.43 sec is which is nothing but 0.01 hours
swimmers take time greater than the upper quartile might be slower
For example, if a swimmer has taken 75 seconds means he has taken more time to complete the event then he cant be a fast swimmer. So definitely, the statement is correct.

d) Swimmers with times between the first and the second quartiles are faster than the first 10 swimmers who finished the swimming event.
Answer:
the above-given quartiles:
Q1=62.05 seconds,
Q₂ = 69.16 seconds
This we cannot say about the faster swimmers. Because we know quartile timings and we don’t know about the first 10 swimmers timings.
In how much time they finished the event. Until we know the finish timings of the first 10 swimmers we cannot tell the Swimmers with times between the first and the second quartiles are faster than the first 10 swimmers who finished the swimming event.

e) The interquartile range shows that the time differences of the middle 50% of the swimmers is not more than 9.38 seconds.
Answer:
interquartile range=Q3-Q1
Interquartile range= 71.43 – 62.05
Interquartile range=9.38
Yes, it is correct.
It is not more than 9.38
therefore, the statement is correct.

f) 50% of the swimmers completed the 100-meter event between 62.05 and 71.43 seconds.
Answer: yes
Nearly 20 people completed the event between 62.05 and 71.43
In the above question, we got the timings of the swimmer that are more than 62.05
definitely, this statement is correct.
g) 25% of the swimmers took 71.43 or more seconds to complete the event.
Answer:
25% means the remaining people completed the event.
25*40/100
=10
10 people took 71.43 or more seconds to complete the event.
h) 10 swimmers completed the event in less than 62.05 seconds.
Answer:
total 40 people have recorded the finishing timings.
20 people took time between 62.05 and 71.43
10 people took time 71.43 or more seconds.
and the other 10 remaining people might take a chance of 62.05 seconds or less than that.
So the statement is correct.

i) The number of swimmers with times between the first and the second quartiles is more than the number of swimmers with times between the second and the third quartiles.
Answer:
The above-given quartiles:
Q₁ = 62.05 seconds, Q₂ = 69.16 seconds, and Q₃ = 71.43 seconds.
This is the record:
total 40 people have recorded the finishing timings.
20 people took time between 62.05 and 71.43
10 people took time 71.43 or more seconds.
and the other 10 remaining people might take a chance of 62.05 seconds or less than that.
By observing the above statements, the given statement is wrong.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 9 Statistics detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 9 Answer Key Statistics

Math in Focus Grade 7 Chapter 9 Quick Check Answer Key

Find the mean of each set of data. Round your answer to 2 decimal places if it is not exact.

Question 1.
9, 11, 6, 29, 5
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

The above-given data: 9, 11, 6, 29, 5
Mean=9+11+6+29+5/5
Mean=60/5
Mean=12.

Question 2.
43, 88, 39, 10, 26, 17, 35
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

The above-given data: 43, 88, 39, 10, 26, 17, 35
Mean=43+88+39+10+26+17+35/7
Mean=258/7
Mean=36.85 (round to nearest number)
Mean=37.

Question 3.
53.6, 36.7, 88.5, 90.6
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

The above-given data: 53.6, 36.7, 88.5, 90.6
Mean=53.6+36.7+88.5+90.6/4
Mean=269.4/4
Mean=67.35
Mean=67 (rounded to nearest number)

Question 4.
0.14, 1.05, 3.1, 7.18, 4.3, 8
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

The above-given data: 0.14, 1.05, 3.1, 7.18, 4.3, 8
Mean=0.14+1.05+3.1+7.18+4.3+8/6
Mean=23.77/6
Mean=3.96
Mean=4

Solve. Show your work.

Question 5.
The heights, in inches, of 10 children are
54, 66, 52, 60.5, 61.25, 55, 58.75, 51.5, 53, 50.
Find the mean height of the children.
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

the above-given data: 54, 66, 52, 60.5, 61.25, 55, 58.75, 51.5, 53, 50.
Mean=54+66+52+60.5+61.25+55+58.75+51.5+53+50/10
Mean=562/10
Mean height of children=56.2

Find the median of each set of numbers.

Question 6.
41,29, 78, 12, 56, 30, 22
Answer: 30
Explanation:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
The formula to calculate the median of the data set is given as follow.
An odd number of observations:
If the total number of observations given is odd, then the formula to calculate the median is:
Median = {(n+1)/2}^thterm
where n is the number of observations
Even number of observations:
If the total number of observations is even, then the median formula is:
Median  = [(n/2)^th term + {(n/2)+1}^th]/2
where n is the number of observations
How to  calculate Median:
To find the median, place all the numbers in the ascending order and find the middle.
Put them in ascending order: 12, 22, 29, 30, 41, 56, 78
The middle number is 30, so the median is 30.

Question 7.
193, 121.5, 162.3, 125, 103.8, 149.6
Answer:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 103.8, 121.5, 125, 149.6, 162.3, 193
Meadian=125+149.6/2
Median=274.6/2
Median=137.3
Thus, the median is 137.3

Question 8.
9, 2, 2, 4, 4, 4, 1, 3, 6, 5, 3, 6
Answer: 4
Explanation:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 1, 2, 2, 3, 3, 4, 4, 4, 5, 6, 6, 9
Median=4+4/2
Median=8/2
Median=4
Thus, the median is 4.

Question 9.
1,011, 1,100, 1,001, 1,010, 1,110, 1,000, 1,011, 100
Answer:
Explanation:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 100, 1000, 1001, 1010, 1011, 1011, 1100, 1110
Median=2021/2
Median=1010.5
Thus, the median is 1010.5

Solve. Show your work.

Question 10.
The daily low temperatures for the past 10 days were
30.6°F, 32.1 °F, 29.5°F, 30.2°F, 26.4°F, 34.3°F, 31.6°F, 32°F, 25.9°F, and 26.4°F.
What was the median daily low temperature?
Answer:
The statistical concept of the median is a value that divides a data sample, population, or probability distribution into two halves. Finding the median essentially involves finding the value in a data sample that has a physical location between the rest of the numbers. Note that when calculating the median of a finite list of numbers, the order of the data samples is important. Conventionally, the values are listed in ascending order, but there is no real reason that listing the values in descending order would provide different results. In the case where the total number of values in a data sample is odd, the median is simply the number in the middle of the list of all values. When the data sample contains an even number of values, the median is the mean of the two middle values. While this can be confusing, simply remember that even though the median sometimes involves the computation of a mean, when this case arises, it will involve only the two middle values, while a mean involves all the values in the data sample. In the odd cases where there are only two data samples or there is an even number of samples where all the values are the same, the mean and median will be the same. Given the same data set as before, the median would be acquired in the following manner:
30.6°F, 32.1 °F, 29.5°F, 30.2°F, 26.4°F, 34.3°F, 31.6°F, 32°F, 25.9°F, and 26.4°F.
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 25.9, 26.4, 26.4, 29.5, 30.2, 30.6, 31.6, 32, 32.1, 34.3
Median=30.2+30.6/2
Median=60.8/2
Median=30.4
Thus, the median is 30.4

Summarize each of the following data sets in a frequency table and draw a dot plot.

Question 11.
The data show the number of misspelled words found in 15 essays.

Answer:
Definition of frequency: Frequency means the number of times a value appears in the data. A table can quickly show us how many times each value appears. If the data has many different values, it is easier to use intervals of values to present them in a table.

A dot chart or dot plot is a statistical chart consisting of data points plotted on a fairly simple scale, typically using filled in circles.

Question 12.
In a game, each child is given 10 balls to hit at moving targets. The data show the number of hits scored by 20 children.

Answer:
Definition of frequency: Frequency means the number of times a value appears in the data. A table can quickly show us how many times each value appears. If the data has many different values, it is easier to use intervals of values to present them in a table.

A dot chart or dot plot is a statistical chart consisting of data points plotted on a fairly simple scale, typically using filled in circles.

This handy Math in Focus Grade 7 Workbook Answer Key Cumulative Review Chapters 6-8 detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Cumulative Review Chapters 6-8 Answer Key

Concepts and Skills

Tell whether each pair of angles is supplementary, complementary, or neither. (Lesson 6.1)

Question 1.
m∠1 = 18° and m∠2 = 82°
Answer:
Given each pair of angles are complementary angles.

Explanation:
The sum of two angles is equal to 90° is called complementary angles. In the given figure if we sum up both the angles that is 18°+ 82°we get a 90° angle.

Question 2.
m∠3 = 103° and m∠4 = 77°
Answer:
Given each pair of angles are supplementary angles.

Explanation:
Supplementary angles are those angles that sum up to 180°.

Question 3.
m∠5 = 95° and m∠6 = 85°
Answer:
Given each pair of angles are supplementary angles.

Explanation:
Supplementary angles are those angles that sum up to 180°. Therefore add 95° and 85° then we get 180°.

Question 4.
m∠7 = 21° and m∠8 = 69°
Answer:
Given each pair of angles are complementary angles.

Explanation:
The sum of two angles is equal to 90° is called complementary angles. In the given figure if we sum up both the angles that are 21°+ 69°we get a 90° angle.

Find the measure of each numbered angle. (Lessons 6.1, 6.2)

Question 5.
\(\overleftrightarrow{P R}\) is a straight line.

Answer:
54°

Explanation:
Given, ∠SQR = 126°
∠QPR = 180°
∠PQS = 1
Here we need to calculate the numbered angle.
To find that we need to subtract ∠QPR from ∠SQR
∠1 = 180°-126°
∠1   = 54°
Hence the numbered angle is 54°

Question 6.
\(\overleftrightarrow{P R}\) is a straight line.

Answer:
88°

Explanation:
The straight-line QPR has 180°.
Given, ∠PQS = 54°
∠QTR = 38°
∠SQT is numbered angle that is ∠2
Now we need to calculate the numbered angle
Therefore QPR = ∠PQS + ∠QTR + ∠2
180°= 54°+38°+ ∠2
180°= 92°+ ∠2
180°-92° = ∠2
∠2 = 88°

Question 7.

Answer:
The numbered angle 3 is 94°

Explanation:
The sum of angles around a point is 360°
40°+125°+101°+3=360°
3= 360°-40°-125°-101°
3= 360° – 266°
3 = 94°
Hence the numbered angle 3 is 94°

Question 8.

Answer:
36°

Explanation:

The sum of angles around a point is 360°
When two lines cross the angles are opposite to each other. Therefore opposite angles are equal to each other.
The angle ∠ABC and ∠DBE are opposite to each other. So ∠ABC = ∠DBE
Given, ∠ABC = 162°, ∠DBE means the numbered angle 4 is also 162°
Now we need to calculate the numbered angle 5
To calculate the angle 5 we need to add ∠ABC + ∠DBE+ 5 = 360°
162°+162°+∠5 = 360°
324°+ ∠5 =360°
∠5 = 360°-324°
∠5 = 36°

\(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\). Find the measure of each numbered angle. (Lesson 6.3)

Question 9.

Answer:
m∠1 = 81°
Vertical ∠s

Question 10.

Answer:
m∠3 = 49°
Alternate interior ∠s

Find the measures of ∠1 and ∠2 in each diagram. (lesson 6.4)

Question 11.
Triangle ABC is an isosceles triangle.

Answer:
32°

Explanation:

1+32°+x =180°
1+x° = 180°-32°
1+x°= 148°
1+x°+2 = 180°
148° +2 = 180°
2 = 180°-148°
2 = 32°
1+x°= 148°
32°+x°= 148°
x°=148°-32°
x°=  116°
1=148-116
1=32°

Question 12.
\(\overleftrightarrow{P R}\) is a straight line.

Answer:
110°+1 = 180°
1 = 180°-110°
1 = 70°

Use the given Information to construct each polygon.

Construct triangle ABC with sides AB = 5.8 cm, BC = 6.6 cm, and AC = 4.5 cm. (Lessons 7.1, 7.2, 7.3)

Step 1:
we are given
AB = 5.8
BC = 6.6
AC = 4.5

Step 2:
Use a ruler to draw the segment 66 cm long. LabeL its endpoints by B and C.

Step 3:
Because AB = 5.8 cm, use the ruLer to set the compass to a radius of 5.8 cm Then using A as center draw an arc of radius 5.8 cm above \(\overline{B C}\).

Step 4:
Because AC = 4.5 cm, use the ruler to set the compass to a radius of 4.5 cm. Then using G as center draw an arc of radius 4.5 cm that intersects the first arc. Label the intersection as A.

Step 5:
Use the ruler to draw \(\overline{A B}\) and \(\overline{A C}\)

a) Measure the angles of the triangle.
Answer:
m∠A ≈ 79°
m∠B ≈ 42°
m∠C ≈ 59°

b) Use a compass and straightedge to bisect angle B.
Answer:

c) Use a compass and straightedge to construct the perpendicular bisector of AB.
Answer:

d) The two bisectors intersect at a point D, label D in your construction.
Answer:

e) Measure the distances BD and CD.
Answer:
BD ≈ 3.2 cm
CD ≈ 3.8 cm

Question 14.
Construct quadrilateral PQRS with PQ = PR = 7.8 cm, PQ ||SR, m∠PQR = 73°, and RS = 4 cm. (Lessons 7.1, 7.2, 7.4)

Step 1:
We are given
PQ = PR = 7.8
PQ || SR
m∠PQR = 73°
RS = 4

Step 2:
Sketch the quadrilateral

Step 3:
First we construct △PRQ using the side-side-angle method.

Step 4:
Determine m∠RPQ
m∠RPQ + 2.73° = 180°
m∠RPQ = 180° – 146°
m∠RPQ = 34°

Step 5:
As PQ || SR, we have
m∠SRP = m∠RPQ = 34°

Step 6:
We construct △PRS, using the side-angle-side method.

a) Measure the length of PS.
Answer:
PS ≈ 5 cm

b) Use a compass and straightedge to construct the angle bisector of ∠S.
Answer:

c) Use a compass and straightedge to construct the perpendicular bisector of \(\overline{P Q}\).
Answer:

d) Label the point where the angle bisector and perpendicular bisector intersect as W.
Answer:

Solve. Show your work. (Lesson 7.5)

Question 15.
Jacob builds a model plane that is 800 millimeters long. If the actual plane is 40 meters, what is the scale of the model? (10 mm = 1 cm)
Answer:
We are given
Length on the plane: 800 mm
Actual length: 40 m

Determine the scale of the model:
\(\frac{\text { Length on the plane }}{\text { Actual length }}=\frac{800 \mathrm{~mm}}{40 \mathrm{~m}}=\frac{80 \mathrm{~cm}}{4,000 \mathrm{~cm}}=\frac{1}{50}\)
= \(\frac{1}{50}\)

Question 16.
The scale of a map is 1 : 1,500. If a square piece of land measures 4 inches on each side on the map, find the actual area of this piece of land to the nearest tenth of an acre. (1 acre = 43,560 ft²)
Answer:

We have:
Step 1:
Map length : Actual length = 1 in. : 1,500 in.
Map area: ActuaL area = 1 in.² : 1, 5002 in²

Step 2:
Let y represent the actual area of the piece of land in square inches.

Step 3:
Write a proportion: \(\frac{\text { Piece area on map }}{\text { Actual area of piece }}=\frac{1}{2,250,000}\)

Step 4:
Substitute:
\(\frac{4^{2}}{y}=\frac{1}{2,250,000}\)

Step 5:
Write the cross products:
y = 16.2,250,000

Step 6:
Simplify
y = 36,000,000 in²

Step 7:
Convert to acres:
1 acre = 43,560 ft = 43. 560. 12² = 6. 272, 640 in²
\(\frac{36,000,000}{6,272,640}\) ≈ 5.7 acres
= 5.7 acres

Match each solid with its net. (Lesson 8.1)

Question 17.

Answer:
The net of a cone consists of a circle and a sector of a circle.
The net b) does not consists a sector of circle, so we eliminate it
The net a) also doesn’t contain a sector of circle, therefore we eliminate it
The net d) is an ellipse sector, not a circle sector, so we eliminate it

The correct answer is (c)

Question 18.

Answer:

The net of a trianguLar prism contains 4 triangles.

The net b) contains 3 triangles and a rectangle, not 4 triangles, so we eliminate it
The net c) contains 4 triangles and a rectangle, not 4 triangles, so we eliminate it
The net d) contains 3 triangles, not 4 triangles, so we eliminate it

The correct answer is (a)

For each solid, sketch and describe the shape of the cross section formed when a plane slices the solid in the direction indicated. (Lesson 8.1)

Question 19.
A cube of length 6 centimeters perpendicular to a base and parallel to two opposite faces.

Answer:

A cross-section perpendicular to a base and parallel to two opposite faces can be drawn starting with a point for example on one of the edges of the top face. We draw a parallel line to two of the edges on the top face other than the one opposite to the one containing the initial point From the ends of the line we draw two vertical lines to the vertical edges. We join the ends of these two paralLels.

Question 20.
A cylinder of radius 3 centimeters parallel to its bases.

Answer:

A cross section parallel to its bases has the shape of a circle of radius 3:

Question 21.
A right triangle base prism parallel to its bases.

Answer:

A cross-section paralleL to its bases can be drawn starting with a point for example on one of the edges other than the bases We draw a parallel line to each of the corresponding edges on the bases. We join the ends of these two parallels.

Question 22.
Copy and complete the table. Use 3.14 as an approximation for π. (Lesson 8.2)

Answer:
Step 1:
We are given the cylinder:
V = 942
h = 12

Step 2:
Determine the radius r:
V = πr²h
942 ≈ 3.14 · r² · 12
942 = 37.68 r²
r² = \(\frac{942}{36.68}\)
r² = 25
r = \(\sqrt{25}\)
r = 5 ft

Step 3:
Determine the diameter:
d = 2 · r = 2 · 5 = 10 ft

Step 4:
Determine the total surface area:
S = 2πr² + 2πrh
= 2π · 5² + 2π · 5 · 12
= 50π + 120π
= 170π
≈170 · 3.14
= 533.8 ft²

Step 5:
We are given the cylinder:
V = 879.2
d = 10

Step 6:
Determine the radius:
r= \(\frac{d}{2}\) = \(\frac{10}{2}\) = 5 cm

Step 7:
Determine the height:
V = πr²h
879.2 ≈ 3.14 · 5² · h 
879.2 = 78.5h
h = \(\frac{879.2}{78.5}\) = 11.2cm

Step 8:
Determine the total surface area:
S = 2πr² -1- 2πrh
= 2π · 5² + 2π · 5 · 11.2
= 50π + 112π
= 162π
≈ 162 · 3.14
= 508.7cm²

Step 9:
We are given the cylinder:
r = 4
S = 251.2

Step 10:
Determine the height:
S = 2πr² + 2πrh
251.2 ≈ 2 · 3.14 · 4² + 2 · 3.14 · 4 · h
251.2 = 100.48 + 25.12h
251.2 – 100.18 = 100.48 + 25.12h – 100.48
150.72 = 25.12h
h = \(\frac{150.72}{25.12}\)
h = 6 m

Step 11:
Determine the volume:
V = πr²h ≈ 3.14 · 4^{2 ·} 6 ≈ 301.4 m³

Step 12:
Determine the diameter:
d = 2r = 2 · 4 = 8 m

Step 13:
Complete the table

Find the exact surface area of each solid cone. (Lesson 8.3)

Question 23.
A cone with radius of 8.7 meters and a slant height of 12.8 meters.
Answer:
We are given the cone:
r = 8.7
l = 12.8
Use the formula for the cone’s surface area:
S = πr² + πrl
Substitute r, l to find the exact surface area:
= π · 8.7² + π · 8.7 · 12.8
= 75.69π + 111.36π
= 187.05 π m²

Question 24.
A cone of diameter 16.8 centimeters and slant height 20.6 centimeters.
Answer:
We are given the cone:
d = 16.8
l = 20.6
First we determine the radius:
r = \(\frac{d}{2}\) = \(\frac{16.8}{2}\) = 8.4
Use the formuLa for the cones surface area:
S = πr² + πrl
Substitute r, ito find the exact surface area:
= π · 8.42 + π · 8.4 · 20.6
= 70.56π + 173.04π
= 243.6π cm²

Find the volume of each solid. Use 3.14 as an approximation for π. Round your answers to the nearest tenth when you can. (Lessons 8.3, 8.4)

Question 25.
A square pyramid with a height of 13 centimeters and a base that is 8 centimeters on each edge.
Answer:
Use the formula:
V = \(\frac{1}{3}\) Bh
Substitute for B and h.
= \(\frac{1}{3}\) · (8²) · 13
Multiply and round:
≈ 277.3 cm³

= 277.3 cm³

Question 26.
An octagonal pyramid with a height of 8.2 inches and a base area of 34.5 square inches.
Answer:
Use the formula:
V = \(\frac{1}{3}\) Bh
Substitute for B and h.
= \(\frac{1}{3}\) · 34.5 · 8.2
Multiply and round:
= 94.3 in³

Question 27.
A cone with a radius of 6.5 centimeters and a height of 14 centimeters.
Answer:
We are given the cone:
r = 6.5
h = 14
Use the formula for the cone’s volume to find the volume:
V = \(\frac{1}{3}\)πr²h
≈ \(\frac{1}{3}\) · 3.14 · 6.5² · 14
≈ 619.1 cm³
= 619.1 cm³

Question 28.
A cone with a diameter of 24.6 feet and a height of 18.5 feet.
Answer:
We are given the cone:
d = 24.6
h = 18.5
Determine the radius:
r = \(\frac{d}{2\) = \(\frac{24.6}{2}\) = 12.3 ft
Use the formula for the cone’s volume to find the volume:
V = \(\frac{1}{3}\)πr²h
≈ \(\frac{1}{3}\) · 3.14 · 12.3² · 18.5
≈ 2,929.5 ft³
= 2,929.5 ft³

Question 29.
A sphere with a radius of 9.6 feet.
Answer:
We are given the sphere:
r = 9.6
Use the formula for the volume of a sphere:
V = \(\frac{4}{3}\)πr³ ≈ \(\frac{4}{3}\) · 3.14 · 9.6³
Multiply:
= 3,704.094…
Round to the nearest tenth:
≈ 3,704.1 ft³
= 3,704.1 ft³

Question 30.
A sphere with a diameter of 39.8 centimeters.
Answer:
We are given the sphere:
d = 39.8
Determine the radius of the sphere:
r = \(\frac{d}{2}\) = \(\frac{39.8}{2}\) = 19.9 cm
Use the formula for the volume of a sphere:
V = \(\frac{4}{3}\)πr³ ≈ \(\frac{4}{3}\) · 3.14 · 19.9³
Multiply:
= 32,993.4411….
Round to the nearest tenth:
≈ 32,993.4 cm³
= 32,993.4 cm³

Problem Solving

Find the value of each variable. (Chapter 6)

Question 31.

Answer:
x°=37°
2x=74°

Explanation:

A 360-degree angle is a complete angle of a point.
∠ABC = 360°-291°
∠ABC = 69°
∠ABC+BCA+CAB = 180°
69°+x°+2x°=180°
3x°+69=180°
3x°=180-69
3x°=111
x°=111÷3
x°=37°
2x°=74°

Question 32.

Answer:

Explanation:

∠DFE = 180°-112°
∠DFE = 68°
y°+3y°+ 68°= 180°
4y°+68° = 180°
4y°= 180°- 68°
4y°=112
y°=112÷4
y°= 28°
3y°= 84°

Question 33.

Answer:
Alternate exterior ∠s:
m∠HIN = 132°
VerticaL ∠
m∠HIN = 3b° + b°
Substitute and simplify:
132 = 4b
Divide by 4:
\(\frac{132}{4}\) = \(\frac{4b}{4}\)
Simplify
b = 33
Straight ∠:
3a° + 3b° + b° = 180°
Substitute and simplify:
3a ° + 4b° = 180°
3a + 4 · 33 = 180
3a + 132 = 180
Subtract 132 from both sides:
3a + 132 – 132 = 180 – 132
Simplify
3a = 48
Divide by 3
\(\frac{3a}{3}\) = \(\frac{48}{3}\)
Simplify
a = 16

a = 16; b = 33

Question 34.

Answer:
35°

Explanation:

∠PTU = c°
∠PRS = 80°
∠PUR = 2c°
∠TPS = 5c°
The sum of angles around the point is 360°
Sum of ∠PTU+ ∠PRS + ∠PUR + ∠TPS = 360°
c°+80°+2c°+5c° = 360°
8c° + 80° = 360°
8c° = 360°- 80°
8c° = 280°
c° = 280÷8
c°= 35°
2c°= 70°
5c°= 175°

Solve. Show your Work.

Question 35.
Segment AB is parallel to segments CD and EF. Segment BC is parallel to segment ED. Find the measure of ∠ABC. (Chapter 6)
Answer:

Question 36.
The diagram shows a triangular pin. Triangle ABC is an isosceles triangle. Line BD is parallel to line EF. m∠AEF = 40° and m∠DBC = 27°.
Find the measures of ∠EBD and ∠BAC. (Chapter 6)

Answer:
We are given:

Corresponding ∠s:
m∠EBD = 40°
Determine m∠ABC:
m∠ABC – m∠EBD + m∠DBC
= 40° – 27°
= 67°
△ABC isosceles:
m∠ACB = m∠ABC = 67°
Sum of ∠s in a triangle:
m∠BAC + m∠ACB + m∠ABC = 180°
Substitute:
m∠BAC + 67° + 67° = 180°
Simplify:
m∠BAC + 134° = 180°
We subtract 134° from both sides:
m∠BAC + 134° – 134° = 180° – 134°
Simplify:
m∠BAC = 46°

m∠EBD = 40°
m∠BAC = 46°

Question 37.
X, Y, and Zare the location of three manufacturing factories linked by three straight roads as shown in the diagram. A water pipe will go from Factory Y, and will be equidistant from the two roads linking Factory X and Factory Z to Factory Y. (Chapter 7)
a) Construct triangle XYZ to represent the locations of the three factories using a scale of 1 centimeter to 100 meters.
Answer:

b) Using a compass and straightedge only, construct the path of the water pipe.
Answer:
The Line whose points are equidistant from \(\overline{Y^{\prime} X^{\prime}}\) and \(\overline{Y^{\prime} Z^{\prime}}\) is the angle bisector of ∠Y’. We construct it:

c) The water department needs to build a temporary construction worksite that must be equidistant from Factory Y and Factory Z and at a distance of 400 meters from Factory X. Label the position of the construction worksite to be set up in your diagram as point W.

Answer:
The tine whose points are equidistant from Y’ and Z’ is the perpendicular bisector of \(\overline{Y^{\prime} Z^{\prime}}\). We construct it and label by W its intersection with the angle bisector from part b):

Question 38.
A model of a gym uses a scale of 1 : 12. The actual gym will be a rectangular prism. It will be 69.6 feet long, 54 feet wide, and 22.8 feet high. Find, in cubic feet, the volume of the model of the gym. (Chapters 7, 8)
Answer:
85691 cubic feet

Explanation:
The actual gym is a rectangular prism.
Given the length of a rectangular prism is 69.6 feet
Width = 54 feet
Height = 22.8 feet
The voulme of a rectangular prism is = Length × Width × Height
Volume = 69.6 × 54 × 22.8
Volume = 85691 cubic feet

Question 39.
Mrs. Sullivan buys a 5-liter bottle of apple juice. She can either use cylindrical glass A or cylindrical glass B to serve the apple juice. Use 3.14 as an approximation for π. (1,000 cm³ = 1 L) (Chapter 8)

We are given:
d_A = 6
h_A = 9
d_B = 5
h_B = 12

a) What is the volume of glass A?
Answer:
Determine the radius of glass A:
r_A = \(\frac{d_{A}}{2}\) = \(\frac{6}{2}\) = 3 cm
Determine the volume of glass A
V_A = πr_A²h_A ≈ 3.14 · 3² · 9 = 254.34 cm³

b) How many glasses of apple juice can she serve if she uses glass A?
Answer:
Determine the number of glasses A:
\(\frac{5 \cdot 1,000}{254.34}\) = \(\frac{5,000}{254.34}\) ≈ 19 glasses

c) What is the volume of glass B?
Answer:
Determine the radius of glass B:
r_B = \(\frac{d_{B}}{2}\) = \(\frac{5}{2}\) = 2.5 cm
Determine the volume of glass B
V_B = πr_B²h_B ≈ 3.14 · 2.5² · 12 = 235.5 cm³

d) How many glasses of apple juice can she serve if she uses glass B?
Answer:
Determine the number of glasses B:
\(\frac{5 \cdot 1,000}{235.5}\) = \(\frac{5,000}{235.5}\) ≈ 21 glasses

e) If Mrs. Sullivan sells one glass of apple juice for $1.80, what is the maximum amount of money she can collect?
Answer:
21 · 1.80 = 37.80

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 8 Review Test Answer Key

Concepts and Skills

For this review, r represents radius and h represents height. You may use a calculator and use 3.14 as an approximation for π. Round your answers to the nearest tenth unless otherwise stated.

Question 1.
Find the volume of each cylinder to the nearest unit. Use the given dimensions,
a) r = 4.2 inches; h = 14 inches
Answer:
775 cubic units

Explanation:
The volume of the cylinder is =π r² h
π = 3.14, r = 4.2, h= 14 inches
volume of the cylinder = 3.14×4.2×4.2×14 = 775 cubic units

b) r = 7 centimeters; h = 12 centimeters
Answer:
1846 cubic units

Explanation:|
The volume of the cylinder is =π r² h
π = 3.14, r = 7, h= 12 centimeters
volume of the cylinder = 3.14×7×7×12 = 1846 cubic units

Question 2.
Find the volume of each cone to the nearest unit. Use the given dimensions.
a) r = 3 centimeters; h = 8 centimeters
Answer:
75 cm³.

Explanation:
The volume of each cone is 1/3hπr²
r = 3 centimeters; h = 8 centimeters
Therefore the volume of each cone is = 1/3×8×3.14×3×3 = 75 cm³.

b) r = 8 inches; h = 15 inches
Answer:
1004 cm³.

Explanation:
The volume of each cone is 1/3hπr².
Given r = 8 inches; h = 15 inches and π=3.14
Therefore the volume of each cone is = 1/3×8×3.14×15 ×15 = 1004 cm³.

Question 3.
Find the volume of each pyramid. Use the given dimensions.

a) A square base with an edge length of 6 centimeters; h = 4 centimeters.
Answer:
We are given the cone:
l = 6
h = 4
Use the formula for the volume of a pyramid:
V = \(\frac{1}{3}\)Bh = \(\frac{1}{3}\)l²h
Substitute for l, h:
= \(\frac{1}{3}\) · 6² · 4
Multiply:
= 48 cm³

b) A rectangular base with length of 6 inches and width = 3.3 inches; h = 7 inches.
Answer:
We are given the cone:
l = 6
w = 3.3
h = 7
Use the formula for the volume of a pyramid:
V = \(\frac{1}{3}\)Bh = \(\frac{1}{3}\)lwh
Substitute for l, w, h:
= \(\frac{1}{3}\) · 6 · 3.3 · 7
Multiply:
= 46.2 in³

Question 4.
Find the volume of each sphere to the nearest unit. Use the given dimensions,
a) r = 9.6 centimeters
Answer:
3705 cm³

Explanation:
The volume of each sphere is = 4/3 πr³
Given r = 9.6 centimeters
Volume of the sphere is = 4/3×3.14×9.6×9.6×9.6 = 3705 cm³.

b) d = 26 centimeters
Answer:
9202 cm³.

Explanation:
The volume of each sphere is = 4/3 πr³
Given diameter is = 13 cm
Here we need a radius to calculate the volume of the sphere hence r = d/2 = 26/2 = 13 cm
Volume of the sphere is = 4/3×3.14×13×13×13=9202 cm³.

Question 5.
Find the exact surface area of each solid.

Answer:
a) We are given the cylinder:
r = 3
h = 6
Use the formula for the surface area of a cylinder.
S = 2πr² + 2πrh
Substitute for r, h:
= 2π · 3² + 2π · 3 · 6
Multiply:
= 18π + 36π
Add:
= 54π ft²

b) We are given the sphere:
d = 28
Determine the radius of the sphere:
r = \(\frac{d}{2}\) = \(\frac{28}{2}\) = 14
Use the formula for the surface area of a sphere:
S = 4πr²
Substitute for r:
= 4π · 14²
Multiply:
= 784π m²

c) We are given the cone:
r = 6
l = 10
Use the formula for the surface area of a cone:
S = πr² + πrl
Substitute for r, l:
= π · 6² + π · 6 · 10
Multiply:
= 36π + 60π
Add:
= 96π in²

Question 6.
Find the volume and surface area of each solid. Round to the nearest tenth.
a) A solid cone with a diameter of 5 feet, a slant height of 7 feet, and a height of 6.5 feet.
Answer:
We are given the cone.
d = 5
l = 7
h = 6.5
Find the radius:
r = \(\frac{d}{2}\) = \(\frac{5}{2}\) = 2.5
Use the formula for the volume of the cone:
V = \(\frac{1}{3}\) πr²h
≈ \(\frac{1}{3}\) · 3.14 · 2.5² · 6.5
≈ 42.5 ft³
Use the formula for the surface area of the cone:
S = πr² + πrl
= π · 2.5² + π · 2.5 · 7
= 6.25π + 17.5π
= 23.75π
≈ 23.75 · 3.14
≈ 74.6 ft²

b) A sphere with a radius of 28 millimeters.
Answer:
We are given the sphere:
r = 28
Use the formula for the volume of the sphere:
V = \(\frac{4}{3}\) πr³ = \(\frac{4}{3}\) · 3.14 · 28³
≈ 91,905.7 mm³
Use the formula for the surface area of the sphere:
S = 4πr² ≈ 4 · 3.14 · 28²
≈ 9,487 mm²

c) A solid cylinder with a radius of 1.4 inches and a height of 4.2 inches.
Answer:
We are given the cylinder:
r = 1.4
h = 4.2
Use the formula for the volume of the cylinder:
V = πr²h ≈ 3.14 · 1.4² · 4.2
≈ 25.8 in³
Use the formula for the surface area of the cylinder:
S = 2πr² + 2πrh
= 2π · 1.4² + 2π · 1.4 · 4.2
= 3.92π + 11.76π
= 15.68π
≈ 15.68 · 3.14
≈ 49.2 in²

Problem Solving

Solve. Show your work.

Question 7.
The volume of a cone is 450 cubic centimeters and the radius of the base is 5 centimeters. What is the height of the cone to the nearest tenth?
Answer:
We are given:
V = 450
r = 5
Use the formula for the volume of the sphere:
V = \(\frac{1}{3}\) · πr² · h
Substitute for V, πn r:
450 = \(\frac{1}{3}\) · 3.14 · 5² · h
Multiply:
450 = \(\frac{78.5}{3}\) · h
Multiply:
450 = \(\frac{78.5}{3}\) · h
Multiply both sides by 3:
3 · 450 = 3 · \(\frac{78.5}{3}\) · h
Multiply:
1,350 = 78.5h
Divide both sides by 78.5
\(\frac{1,350}{78.5}\) = \(\frac{78.5h}{78.5}\)
Simplify
h ≈ 17.2 cm

Question 8.
The surface area of a sphere is 498.96 square centimeters. What is the radius of the sphere to the nearest tenth?
Answer:
We are given the sphere:
S = 498.96
Use the formula for the surface area of a sphere:
S = 4πr²
Substitute S and π:
498.96 = 4 · 3.14 · r²
Multiply:
498.96 = 12.56 · r²
Divide each side by 12.56
\(\frac{498.96}{12.56}=\frac{12.56 r^{2}}{12.56}{78.5}\)
Evaluate:
39.7261 ≈ r²
Find the square root of each side:
\(\sqrt[2]{39.7261}\) ≈ r²
Round to the nearest tenth:
r ≈ 6.3 cm

Question 9.
A cone with a height of 6 inches and a slant height of 7.5 inches has a lateral surface with an area of approximately 106 square inches.

a) What is the radius? Round to the nearest tenth.
Answer:
We are given:
h = 6
l = 7.5
A = 106
Use the formula for the cone’s lateral surface area:
A = πrl
Substitute A, l, π to find r:
106 = 3.14 · r · 7.5
Multiply:
106 = 23.55r
Divide both sides by 23.55
\(\frac{106}{23.55}=\frac{23.55 r}{23.55}\)
r ≈ 4.5 in

b) What is the volume of the cone? Round to the nearest tenth.
Answer:
Use the formula for the cone’s volume:
V = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) · 3.14 · 4.5² · 6
≈ 127.2 in³

Question 10.
A cylinder, a cone, and a sphere are shown below. Each solid has a radius of 1 inch and a height of 2 inches. Which of them has the greatest volume? Justify your answer.

Answer:
We are given:
r = 1
h = 2
Determine the cylinder’s volume:
V_cylinder = πr²h = π · 1² · 2 = 2π
Determine the cone’s volume:
V_cone = \(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) · π · 1² · 2
= \(\frac{2 \pi}{3}\)
Determine the sphere’s volume:
V_sphere = \(\frac{4}{3}\) πr³ = \(\frac{4}{3}\) · π · 1³
= \(\frac{4 \pi}{3}\)
We have:
\(\frac{2 \pi}{3}\) < \(\frac{4 \pi}{3}\) < 2π
V_cone < V_sphere < V_cylinder
The greatest volume is the cylinder’s

Question 11.
Two metal cubes each have an edge length of 4 centimeters. They are melted and recast into a square pyramid with a height of 5 centimeters. Find the area of the base of the pyramid.
Answer:
Determine the volume of the two cubes:
V = 2 · 4³ = 2 · 64 = 128 cm³
Use the formula of a pyramids volume:
V = \(\frac{1}{3}\) Bh
Substitute for V, h:
128 = \(\frac{1}{3}\) · B · 5
Multiply both sides by 3:
3 · 128 = 3 · \(\frac{5B}{3}\)
384 = 5B
Divide by 5:
\(\frac{384}{5}\) = \(\frac{5B}{5}\)
B = 76.8 cm²

Question 12.
A conical party hat has a diameter of 7 centimeters and a slant height of 14.5 centimeters. Paul wants to wrap the lateral surface of the party hat with plastic wrap. How much plastic wrap will he use? Round to the nearest square centimeter.
Answer:
We are given the cone:
d = 7
l = 14.5
Determine the radius of the cone:
r = \(\frac{d}{2}\) = \(\frac{7}{2}\) = 3.5
Determine the lateral surface of the cone:
A = πrl ≈ 3.14 · 3.5 · 14.5 ≈ 159 cm²
= 159 cm²

Question 13.
The composite solid shown is made up of a hemisphere attached to the top of a cube. The diameter of the hemisphere is the same as the edge length of the cube. Find the volume of the composite solid. Round to the nearest tenth.
Answer:

Determine the radius of the hemisphere:
r = \(\frac{9}{2}\) = 4.5
Determine the volume of the hemisphere:
V_hemisphere = \(\frac{1}{2}\) · \(\frac{4}{3}\) πr³
≈ \(\frac{2}{3}\) · 3.14 · 4.5³
≈ 190.8 cm³
Determine the volume of the cube:
V_cube = 9³ = 729 cm³
Determine the volume of the composite solid:
V_solid = V_hemisphere + V_cube
= 190.8 + 729
= 919.8 cm³

Question 14.
The circumference of a fully inflated beach ball is 12π, or about 37.68 inches. What is the radius of the beach ball? What is the volume of
the beach ball? Round to the nearest tenth if necessary.
Answer:
The radius of the beach ball is 6 inches.
The volume of the beach ball is 151 inches.

Explanation:
The circumference of a fully inflated beach ball is 12π.
Circumference of a circle = 2πr
12π=2πr
r=6
Hence radius of the beach ball is 6 inches.
Now we need to calculate the volume of a beach ball
The volume of the sphere is 4/3 πr³
Volume = 4/3×3.14×6×6×6
The volume of the beach ball = 904 inches.

Question 15.
A pyramid made of clay has a square base of length 8 inches and a height of 12 inches. The pyramid is reshaped into a cylinder with a radius of 8 inches. What is the height of the cylinder? Round to the nearest inch
Answer:
The height of the cylinder is 1.27 inches

Explanation:
Given: The base(b) of the square pyramid as 8 inches length and height(h) of the square pyramid as 12 inches.
Now we need to calculate Volume of square pyramid to know the height of cylinder.
The volume of the square pyramid is given by = 1/3(b)²h
Volume = 1/3 (8×8)²×12
Volume = 1/3 (64)×12
Volume = 256 inches
Now pyramid is reshaped into a cylinder with a radius = 8 inches
Volume of cylinder is π r² h
Volume = 3.14×8×8×h
Volume is 256 inches
256 = 3.14×8×8×h
h = 1.27 inches

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Lesson 8.4 Finding Volume and Surface Area of Spheres detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 8 Lesson 8.4 Answer Key Finding Volume and Surface Area of Spheres

Math in Focus Grade 7 Chapter 8 Lesson 8.4 Guided Practice Answer Key

Solve.

Question 1.
The diameter of a sphere is 8.8 meters. What is the volume of the sphere? Round your answer to the nearest hundredth. Use 3.14 as an approximation for π.

Radius of the sphere:
8.8 ÷ = m The radius is half the diameter.
Volume of the sphere:
\(\frac{4}{3}\)πr³ ≈ \(\frac{4}{3}\) ∙ ∙ ∙ ∙ Use the formula for the volume of a sphere.
= Multiply.
≈ m³ Round to the nearest hundredth.
The volume of the sphere is about cubic meters.
Answer: The volume of the sphere is about 356.63 cubic meters.

Explanation:
Given, diameter of a sphere is 8.8 meters.
Then r = 4.4 m
We know that the volume of the sphere is V =  \(\frac{4}{3}\)πr³
V = \(\frac{4}{3}\) × 3.14 × 4.4 × 4.4 × 4.4
V = \(\frac{1069.9}{3}\)
V = 356.63 m³
So, The volume of the sphere is about 356.63 cubic meters.

Question 2.
Diane bought a spherical ball made of quartz at a garage sale. The volume of the ball is 1,450 cubic centimeters. What is the radius of the ball to the nearest centimeter? Use 3.14 as an approximation for π.

The radius of the spherical ball is about centimeters.
Answer: The radius of the sphere is  7 meters.

Explanation:
Given, The volume of the sphere is about 1450 cubic meters.
Then r = ?
We know that the volume of the sphere is V =  \(\frac{4}{3}\)πr³
1450 = \(\frac{4}{3}\) × 3.14 ×r³
r³ = \(\frac{1450 × 3}{3.14 × 4}\)
r³ = \(\frac{4350}{12.56}\)
r³= 346.33
r = 7
So, The radius of the sphere is  7 meters.

Hands-On Activity

Materials:

• string
• hemisphere
• cylinder that has the same radius and height twice the radius of the hemisphere

Work in pairs.

STEP 1: Wrap the string tightly around the hemisphere as shown in the diagram. Measure the length of the string used, and then use the same length of string to wrap around the curved surface of the cylinder.

STEP 2: How much of the curved surface of the cylinder does the string cover? If you wrap the whole sphere with string, how much of the curved surface of the cylinder would the same string cover?

STEP 3: How is the surface area of the sphere related to the area of the curved surface of the cylinder?

STEP 4: Copy and complete the following to write a formula for the surface area of a sphere. (Hint: The height of a cylinder is twice the radius of a sphere.)
Surface area of a sphere = Area of cylinder’s curved surface

Solve.

Question 3.
What is the surface area of a sphere with a radius of 6 centimeters? Use 3.14 as an approximation for π.
Answer: The surface area of a sphere is  452.16 cm²

Explanation:
Given, r = 6 cm,
The surface area of a sphere is A = 4πr²
= 4 × 3.14 × 6 × 6
= 452.16 cm²
So, The surface area of a sphere is  452.16 cm²

Question 4.
A spherical rubber ball has a surface area of 3,215.36 square centimeters. What is the radius of the rubber ball to the nearest centimeter? Use 3.14 as an approximation for π.

The radius of the rubber ball is about centimeters.
Answer: The radius of a sphere is 16 cm.

Explanation:
rubber ball has a surface area of 3,215.36 square centimeters.
Given, A = 3,215.36 cm²,
The surface area of a sphere is A = 4πr²
3,215.36 = 4 × 3.14 × r²
r² = \(\frac{3,215.36}{3.14 × 4}\)
r² = 256
r = 16
So, The radius of a sphere is 16 cm.

Math in Focus Course 2B Practice 8.4 Answer Key

For this practice, you may use a calculator and use 3.14 as an approximation for π. Round your answers to the nearest tenth when you can.

Solve.

Question 1.
What is the volume of a sphere with a radius of 5 centimeters?
Answer: The volume of the sphere is about 523.33 cm³.

Explanation:
Given,  r = 5 cm
We know that the volume of the sphere is V =  \(\frac{4}{3}\)πr³
V = \(\frac{4}{3}\) × 3.14 × 5 × 5 × 5
V = \(\frac{1570}{3}\)
V = 523.33 cm³
So, The volume of the sphere is about 523.33 cm³.

Question 2.
What is the surface area of a sphere with a radius of 4.4 centimeters?
Answer: The surface area of a sphere is  243.16 cm²

Explanation:
Given, r = 4.4 cm,
The surface area of a sphere is A = 4πr²
= 4 × 3.14 × 4.4 × 4.4
= 243.16 cm²
So, The surface area of a sphere is  243.16 cm²

Question 3.
A fully inflated beach ball has a radius of 10 inches.
a) What is the surface area of the beach ball?
Answer: The surface area of a sphere is  1256 in²

Explanation:
Given, r = 10 in,
The surface area of a sphere is A = 4πr²
= 4 × 3.14 ×10 × 10
= 1256 in²
So, The surface area of a sphere is  1256 in²

b) What is the volume of air inside the beach ball?
Answer:  The volume of the air inside the ball is about 4186.6 in³.

Explanation:
Given,  r = 10 in
We know that the volume of the sphere is V =  \(\frac{4}{3}\)πr³
V = \(\frac{4}{3}\) × 3.14 × 10 × 10 × 10
V = 4186.6 in³
So, The volume of the air inside the ball is about 4186.6 in³.

Question 4.
What is the surface area of a sphere.
a) with a diameter of 28.6 centimeters?
Answer: The surface area of a sphere is  2,568 cm²

Explanation:
Given, d = 28.6
Then, r = 14.3 cm,
The surface area of a sphere is A = 4πr²
= 4 × 3.14 × 14.3 × 14.3
= 2,568 cm²
So, The surface area of a sphere is  2,568 cm²

b) with a volume of 3,680 cubic centimeters?
Answer:  The surface area of a sphere is  1133.54 cm²

Explanation:
Given, The volume of the sphere is about 3680 cubic meters.
Then r = ?
We know that the volume of the sphere is V =  \(\frac{4}{3}\)πr³
3680 = \(\frac{4}{3}\) × 3.14 ×r³
r³ = \(\frac{3680× 3}{3.14 × 4}\)
r³ = \(\frac{11040}{12.56}\)
r³= 878.9
r = 9.5
So, The radius of the sphere is  9.5 cm.

Then, r = 9.5 cm,
The surface area of a sphere is A = 4πr²
= 4 × 3.14 × 9.5 × 9.5
= 1133.54 cm²
So, The surface area of a sphere is  1133.54 cm²

Question 5.
A billiard ball has a surface area of 84 square centimeters. What is the radius of the billiard ball?
Answer: The radius of a billiard ball is 2..5 cm.v

Explanation:
billiard ball has a surface area of 84 square centimeters.
Given, A = 84 cm²,
The surface area of a sphere is A = 4πr²
84 = 4 × 3.14 × r²
r² = \(\frac{84}{3.14 × 4}\)
r² = 6.68
r = 2.5
So, The radius of a billiard ball is 2..5 cm.

Question 6.
A bowl is in the shape of a hemisphere. The radius of the bowl is 10 centimeters. How many liters of water can the bowl hold? (1,000 cm³ = 1 liter)

Answer: the bowl can hold 0.020 liters of water

Explanation:
Volume of the hemisphere  V = \(\frac{2}{3}\)πr
Given, r = 10 cm
V = \(\frac{2}{3}\) × 3.14 × 10
V = 20.9 cm³
The volume of the hemisphere is 20.9 cm³

Then, Given that 1000 cm = 1 liter, and
we want to know how many liters the bowl can hold.
= \(\frac{20.93}{1000}\)
= 0.020 liters
So, the bowl can hold 0.020 liters of water

Question 7.
Math Journal
a) What dimension of a sphere do you need to find its surface area and volume?
Answer: We need the dimension of radius of a sphere to find the surface area and its volume

b) Suppose a sphere has a radius greater than 1 unit. If you double the radius, which value will increase by a greater amount, the volume or the surface area of the sphere? Explain your thinking.
Answer: The surface area increases. with the square of the radius.
If you double the radius you quadruple the surface area

Explanation:
If r is the radius of a sphere its surface area = 4πr².
If the radius is doubled then its surface area = 4π(2r)² = 4 × 4πr².
So the surface area becomes 4-fold or the increase in surface area is 300%.

Question 8.
A basketball is shipped in a cube-shaped box. The basketball just touches the sides of the box, as shown.

a) What is the radius of the basketball?
Answer: the radius of the basketball is  4.8 in.

Explanation:
Given, diameter of a sphere is 9.6 in.
r = \(\frac{d}{2}\)
= \(\frac{9.6}{2}\)
= 4.8 in.
Then r = 4.8 in.
So, the radius of the basketball is  4.8 in.

b) What is the volume of the basketball?
Answer: The volume of the basketball is about 463 in³.

Explanation:
Given,  r = 4.8 in
We know that the volume of the sphere is V =  \(\frac{4}{3}\)πr³
V = \(\frac{4}{3}\) × 3.14 × 4.8 × 4.8 × 4.8
V = \(\frac{1389}{3}\)
V = 463 in³
So, The volume of the basketball is about 463 in³.

c) About what percent of the space in the cube is occupied by the basketball?
Answer: 52.3 % of the cube is occupied by the baseball

Explanation:
Given, the cube side a = 9.6 in
The volume of a cube  V = a³
V = 9.6 × 9.6 × 9.6
V = 884.73 in³
The volume of a cube is 884.73 in³

The volume of the basketball is about 463 in³.
So, around 52.3 % of the cube is occupied by the baseball

Question 9.
A solid metal ball with a radius of 10 inches is melted and made into smaller spherical metal balls with a radius of 2 inches each. How many smaller spherical balls can be made?
Answer: total of 125 smaller spherical metal balls can be made

Explanation:
r = 10 in
Volume of the metal ball  V =  \(\frac{4}{3}\)πr³
V = \(\frac{4}{3}\) × 3.14 × 10 × 10 × 10
V = 4186.6 in³
So, The volume of the sphere is about 4186.6 in³

Volume of the smaller spherical metal balls , with r = 2 in.
V =  \(\frac{4}{3}\)πr³
V = \(\frac{4}{3}\) × 3.14 × 2 × 2 × 2
V = 33.49 in³

So, divide the volumes, we get
= \(\frac{4186.6}{33.49}\)
= 125
Hence,  total of 125 smaller spherical metal balls can be made

Question 10.
Nathan cuts a clay sphere in half to get two hemispheres. He measures the circumference of the hemispheres to be 175.84 centimeters.

a) What is the radius of each hemisphere?
Answer: The radius of each hemisphere is 28 cm

Explanation:
Given, C = 175.84 cm
The circumference of the hemisphere is C = 2πr
175.84 = 2 × 3.14 × r
r = \(\frac{175.84}{6.28}\)
r = 28 cm
So, The radius of each hemisphere is 28 cm

b) What is the total surface area of each solid hemisphere?
Answer: The  total surface area of each hemisphere is  7,385.28 cm²

Explanation:
the total surface area of each hemisphere is A = 3πr²
= 3 × 3.14 × 28 × 28
= 7,385.28 cm²
So, The  total surface area of each hemisphere is  7,385.28 cm²

Question 11.
The volume of a sphere is 3,052.08 cubic meters.
a) What is the radius of the sphere?
Answer: The radius of the sphere is  9 cm.

Explanation:
Given, The volume of the sphere is about 3,052.08 cubic meters.
Then r = ?
We know that the volume of the sphere is V =  \(\frac{4}{3}\)πr³
3,052.08 = \(\frac{4}{3}\) × 3.14 ×r³
r³ = \(\frac{3,052.08× 3}{3.14 × 4}\)
r³ = \(\frac{91529}{12.56}\)
r³= 729
r = 9
So, The radius of the sphere is  9 cm.

b) What is the surface area of the sphere?
Answer: The surface area of a sphere is  1017.36 cm²

Explanation:
Then, r = 9 cm,
The surface area of a sphere is A = 4πr²
= 4 × 3.14 × 9 × 9
= 1017.36 cm²
So, The surface area of a sphere is  1017.36 cm²

Question 12.
Once you know how to find the surface area of a sphere, you can use the surface area formula to see why the volume formula works.
Think of a sphere as being divided up into hundreds of “pyramids” that have a common vertex at the center of the sphere. The surface of the sphere is made up of the bases of all these pyramids. The height of each pyramid is r, and you can call the areas of the bases B₁, B₂, B₃, and so on. To find the surface area of the sphere, you can find the sum of the areas of all the bases.

Show why the volume formula works by supplying a reason for each step below.

Answer:
Add the volumes of the pyramids to get the volume of the sphere:
V = \(\frac{1}{3}\) · r · B₁ + \(\frac{1}{3}\) · r · B₂ + \(\frac{1}{3}\) · r · B₃ + ………..
Factor out \(\frac{1}{3}\)r:
= \(\frac{1}{3}\)r(B₁ + B₂ + B₃)
The sum of the pyramids bases form the surface area of the sphere:
= \(\frac{1}{3}\)r(4πr²)
Use the commutative property of multiplication:
= \(\frac{4}{3}\) πr³

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Lesson 8.5 Real-World Problems: Composite Solids detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 8 Lesson 8.5 Answer Key Real-World Problems: Composite Solids

Math in Focus Grade 7 Chapter 8 Lesson 8.5 Guided Practice Answer Key

Solve.

Question 1.
A composite solid is made up of a cone and a cylinder. The slant height of the cone is 25 centimeters. The height of the cylinder is 15 centimeters and its radius is 12 centimeters. The height of the solid is 37 centimeters. Use 3.14 as an approximation for π.

a) Find the volume of the composite solid to the nearest cubic centimeter.
Volume of the cylinder:
πr²h = ∙ ∙ Use the formula for volume of a cylinder.
= cm³ Evaluate .

Volume of the cone:
Height of cone = –
= cm
\(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) ∙ ∙ ∙ ∙ Use the formula for the volume of a cone.
= cm³ Multiply.
Volume of the composite solid:
+ = Add the volumes of the cylinder and the cone.
≈ cm³ Multiply and round.
The volume of the composite solid is about cubic centimeters.
Answer:  The volume of the composite solid is about 10,098 cm³
We are given:
l = 25
h_cylinder = 15
H = 37
r = 12
Use the formula for the volume of a cylinder:
V_cylinder = πr²h_cylinder
= π · 12² · 15
Evaluate:
= 2,160π cm³
The height of the cone:
h_cone = H – h_cylinder
= 37 – 15
= 22 cm
Use the formula for the volume of a cone:
V_cone = \(\frac{1}{3}\)πr²h_cone
= \(\frac{1}{3}\) · π · 12² · 22
Multiply:
= 1,056π cm³
Find the volume of the composite solid by adding the volumes of the cylinder and the cone:
V = V_cylinder + V_cone
= 2,160π + 1,056π
= 3,216π cm³
Multiply and round:
≈ 10,098 cm³

b) Find the surface area of the composite solid to the nearest square centimeter.

The surface area of the composite solid is about square centimeters.
Answer: The surface area of the composite solid is about  2,525 cm^2.
Use the formuLas for the surface areas:
S = πrl + 2πrh_cylinder + πr²
Substitute for r, l, h_cylinder
= π · 12 · 25 + 2π · 12 · 15 + π · 12²
Evaluate each term:
= 300π + 360π + 144π
Add:
= 804π
Multiply and round:
≈ 804 · 3.14 ≈ 2,525 cm²

Question 2.
A birdhouse looks like a cube with a square pyramid on top. As shown, the birdhouse has a circular entrance with a diameter of 4 inches. Find the exterior surface area of the birdhouse.

Answer: The exterior surface area of the birdhouse is 647.4 in²
Determine the Lateral surface area of the pyramid:
S₁ = 4 · \(\frac{1}{2}\) · 10 · 13 = 260 in²
Determine the radius of the entrance:
r = \(\frac{4}{2}\) = 2
Determine the lateral surface area of the house:
S₂ = 4 · 10 · 10 – π · 2² ≈ 387.4 in²
Compute the exterior surface area of the house:
S = S₁ + S₂ = 260 + 387.4 = 647.4 in²
= 647.4 in²

Question 3.
The solid shown is a cylinder with a cone-shaped hole. The diameter of the cylinder is 22 centimeters. Its height is 15 centimeters. The radius of the cone-shaped hole is 7 centimeters and the height is 10 centimeters. Find the volume of the solid. Use 3.14 as an approximation for π. Round your answer to the nearest cubic centimeter.

Answer: The volume of the solid is  5. 187 cm³
Determine the radius R of the cylinder:
R = \(\frac{22}{2}\) = 11 cm
Determine the volume of the cylinder
V_cylinder = πR²h = π · 11² ·  15 = 1, 815π cm²
Determine the volume of the cone:
V_cone = \(\frac{1}{3}\)πR²h = \(\frac{1}{3}\) · π · 7² · 10 ≈ 163π cm³
Determine the volume of the solid:
V = V_cylinder – V_cone
= 1,815π – 163π
= 1,652π
≈ 3.14 · 1,652
≈ 5. 187 cm³

Math in Focus Course 2B Practice 8.5 Answer Key

For this practice, you may use a calculator and use 3.14 as an approximation for π. Round your answers to the nearest tenth when you can.

Solve.

Question 1.
Jack has a cylindrical block that has a radius of 0.6 cm and is 22 centimeters long. He puts together 8 such blocks to form the composite solid shown. What is the volume of the composite solid?

Answer: The volume of the composite solid is 199 cm³
We are given the cylinder:
r = 0.6
h = 22
Determine the volume of one cylinder:
V_cylinder = πr²h ≈ 3.14 · 0.6² · 22
≈ 24.87 cm³
Determine the volume of the composite solid:
V_solid = 8V_cylinder = 8 · 24.87 ≈ 199 cm³

Question 2.
The trophy for a basketball tournament is made up of a miniature basketball attached to a rectangular prism. The radius of the basketball is 5 centimeters. The prism measures 8 centimeters by 5 centimeters by 15 centimeters. What is the volume of the trophy to the nearest cubic centimeter?

Answer: The volume of the trophy is 1,123 cm³
We are given:
r = 5
l = 8
w = 5
h = 15
Determine the volume of the sphere:
V_sphere = \(\frac{4}{3}\)πr³
≈ \(\frac{4}{3}\) · 3.14 · 5³
≈ 523 cm³
Determine the volume of the prism:
V_prism = lwh = 8 · 5 · 15 = 600 cm³
Determine the volume of the trophy:
V_trophy = V_sphere + V_prism
= 523 + 600
= 1,123 cm³

Question 3.
A necklace is made up of 50 spherical beads. Each bead has a radius of 8 millimeters.
a) What is the volume of the necklace?
Answer: The volume of the necklace is 107, 178.7 mm³
We are given the sphere:
r = 8
Determine the volume of a spherical bead:
V_bead = \(\frac{4}{3}\)πr³
≈ \(\frac{4}{3}\) · 3.14 · 8³
≈ 2,143.573
Determine the volume of the necklace:
V_necklace = 50 · V_bead
≈ 50 – 2. 143.573 ≈ 107, 178.7 mm³

b) What is the surface area of the necklace?
Answer: The surface area of the necklace is 40,192 mm²
Determine the surface area of a spherical bead:
S_bead = 4πr² ≈ 4 · 3.14 · 8²
≈ 803.84 mm²
Determine the surface area of the necklace:
S_necklace = 50 · S_bead
= 50 · 803.84
≈ 40,192 mm²

Question 4.
A crystal trophy is made up of a rectangular pyramid whose base is attached to the top of a rectangular prism. The base of the pyramid and the top of the prism are each 4 inches long and 3 inches wide. The height of the pyramid is 2.5 inches and the height of the prism is 9.5 inches. What is the volume of the crystal trophy?
Answer: The volume of the crystal trophy is 124 in³
We are given:
l = 4
w = 3
h_pyramid = 2.5
h_prism = 9.5
Determine the volume of the pyramid:
V_pyramid =\(\frac{1}{3}\) · lwh_pyramid 
= \(\frac{1}{3}\) · 4 · 3 · 2.5
= 10 in³
Determine the volume of the prism:
V_prism = lwh_prism = 4 · 3 · 9.5
= 114 in³
Determine the volume of the trophy:
V_trophy = V_pyramid + V_prism
= 10 + 114
= 124 in³

Question 5.
At a food stand, you can buy a paper cone filled with slush made of frozen juice. The slush forms a hemisphere on top of the cone, as shown. What is the volume of the cone of slush?

Answer: The volume of the cone of slush is 314.9 cm³
We are given:
d = 8
h = 10.8
Determine the radius:
r = \(\frac{d}{2}\) = \(\frac{8}{2}\) = 4
Determine the volume of the hemisphere:
V_hemisphere = \(\frac{1}{2}\) · \(\frac{4}{3}\) πr³
≈ \(\frac{2}{3}\) · 3.14 · 4³
≈ 134 cm³
Determine the volume of the cone:
V_cone = \(\frac{1}{3}\) πr²h
≈ \(\frac{1}{3}\) · 3.14 · 4² · 10.8
≈ 180.9 cm³
Determine the volume of the cone of slush:
V = V_hemisphere + V_cone
= 134 + 180.9
= 314.9 cm³

Question 6.
A wooden paper towel holder is composed of two cylinders. The diameter of the base is 12 .centimeters and its height is 2 centimeters. The combined height of the two cylinders is 30 centimeters. What is the volume of the paper towel holder?

Answer: The volume of the paper towel holder is 438 cm³
We are given:
d₁ = 3
h₁ = 30
d₂ = 12
h₂ = 2
Determine the radius of the upper cylinder:
r₁ = \(\frac{d_{1}}{2}\) = \(\frac{3}{2}\) = 1.5
Determine the volume of the upper cylinder:
V₁ = πr₁²h₁ = π · 1 5² · 30 = 67.5π cm³
Determine the radius of the lower cylinder:
r₂ = \(\frac{d_{2}}{2}\) = \(\frac{12}{2}\) = 6
Determine the volume of the lower cylinder:
V₂ = πr₂²h₂ = π · 6² · 2 = 72π cm³
Determine the volume of the paper tower holder:
V = V₁ + V₂ = 67.5π + 72π
≈ 139.5π
≈ 139.5 · 3.14
≈ 438 cm³

Question 7.
The edge of the base of a square pyramid is 11 inches. The pyramid has a height of 14 inches. What is the volume of the composite solid formed when two such pyramids are joined at the base, as shown in the diagram?

Answer: The volume of the composite solid is  1,129.3 in³
We are given:
L = 11
h = 14
Determine the volume of one pyramid:
V₁ = V₂ = \(\frac{1}{3}\)l²h = \(\frac{1}{3}\) · 11² · 14 ≈ 564.67 in³
Determine the volume of the composite solid:
V = V₁ + V₂ = 2V₁ = 2 · 564.67
≈ 1,129.3 in³

Question 8.
A clock in the shape of an hour glass is made up of two identical cones connected at their vertices. The radius of each cone is 7 centimeters. The combined height of the two cones is 11.8 centimeters. What is the volume of the clock?

Answer: The volume of the clock is 605.2 cm³
We are given:
r = 7
H = 11.8
Determine the height of each cone:
h = \(\frac{H}{2}\) = \(\frac{11.8}{2}\) = 5.9
Determine the volume of one cone:
V₁ = V₂ = \(\frac{1}{3}\)πr²h =  \(\frac{1}{3}\) · π · 7² · 59
≈ 96.3677π
Determine the volume of the clock:
V = V₁ + V₂ = 2V₁ = 2 · 96.367π
≈ 192.734 · 3.14
≈ 605.2 cm³

Question 9.
Jason made a long pole by joining three different lengths of cylindrical poles together. Each pole has the same diameter of 18 cm. The diagram below is not drawn to scale.

a) What is the volume of the long pole?
Answer: The volume of the long pole is 8,393.2 cm³
We are given:
d = 18
h₁ = 7
h₂ = 11
h₃ = 15
Determine the radius
r = \(\frac{d}{2}\) = \(\frac{18}{2}\) = 9
Determine the volume of the tong pole:
V = V₁ + V₂ + V₃
= πr²h₁ + πr²h₂ + πr² h3
= πr²(h₁ + h₂ + h₃)
≈ 3.14 · 9² · (7 + 11 + 15)
≈ 8,393.2 cm³

b) What is the surface area of the long pole?
Answer: The surface area of the long pole is  2,373.8 cm²
Determine the surface area of the long pole
= 2πr² + 2πr(h₁ + h₂ + h₃)
= 2πr(r + h₁ + h₂ + h₃)
≈ 2 · 3.14 · 9(9 + 7 + 11 + 15)
≈ 2,373.8 cm²

Brain @ Work

How can you make the following cross-sections by slicing a cube? Use a computer drawing program or pencil and paper to show your answers for a) and b).
a) An isosceles triangle
Answer:
We take a point on one edge, then take two congruent segments from one of the vertices on the same edge on the other two edges
We draw the cross-section in the cube in the shape of an isosceles triangle:

b) A regular hexagon
Answer:
To get a hexagon, we slice the cube with a plane going through alt six faces of the cube.
We draw the cross-section in the cube in the shape of a regular hexagon:

c) What other polygons can be cross-sections of a cube?
Answer:
The cross-section can be a quadrilateral when a plane crosses 4 faces of the cube:

The cross-section can also be a pentagon when a plane crosses 5 faces of the cube:

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Lesson 8.3 Finding Volume and Surface Area of Pyramids and Cones detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 8 Lesson 8.3 Answer Key Finding Volume and Surface Area of Pyramids and Cones

Hands-On Activity

Materials:

• nets for a pyramid, a triangular prism, a cylinder, and a cone
• scissors
• ruler
• tape
• rice grains

EXPLORE VOLUME RELATIONSHIPS BETWEEN A PYRAMID AND A PRISM, AND BETWEEN A CONE AND A CYLINDER

Work in pairs.

STEP 1: Cut out the nets for a triangular pyramid and triangular prism.

STEP 2: Fold the nets to make a triangular pyramid and triangular prism. Each solid will be open at one end. Tape the edges to hold them in place.

STEP 3: Do the pyramid and prism have the same height? Do the base of the pyramid and the base of the prism have about the same area?

STEP 4: Fill the pyramid completely with rice grains and pour them into the prism. Keep doing this until the prism is filled with rice grains. Flow many pyramids of rice grains are needed to fill the prism?

Math Journal What is the relationship between the volume of a triangular pyramid and the volume of a triangular prism with the same base and height? Suggest a formula for the volume of a pyramid.

STEP 5: Cut out the nets for a cylinder and cone.

STEP 6: Use the cutouts of the rectangle and circle to make a cylinder. Tape the edges to hold them in place. The cylinder will be open at one end.

STEP 7: Fold the cutout of the sector of the circle to make a cone. Tape the edges to hold them in place. The cone will be open at its wide end.

STEP 8: Are the cylinder and the cone about the same height? Do the base of the cylinder and the base of the cone have about the same area?

STEP 9: Fill the cone completely with rice grains and pour them into the cylinder. Keep doing this until the cylinder is filled with rice grains. How many cones of rice grains are needed to fill the cylinder?

Math Journal What is the relationship between the volume of a cone and the volume of a cylinder with the same base and height? Suggest a formula for the volume of a cone.
Answer: The relationship between the volume of a cone and the volume of a cylinder with the same base and height is the volume of the cone is one-third the volume of the cylinder.

Explanation:
We know that the volume of the cylinder is  V = πr²h
the volume of the cone V=1/3hπr².
Then, The relationship between the volume of a cone and the volume of a cylinder,
with the same base and height is the volume of the cone is one-third the volume of the cylinder.
And, \(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) πr²h
Hence, The volume of the cone and the volume of the cylinder is equal when their respective base and height are equal.

Math in Focus Grade 7 Chapter 8 Lesson 8.3 Guided Practice Answer Key

Solve.

Question 1.
The base of a pyramid is a right triangle. The triangle has a base of 5 centimeters and a height of 3 centimeters. The pyramid has a height of 6 centimeters. What is the volume of the pyramid?

Use the formula for the volume of a pyramid.
Volume of pyramid
= \(\frac{1}{3}\) Bh Use the formula.
= \(\frac{1}{3}\) ∙ ( ∙ ∙ ) ∙     Substitute for B and h.
= cm³ Multiply.
The volume of the pyramid is cubic centimeters.
Answer:  The volume of the pyramid is 10 cm³

Explanation:
The volume of the pyramid is \(\frac{1}{3}\) bh
Here b = 5 cm, h = 6 cm
V = \(\frac{1}{3}\) (\(\frac{1}{2}\) × 3 × 5) 6
V = 15 cm³
So, The volume of the pyramid is 15 cm³.

Question 2.
A square pyramid has a height of 12 meters. If a side of the base measures 5.5 meters, what is the volume of the pyramid?
Answer: The volume of the Square pyramid is 121 cm³.

Explanation:
The volume of the Square pyramid is \(\frac{1}{3}\) a²h
Here a = 5.5 m, h = 12 m
V = \(\frac{1}{3}\) (5.5 × 5.5 × 12)
= 30.25 × 4
V = 121 m³
So, The volume of the Square pyramid is 121 cm³.

Question 3.
A square pyramid has a volume of 400 cubic centimeters. The length of the base is 10 centimeters. What is the height of the pyramid?
Let the height of the pyramid be h centimeters.
Volume of pyramid = \(\frac{1}{3}\) Bh Use the formula.
= \(\frac{1}{3}\) ∙ ( ∙ ) ∙ h Substitute for the volume and dimensions of the base.
= ∙ h Multiply both sides by .
= Divide both sides by
= h Simplify.
The height of the pyramid is centimeters.
Answer: The height of the Square pyramid is 12 cm.

Explanation:
The volume of the Square pyramid is 400 cm³
Here a = 10 cm, h = ?
V = \(\frac{1}{3}\) a²h
400 = \(\frac{1}{3}\) 10 × 10 × h
h = \(\frac{400 × 3}{10 × 10}\)
= 12 cm
So, The height of the Square pyramid is 12 cm.

Question 4.
A rectangular pyramid has a volume of 58 cubic meters. If the sides of the base measure 3 meters by 5 meters, what is the height of the pyramid?
Answer: The height of the Rectangular pyramid is 11.6 m.

Explanation:
The volume of the Rectangular pyramid is 58 m³
Here l = 5 m, b = 3 m , h = ?
V = \(\frac{1}{3}\) lbh
58 = \(\frac{1}{3}\) 5 × 3 × h
h = \(\frac{58 × 3}{5 × 3}\)
= 11.6 m
So, The height of the Rectangular pyramid is 11.6 m.

Question 5.
A party hat is in the shape of a cone. Find the exact volume of the party hat. Use 3.14 as an approximation for π to find the approximate volume of the cone. Round your answer to the nearest tenth.

Use the formula for the volume of a cone, V = \(\frac{1}{3}\)Bh
Volume of the party hat = \(\frac{1}{3}\) ∙ πr² ∙ h Use the formula.
= \(\frac{1}{3}\) ∙ ∙ ∙ Substitute for π, r, and h.
= Multiply.
≈ in³ Round to the nearest tenth.
The exact volume of the party hat is cubic inches.
An approximate volume is cubic inches.
Answer: The volume of the party hat is 504 in³.

Explanation:
the volume of the cone V=1/3hπr².
Here r = 6 in. , h = 14 in.
V = \(\frac{1}{3}\) 14 × 6 × 6 × 3.14
V = 504 in³.
So, The volume of the party hat is 504 in³.

Question 6.
The diagram shows a cone-shaped container. Find the exact volume of the container. Use 3.14 as an approximation for π to find the approximate volume of the container. Round your answer to the nearest tenth.

Answer: The volume of the container is 283 cm³.

Explanation:
the volume of the cone V=1/3hπr².
Here d= 9 cm , then
r = 4.5 cm , h = 14 cm.
V = \(\frac{1}{3}\) 14 × 4.5 × 4.5 × 3.14
V = 283.5 cm³.
So, The volume of the container is 283 cm³.

Question 7.
A cone has a height of 57 centimeters and a volume of 2,923.34 cubic centimeters. What is the radius of the cone? Use 3.14 as an approximation for π.
Let the height of the cone be h centimeters.

The radius of the cone is about centimeters.
Answer: The radius of the cone  is 7.16 cm.

Explanation:
the volume of the cone V=1/3hπr².
Here V = 2923.34 cm³
r = ? , h = 57 cm.
2923.34 = \(\frac{1}{3}\) 57 × r² × 3.14
r² = \(\frac{2923.34 × 3}{57 × 3.14}\)
r² =51.28 cm.
r = 7.16 cm
So, The radius of the cone  is 7.16 cm.

Question 8.
A cone has a height of 7.2 inches and a volume of 188.4 cubic inches. What is the radius of the cone? Use 3.14 as an approximation for π.
Answer: The radius of the cone  is 5.11 in.

Explanation:
the volume of the cone V=1/3hπr².
Here V = 188.4 in³
r = ? , h =7.2 in.
188.4 = \(\frac{1}{3}\) 7.2 × r² × 3.14
r² = \(\frac{188.4 × 3}{7.2 × 3.14}\)
r² =26.16 in.
r = 5.11 in
So, The radius of the cone  is 5.11 in.

Question 9.
Jean wants to sell lemonade in cone-shaped paper cups. Each cup has a diameter of 4 centimeters and a height of 8 centimeters. She wants to make enough lemonade for 50 cups. How much lemonade does Jean need to make? Round your answer to the nearest tenth. Use 3.14 as an approximation for π.

Answer: The lemonade jean needs to made is 1600 ml

Explanation:
the volume of the cone V=1/3hπr².
Here d= 4 cm , then
r = 2 cm , h = 8 cm.
V = \(\frac{1}{3}\) 8 × 2 × 2 × 3.14
V = 32 cm³.
So, The volume of the cup is 32 cm³.

Given, She wants to make enough lemonade for 50 cups.
= 50 × 32
= 1600 ml
So, The lemonade jean needs to made is 1600 ml

Hands-On Activity

FIND A FORMULA FOR THE AREA OF THE CURVED SURFACE OF A CONE

Work in pairs.

STEP 1: Copy and complete the table. For each pyramid, the base has edges of length b. The slant height is h.

STEP 2: In general, what is the formula for the area of all the lateral triangular faces of a pyramid?

STEP 3: Suppose the number of sides of the base of the pyramid increases as shown below. Name the solid that is eventually formed.

STEP 4: The formula for the curved surface area of a cone is related to the formula for the surface area of the triangular faces of a pyramid. Copy and complete the formula for the curved surface area of a cone. Let r represent the radius of the base of a cone. Let € represent the slant height of the cone.

Question 10.
A solid cone has a radius of 7 inches and a slant height of 14 inches.

a) What is the exact area of the cone’s curved surface?
πrl = ∙ ∙ Use the formula for the lateral surface area of a cone.
= in² Multiply.
The exact area of the cone’s curved surface is square inches.
Answer: The exact area of the cone’s curved surface is 307.72 in²

Explanation:
The curved surface area of a cone  S =  πrl
Here r = 7 in, l = 14 in
S = 3.14 × 7 × 14
S = 307.72 in²
So, The exact area of the cone’s curved surface is 307.72 in²

b) What is the total surface area of the cone? Find both the exact value and an approximate value. Use \(\frac{22}{7}\) an approximation for π.
Total surface area of the cone:
Area of base + Area of curved surface = πr² + The lateral surface area is .
= ∙ ∙ + Substitute for r.
= Multiply.
= Add.
= in² Substitute for π.
The total surface area of the cone is exactly square inches, and approximately square inches.
Answer: The total surface area of the cone is exactly 462 in²

Explanation:
The total surface area of the cone is T = πr(r +l)
Here r = 7 in, l = 14 in
T = \(\frac{22}{7}\) × 7 ( 7 + 14 )
= 21 × 22
T = 462 in²
So, The total surface area of the cone is exactly 462 in²

Question 11.
The radius of a cone is 3 inches, and the slant height is 12 inches.

a) What is the exact area of the cone’s curved surface?
Answer: The exact area of the cone’s curved surface is 113.04 in²

Explanation:
The curved surface area of a cone  S =  πrl
Here r = 3 in, l = 12 in
S = 3.14 × 3 × 12
S = 113.04 in²
So, The exact area of the cone’s curved surface is 113.04 in²

b) What is the total surface area of the cone? Find both the exact value and an approximate value. Use 3.14 as an approximation for π.
Answer: The total surface area of the cone is exactly 141.3 in²

Explanation:
The total surface area of the cone is T = πr(r +l)
Here r = 3 in, l = 12 in
T = 3.14 × 3 ( 3 + 12 )
= 3.14 × 45
T = 141.3 in²
So, The total surface area of the cone is exactly 141.3 in²

Question 12.
A conical straw hat has a diameter of 16 inches and a lateral surface area of 251.2 square inches. Find the approximate slant height of the straw hat. Use 3.14 as an approximation for π.

Let the lateral height of the hat be l inches.
Area of lateral surface of hat = πrl Use the formula for the lateral surface area of a cone.

The slant height of the hat is about inches.
Answer: The slant height of the hat is about 10 in.

Explanation:
Area of lateral surface of hat = πrl
Here, d = 16 in , S = 251.2 in²
Then r = 8 in
251.2 = 3.14 × 8 × l
l = \(\frac{251.2}{3.14 × 8}\)
l = 10 in
So, The slant height of the hat is about 10 in.

Question 13.
Jessica makes a cone-shaped paper filter to line a cone-shaped funnel. The funnel has a radius of 5 inches and a slant height of 10 inches. Suppose Jessica wants to make cone-shaped filters for 25 such funnels. About how many square inches of filter paper will she need? Use 3.14 as an approximation for π.
Answer: The filter paper she needs is  3925 in²

Explanation:
The lateral surface area of the funnel S = πrl
Here r = 5 in, l = 10 in
S = 3.14 × 5 × 10
S = 157 in²
Given, Jessica wants to make cone-shaped filters for 25 such funnels.
157 × 25 = 3925 in²
So, The filter paper she needs is  3925 in²

Math in Focus Course 2B Practice 8.3 Answer Key

For this practice, you may use a calculator and use 3.14 as an approximation for π. Round your answers to the nearest tenth when you can.

Find the volume of each pyramid.

Question 1.

Base area = 72 cm²
Answer:  The volume of the pyramid is 48 cm²

Explanation:
The volume of a pyramid is V = \(\frac{1}{3}\)bh
Given, h = 15 cm, A = 72 cm²
A = \(\frac{1}{2}\)bh
b = \(\frac{72 × 2}{15}\)
b = 9.6 cm

V = \(\frac{1}{3}\) 9.6 × 15
V = 48cm²
So, The volume of the pyramid is 48 cm²

Question 2.

Answer: The volume of the pyramid is 32 cm³.

Explanation:
The volume of the pyramid is \(\frac{1}{3}\)bh
Here b = 8 cm, h = 12 cm
V = \(\frac{1}{3}\) 8 × 12
= 8 × 4
V = 32 cm³
So, The volume of the pyramid is 32 cm³.

Question 3.
A rectangular pyramid with a height of 10 inches and a base that measures 9 inches by 7 inches.
Answer: The volume of the rectangular  pyramid is 210 in³.

Explanation:
The volume of the rectangular pyramid is \(\frac{1}{3}\)lbh
Here b = 7 in, h = 10 in, l = 9 in
V = \(\frac{1}{3}\) 9 × 7 × 10
= 30 × 7
V = 210 in³
So, The volume of the rectangular  pyramid is 210 in³.

Question 4.
A square pyramid with a height of 18 feet and a base that is 12 feet on each edge.
Answer: The volume of the square  pyramid is 864 ft³.

Explanation:
The volume of the square pyramid is \(\frac{1}{3}\)a²h
Here h = 18 ft, l = 12 ft
V = \(\frac{1}{3}\) 12 × 12 × 18
= 144 × 6
V =864 ft³
So, The volume of the square  pyramid is 864 ft³.

Find the exact and approximate volume for each cone.

Question 5.

Answer: The exact volume of the cone is 16.74 cm³.
The approximate volume of the cone is 16 cm³

Explanation:
the volume of the cone V=1/3hπr².
Here r = 2 cm. , h = 4 cm.
V = \(\frac{1}{3}\) 4 × 2 × 2 × 3.14
V = 16.74 cm³.
So, The exact volume of the cone is 16.74 cm³.
The approximate volume of the cone is 16 cm³

Question 6.

Answer: The exact volume of the cone is 512.86 cm³.
The approximate volume of the cone is 512 cm³

Explanation:
the volume of the cone V=1/3hπr².
Here r = 7 cm. , h = 10 cm.
V = \(\frac{1}{3}\) 10 × 7 × 7 × 3.14
V = 512.86 cm³.
So, The exact volume of the cone is 512.86 cm³.
The approximate volume of the cone is 512 cm³

Question 7.
A solid cone with a diameter of 10 centimeters and a height of 8 centimeters.
Answer: The exact volume of the cone is 209.33 cm³.
The approximate volume of the cone is 209 cm³

Explanation:
the volume of the cone V=1/3hπr².
Here, d = 10 cm,
r = 5 cm. , h = 8 cm.
V = \(\frac{1}{3}\) 8 × 5 × 5 × 3.14
V = 209.33 cm³.
So, The exact volume of the cone is 209.33 cm³.
The approximate volume of the cone is 209 cm³

Question 8.
A cone with a radius of 4.9 centimeters and a height of 6.9 centimeters.
Answer:  The exact volume of the cone is 173.4 cm³.
The approximate volume of the cone is 173 cm³

Explanation:
the volume of the cone V=1/3hπr².
Here r = 4.9 cm. , h = 6.9 cm.
V = \(\frac{1}{3}\) 6.9 × 4.9 × 4.9 × 3.14
V = 173.4 cm³.
So, The exact volume of the cone is 173.4 cm³.
The approximate volume of the cone is 173 cm³

Find the exact and an approximate surface area for each solid cone. Round your approximation to the nearest square unit.

Question 9.
A cone with a radius of 2.5 centimeters and a slant height of 5.6 centimeters.
Answer:  The exact area of the cone’s curved surface is 43.96 cm²
The  approximate area of the cone’s curved surface is 44 cm²

Explanation:
The curved surface area of a cone  S =  πrl
Here r = 2.5 cm , l = 5.6 cm
S = 3.14 × 2.5 × 5.6
S = 43.96 cm²
So, The exact area of the cone’s curved surface is 43.96 cm²
The  approximate area of the cone’s curved surface is 44 cm²

Question 10.
A cone with a diameter of 72 centimeters and a slant height of 48 centimeters.
Answer:  The exact area of the cone’s curved surface is 5425.92 cm²
The  approximate area of the cone’s curved surface is 5426 cm²

Explanation:
The curved surface area of a cone  S =  πrl
Here, d = 72 cm then,
r = 36 cm , l = 48 cm
S = 3.14 × 36 × 48
S = 5425.92 cm²
So, The exact area of the cone’s curved surface is 5425.92 cm²
The  approximate area of the cone’s curved surface is 5426 cm²

Question 11.
A cone with a diameter of 18 meters and a slant height of 22 meters.
Answer:  The exact area of the cone’s curved surface is 621.72 m²
The  approximate area of the cone’s curved surface is 622 m²

Explanation:
The curved surface area of a cone  S =  πrl
Here, d = 18 m then,
r = 9 m , l =  22 m
S = 3.14 × 9 × 22
S = 621.72 m²
So, The exact area of the cone’s curved surface is 621.72 m²
The  approximate area of the cone’s curved surface is 622 m²

Solve.

Question 12.
A square pyramid has a volume of 605 cubic centimeters and a height of 15 centimeters.
a) What is the area of the base of the pyramid?
Answer: the area of the base of the Square pyramid is 121 cm².

Explanation:
the area of the base of the Square pyramid A = a²
Given, V = 605 cm³ , h = 15 cm
The volume of the Square pyramid is \(\frac{1}{3}\) a²h
605 = \(\frac{1}{3}\)  × 15 × a²
a² = \(\frac{605}{5}\)
a²= 121 cm
a = 11 cm

Then, A = 11 × 11 = 121 cm²
So, the area of the base of the Square pyramid is 121 cm².

b) What is the length of an edge of the base?
Answer:  the length of an edge of the base is 11 cm

Explanation:
Given, V = 605 cm³ , h = 15 cm
The volume of the Square pyramid is \(\frac{1}{3}\) a²h
605 = \(\frac{1}{3}\)  × 15 × a²
a² = \(\frac{605}{5}\)
a²= 121 cm
a = 11 cm
So, the length of an edge of the base is 11 cm

Question 13.
The volume of a square pyramid is 333 cubic centimeters. The length of an edge of the base of the pyramid is 10 centimeters. What is the height of the pyramid rounded to the nearest centimeter?
Answer: The height of the Square pyramid is 10 cm.

Explanation:
The volume of the Square pyramid is \(\frac{1}{3}\) a²h
Here V = 333 cm³, a = 10 cm
h = \(\frac{V × 3}{a²}\)
= \(\frac{333 × 3}{10 × 10}\)
h =9.99 cm = 10 cm.
So, The height of the Square pyramid is 10 cm.

Question 14.
A candle in the shape of a cone has a radius of 5 centimeters. The slant height is 15 centimeters.
a) What is the area of the base of the candle?
Answer: the area of the base of the candle is 78.5 cm²

Explanation:
the area of the base of the cone is A =πr²
Here , r = 5 cm,
A =3.14 × 5 × 5
A = 78.5 cm²
So, the area of the base of the candle is 78.5 cm²

b) What is the area of the curved surface of the candle?
Answer: The area of the curved surface of the candle  is 235.5 cm²

Explanation:
The curved surface area of a cone  S =  πrl
Here r = 5 cm, l = 15 cm
S = 3.14 × 5 × 15
S = 235.5 cm²
So, The area of the curved surface of the candle  is 235.5 cm²

c) Suppose the candle is wrapped in plastic. If there is no overlap, how much plastic is needed?
Answer: The area of the surface of the cone and the wrapped plastic area without overlapping will be equal

Explanation:
The curved surface area of a cone  S =  πrl
Here r = 5 cm, l = 15 cm
S = 3.14 × 5 × 15
S = 235.5 cm²
So, The area of the curved surface of the candle  is 235.5 cm²
Hence, The area of the surface of the cone and the wrapped plastic area without overlapping will be equal

Question 15.
A cone has a slant height of 8.5 centimeters. The height of the cone is 7.5 centimeters and the radius is 4 centimeters.
a) What is the area of the lateral surface?
Answer: the area of the lateral surface is 94.2 cm²

Explanation:
Area of lateral surface of hat = πrl
Here, r= 4 cm , l = 7.5 cm
S = 3.14 × 4 × 7.5
S = 94.2 cm²
So, the area of the lateral surface is 94.2 cm²

b) What is the volume of the cone?
Answer: The volume of the cone is125.6 cm³.

Explanation:
the volume of the cone V=1/3hπr².
Here , r = 4 cm , h = 7.5 cm.
V = \(\frac{1}{3}\) 7.5 × 4 × 4 × 3.14
V = 125.6  cm³.
So, The volume of the cone is125.6 cm³.

Question 16.
One of the entrances to the Louvre Museum in France is in the shape of a pyramid. The entrance has a height of about 70 feet and a volume of about 233,330 cubic feet. What is the area of the base of the pyramid to the nearest foot?

Answer: the area of the base of the Square pyramid is 10,000 ft².

Explanation:

the area of the base of the Square pyramid A = a²
Given, V = 233,330 ft³ , h = 70 ft
The volume of the Square pyramid is \(\frac{1}{3}\) a²h
233,330 = \(\frac{1}{3}\)  × 70 × a²
a² = \(\frac{233,330 × 3}{70}\)
a²= 9999.85 ft = 10,000
a = 100 ft

Then, A = 100 × 100 = 10,000 ft²
So, the area of the base of the Square pyramid is 10,000 ft².

Question 17.
Math Journal Cylinder P and cone Q have the same radius and height. The volume of cylinder P is 393 cubic centimeters. Joseph says that the volume of cone Q is 131 cubic centimeters. Explain how Joseph arrived at his answer.

Answer:  The volume of the cone and the volume of the cylinder is equal when their respective base and height are equal.

Explanation:
We know that the volume of the cylinder is  V = πr²h
the volume of the cone V=1/3hπr².
Then, The relationship between the volume of a cone and the volume of a cylinder,
with the same base and height is the volume of the cone is one-third the volume of the cylinder.
And, \(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) πr²h
Hence, The volume of the cone and the volume of the cylinder is equal when their respective base and height are equal.

Question 18.
A circle has a radius of 61 millimeters. Three-quarters of the circle is used to form a net for the curved part of a cone. The net is taped together to form the cone without any overlap. The height of the finished cone is 40 millimeters.

a) What is the circumference of the base of the cone?
Answer: The circumference of the base of the cone is  383.08 mm

Explanation:
The circumference of the cone is 2πr
Here d = 61 mm,
Then r = 30.5 mm
C = 2 × 3.14 × 30.5
= 191.54 mm
So, The circumference of the base of the cone is  191.54 mm

b) What is the radius of the cone?
Answer: r = 30.5 mm

Explanation:
The circumference of the cone is 2πr
Here d = 61 mm,
Then r = 30.5 mm

c) What is the volume of the cone?
Answer: The volume of the cone is 38,946 mm³

Explanation:
the volume of the cone V=1/3hπr².
Here , r = 30.5 mm , h = 40 mm.
V = \(\frac{1}{3}\) 40 × 30.5 × 30.5 × 3.14
V = 38,946 mm³.
So, The volume of the cone is 38,946 mm³

d) What is the area of the curved surface of the cone?
Answer: The area of the cone’s curved surface is 3,830 mm²

Explanation:
The curved surface area of a cone  S =  πrl
Here r =  30.5 mm , l = 40 mm
S = 3.14 × 30.5 × 40
S = 3,830 mm²
So, The area of the cone’s curved surface is 3,830 mm²

Question 19.
Martha used a filter in the shape of a cone to filter sand from a liquid. The volume of liquid that the filter can hold is 66 cubic centimeters. The height of the filter is 6 centimeters. What is the diameter of the filter? Round your answer to the nearest tenth.
Answer: The diameter of the filter is  7 cm.

Explanation:
Given, V = 66 cm³
the volume of the cone V=1/3hπr².
r = ? , h = 6 cm.
66 = \(\frac{1}{3}\) 6 × r² × 3.14
r² = \(\frac{66 × 3}{6 × 3.14}\)
r² =10.5 cm
r = 3.24 cm
The radius of the filter  is 3.24 cm

Then, the diameter d = 2r = 2 × 3.24 = 6.48 cm = 7 cm
So, The diameter of the filter is  7 cm.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Lesson 8.2 Finding Volume and Surface Area of Cylinders detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 8 Lesson 8.2 Answer Key Finding Volume and Surface Area of Cylinders

Math in Focus Grade 7 Chapter 8 Lesson 8.2 Guided Practice Answer Key

Use the given dimensions to find the volume of each cylinder. Use 3.14 as an approximation for π. Round your answer to the nearest tenth.

Question 1.
Radius = 5 cm, Height = 7.5 cm
Answer: The volume of the cylinder is 588.75 cm³

Explanation:
We know that , the volume of the cylinder V = πr²h
Here r = 5 cm , h = 7.5 cm,
V = πr²h
= 3.14 × (5)² × 7.5
= 5.8875 cm³
So, The volume of the cylinder is 588.75 cm³

Question 2.
Diameter = 7 in., Height = 5 in.
Answer: The volume of the cylinder is 192.3 in³

Explanation:
We know that , the volume of the cylinder V = πr²h
Here d = 7 in , h = 5 in,
then, r = 3.5
V = πr²h
= 3.14 × (3.5)² × 5
= 192.3 cm³
So, The volume of the cylinder is 192.3 cm³

Complete. Use 3.14 as an approximation for π.

Question 3.
The volume of a cylindrical tank of water is 1,808.64 cubic meters. The radius is 12 meters. What is the height of the cylindrical tank?
Use the formula for volume to find the height, h, of the cylinder.

The height of the cylindrical tank is about meters.
Answer:  The height of the cylindrical tank is about 4 meters.

Explanation:
Given, the volume of the tank V = 1,808.64 cubic meters, radius r = 12 meters
We know that , the volume of the cylinder V = πr²h
here π = 3.14
V = πr²h
h = \(\frac{V}{πr²}\)
= \(\frac{1808.64}{3.14 × 12 × 12}\)
= \(\frac{1808.64}{452.16}\)
= 4
So, The height of the cylindrical tank is about 4 meters.

Hands-On Activity

Materials:

• a tin can or any other cylindrical object
• paper
• scissors

Work in pairs.

STEP 1: What is the shape of the bases of your cylinder?
Write an expression for the area of one base in terms of r.

STEP 2: Cut a piece of paper to cover the curved surface of the cylinder. One end of the paper should meet the other end without overlapping.

STEP 3: Lay the paper flat on the table to form a rectangle. Use the information below to write expressions for its length and width in terms of r and h.
Length of rectangle = Circumference of the base =
Width of rectangle = Height of cylinder =
Now write an expression for the area of the curved surface in terms of rand h.

STEP 4: Use the expressions written in STEP 1 and STEP 3 to write an expression for the surface area of a cylinder.
Surface area = Area of bases + Area of curved surface

Solve. Use 3.14 as an approximation for π. Round to the nearest tenth.

Question 4.
A cylinder has a radius of 4 inches and a height of 7 inches. Find the surface area of the cylinder.
Answer: The surface area of the cylinder is 276 in².

Explanation:
The formula for the surface area of the cylinder A = 2πrh + 2πr²
Here r = 4 in. , h = 7 in. ,
A = (2 × 3.14 × 4 × 7) + ( 2 × 3.14 × 4 × 4)
= 175.84 + 100.48
= 276.32 in²
≈ 276 in²
So, The surface area of the cylinder is 276 in².

Solve.

Question 5.
The area of the curved surface of a cylindrical can is l62π square centimeters and its height is 9 centimeters. What is the diameter of the can?
Area of curved surface = 2πrh Use the formula for the area of the curved surface.
= 2πr • Substitute the surface area and height.
= 2 • • πr Multiply 2 by
= Simplify.
\(\frac{?}{?}=\frac{?}{?}\) Divide both sides by .
= r Simplify.
The diameter of the can is centimeters.
Answer: The diameter of the can is 18 cm

Explanation:
Given, The area of the curved surface of a cylindrical can is l62π square centimeters and h = 9 cm,
We know that Area of curved surface = 2πrh
Now equate the given data to the area formula we get,
2πrh = l62π
2 × r × 9 = 162
r = \(\frac{162}{18}\)
r = 9 cm
Then, d = 2r = 2 × 9 = 18cm,
So, The diameter of the can is 18cm

Math in Focus Course 2B Practice 8.2 Answer Key

For this practice, you may use a calculator. Use 3.14 as an approximation for π. Round your answers to the nearest tenth when you can.

Find the volume of each cylinder.

Question 1.

Answer: The volume of the cylinder is 263 cm³

Explanation:
We know that , the volume of the cylinder V = πr²h
Here r = 3.8 cm , h = 5.8 cm,
V = πr²h
= 3.14 × (3.8)² × 5.8
= 3.14 × 83.75
= 262.98 cm³
≈ 263 cm³
So, The volume of the cylinder is 263 cm³

Question 2.

Answer: The volume of the cylinder is 12 m³

Explanation:
We know that , the volume of the cylinder V = πr²h
Here d = 1.2 m , h = 10.6 m,
then, r = 0.6 m
Because r = \(\frac{d}{2}\):
V = πr²h
= 3.14 × (0.6)² × 10.6
= 3.14 × 3.816
= 11.98 m³
≈ 12 m³
So, The volume of the cylinder is 12 m³

Solve.

Question 3.
A cylinder has a radius of 6 centimeters and a height of 28 centimeters. What is the volume?
Answer: The volume of the cylinder is 3165 cm³

Explanation:
We know that , the volume of the cylinder V = πr²h
Here r = 6 cm , h = 28 cm,
V = πr²h
= 3.14 × (6)² × 28
= 3.14 × 1008
= 3165.12 cm³
≈ 3165 cm³
So, The volume of the cylinder is 3165 cm³

Question 4.
A cylinder has a volume of 239 cubic centimeters and a height of 6 centimeters. What is the radius?
Answer:  The radius of the cylinder is 3.5cm

Explanation:
We know that , the volume of the cylinder V = πr²h
Here v = 239 cm³ , h = 6 cm,
V = πr²h
r² = \(\frac{V}{πh}\)
= \(\frac{239}{3.14 × 6}\)
r²= 12.68
r = 3.56
≈ 3.5 cm
So, The radius of the cylinder is 3.5cm

Question 5.
Jenny is making a cylindrical pencil holder in shop class. It will be 14 centimeters high and 8 centimeters across. The bottom and sides of the container will be made of metal.

a) What is the area of the base?
Answer: The area of the base is  100.48 cm²

Explanation:
The area of the base = 2πr²
Given, d = 8cm,
Then r = 4 cm
2πr² = 2 × 3.14 × 4 × 4
= 100.48 cm²
So, The area of the base is  100.48 cm²

b) What is the area of the curved surface of the pencil holder?
Answer: The Area of curved surface is 351.68 cm²

Explanation:
We know that Area of curved surface = 2πrh
Here, h = 14 cm , r = 4 cm
2πrh  =  2 × 3.14 × 4 × 14
= 351.68 cm²
So, The Area of curved surface is 351.68 cm²

c) What is the total surface area of the pencil holder?
Answer: The total surface area of the pencil holder is 452 cm².

Explanation:
The formula for the total surface area of the cylinder A = 2πrh + 2πr²
Here r = 4 cm, h = 14 cm ,
A = (2 × 3.14 × 4 × 14) + ( 2 × 3.14 × 4 × 4)
= 351.68 + 100.48
= 452.16 cm²
≈ 452 cm²
So, The total surface area of the pencil holder is 452 cm².

Question 6.
You want to make a tube with a height of 8 inches and a radius of 5 inches out of cardboard. The tube will be open at both ends. How much cardboard will you need to make the tube?
Answer: The cardboard needed  to make the tube will be 753 in.

Explanation:
Given, h = 8 in. , r = 5 in.
cardboard needed  to make the tube will be the curved surface area of the tube,
The  Area of curved surface = 2πrh
2πrh  =  2 × 3.14 × 8 × 15
= 753.6 in²
So, The Area of curved surface is 753 in².

Question 7.
The volume of a soup can is 125.6 cubic inches. The diameter of the can is 8 inches. What is the height of the soup can?

Answer: The height of the can is about 2.5inches .

Explanation:
Given, the volume of a soup can V = 125.6 cubic in., diameter of the can is 8 in.
Then, r = 4 in.,
We know that , the volume of the cylinder V = πr²h
here π = 3.14
V = πr²h
h = \(\frac{V}{πr²}\)
= \(\frac{125.6}{3.14 × 4 × 4}\)
= \(\frac{125.6}{50.24}\)
= 2.5 in.
So, The height of the can is about 2.5inches .

Question 8.
Mrs. Lavender bought a cylindrical lampshade with height 14 inches and diameter-10 inches. As shown in the diagram, the lampshade is open at the top and bottom. Find the surface area of the lampshade.
Answer:  the surface area of the lampshade is  439 in²

Explanation:
We know that Area of curved surface = 2πrh
Here, h = 14 in , d = 10 in.,
Then r = 5 in,
2πrh  =  2 × 3.14 × 5 × 14
= 439.6 in²
So, The Area of curved surface is 439 in².

Question 9.
The volume of a cylinder is 121 π cubic inches and its height is 4 inches.

a) What is the radius of the cylinder?
Answer: The radius of the cylinder is 5 in.

Explanation:
We know that , the volume of the cylinder V = πr²h
Here v = 121 π cubic inches , h = 4 inches.
V = πr²h
r² = \(\frac{V}{πh}\)
= \(\frac{121 × 3.14}{3.14 × 4}\)
r²= 30.25
r = 5.5
≈ 5 in
So, The radius of the cylinder is 5 in.

b) What is the surface area of the cylinder? Give your answer in terms of π.
Answer: the surface area of the cylinder is  40π in².

Explanation:
We know that Area of curved surface = 2πrh
Here, h = 4 in , r = 5 in
2πrh  =  2 × π × 4 × 5
= 40π in²
So, The Area of curved surface is 40π in².

Question 10.
The diagram shows a mallet made by attaching two solid cylinders together. What is the volume of the mallet?

Answer: The volume of the mallet is V = 8024 cm³.

Explanation:
We know that , the volume of the cylinder V = πr²h
The volume of first cylinder is
Here d = 22 cm  , h = 18cm,
then, r = 11 cm
V = πr²h
= 3.14 × (11)² × 18
= 6838.92 cm³
So, The volume of the first cylinder is 6838 cm³

The volume of second cylinder is
Here d = 6 cm  , h = 42 cm,
then, r = 3 cm
V = πr²h
= 3.14 × (3)² × 42
= 1186.92 cm³
So, The volume of the second cylinder is 1186 cm³

Then, The volume of the mallet is  volume of first cylinder + volume of second cylinder
= 6836 + 1186
V = 8024 cm³.
So, The volume of the mallet is V = 8024 cm³.

Question 11.
A company makes cylindrical cans for peaches. Each can has a radius of 4 centimeters and a height of 12 centimeters. The company plans to increase the volume of each can by 25%.
a) What will be the height of the new can if the radius remains the same?
Answer: The height of the new can is about 15 cm

Explanation:
We know that , the volume of the cylinder V = πr²h
Here r = 4 cm , h =12 cm,
V = πr²h
= 3.14 × (4)² × 12
= 602.88 cm³
So, The volume of the cylinder is 602 cm³

Given, increase the volume of each can by 25%.
The 25% of 602 = 150
Then V = 602 + 150 = 752 cm³

The new volume of the can is 752 cm³,
the height of the new can if the radius remains the same will be
V = πr²h
h = \(\frac{V}{πr²}\)
= \(\frac{752}{3.14 × 4 × 4}\)
= \(\frac{752}{50.24}\)
= 14.9 cm.
So, The height of the new can is about 15 cm .

b) What will be the radius of the new can if the height remains the same?
Answer: The radius of the new can is 4 cm.

Explanation:
The new volume of the can is 752 cm³,
the radius of the new can if the height remains the same will be
V = πr²h
r² = \(\frac{V}{πh}\)
= \(\frac{752}{3.14 × 12}\)
r²= 19.95
r = 4.4
≈ 4 cm
So, The radius of the new can is 4 cm

Question 12.
A cylindrical water tank has a radius of 5 feet and a height of 10 feet. The volume of water in the tank is 565.2 cubic feet.
a) What is the height of the water in the tank?
Answer: The height of the water in the tank is 7.2 feet.

Explanation:
Given, V = 565.2 cubic feet, r = 5 feet
We know that , the volume of the cylinder V = πr²h
here π = 3.14
V = πr²h
h = \(\frac{V}{πr²}\)
= \(\frac{565.2}{3.14 × 5 × 5}\)
= \(\frac{565.2}{78.5}\)
= 7.2 feet.
So, The height of the water in the tank is 7.2 feet.

b) What percent of the tank’s volume is filled with water?
Answer: The percent of the tank’s volume is filled with water is 72%

Explanation:
The height of the tank is 10 feet and
the height of the water in the tank is 7.2 feet
Then, 7.2 is 72% of 10
So, The percent of the tank’s volume is filled with water is 72%

Question 13.
Eric is painting 8 wooden cylinders. Each cylinder has a radius of 6.2 inches and a height of 12.4 inches. Eric can paint 50 square inches of wood using one pint of paint. How much paint will Eric need to paint all the wooden cylinders?
Answer:  The paint will Eric need to paint all the wooden cylinders is 4 liters

Explanation:
Given, Eric is painting 8 wooden cylinders.
Each cylinder has a radius of 6.2 inches and a height of 12.4 inches.
Eric can paint 50 square inches of wood using one pint of paint.
= 8 × 50
= 400 in²
The paint will Eric need to paint all the wooden cylinders is 4 liters

Question 14.
The area of the curved surface of a cylindrical jar is 1,584 square centimeters. The height of the jar is 28 centimeters.
a) What is the circumference of the jar?
Answer: The circumference of the jar is 175 cm

Explanation:
Given, The area of the curved surface of a cylindrical jar is1 584 square centimeters and h = 28 cm,
We know that Area of curved surface = 2πrh
Now equate the given data to the area formula we get,
2πrh = 1584
2 × r × 28 = 1584
r = \(\frac{1584}{2 × 28}\)
r =28.28 cm
Then, d = 2r = 2 × 28 = 56 cm,
So, The diameter of the jar is 56cm

Circumference of the jar is C = πd
C = 3.14 × 56 = 175.84 cm
So, The circumference of the jar is 175 cm

b) What is the radius of the jar?
Answer:  The radius of the jar is 28.28 cm

Explanation:
Given, The area of the curved surface of a cylindrical jar is1 584 square centimeters and h = 28 cm,
We know that Area of curved surface = 2πrh
Now equate the given data to the area formula we get,
2πrh = 1584
2 × r × 28 = 1584
r = \(\frac{1584}{2 × 28}\)
r =28.28 cm
So, The radius of the jar is 28.28 cm.

Question 15.
Math journal Joyce uses the formula 5 = 2πr(r + h) to find the surface area of a cylinder. Assuming she uses the correct values for r and h, will she get the correct volume? Explain your thinking.
Answer: Yes, she will get the correct volume.

Explanation:
We know that , Surface area = Area of bases + Area of curved surface
She used the formulas 2πrh + 2πr²
the volume of the cylinder V = πr²h
she uses the correct values for r and h
So, yes she will get the correct volume.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Lesson 8.1 Recognizing Cylinders, Cones, Spheres, and Pyramids detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 8 Lesson 8.1 Answer Key Recognizing Cylinders, Cones, Spheres, and Pyramids

Hands-On Activity

Materials:

• clay
• string

FIND THE SHAPE OF CROSS SECTIONS OF SOLIDS

Work in pairs.

STEP 1: Make three clay cubes. Use the string to slice a cube vertically so that the cross section is parallel to one face, as shown. Sketch the cross section and describe its shape.

STEP 2: Use a string to slice another cube diagonally, as shown. Sketch the cross section and describe its shape.

STEP 3: Use a string to slice the last cube through the midpoints of each of three edges that share a common vertex, as shown. Sketch the cross section and describe its shape.

Math Journal Are you able to slice a cube in other ways to form the cross-section in STEP 1 to STEP 3?
Answer:

Math in Focus Grade 7 Chapter 8 Lesson 8.1 Guided Practice Answer Key

For each solid, name the shape of the cross-section formed when the solid is sliced by the plane shown.

Question 1.

Answer: Rectangle

Explanation:

The shape of the cross-section formed when the solid is sliced by the plane is a  Rectangle.
Because, It  forms a closed figure having four sides of which two of the opposite sides are equal.
So, it forms a Rectangle.

Question 2.

Answer: Rectangle

Explanation:

The shape of the cross-section formed when the solid is sliced by the plane is a  Rectangle.
Because, It  forms a closed figure having four sides of which two of the opposite sides are equal.
So, it forms a Rectangle.

Question 3.

Answer: Square

Explanation:

The shape of the cross-section formed when the solid is sliced by the plane is a  Square.
Because, It  forms a closed figure having four sides of which all the sides are equal.
So, it forms a Square

Question 4.

Answer: Triangle

Explanation:

The shape of the cross-section formed when the solid is sliced by the plane is a  Triangle.
Because, It  forms a closed figure having three sides .
So, it forms a Triangle.

Math in Focus Course 2B Practice 8.1 Answer Key

Match each set to its net.

Answer: The combinations of the figures are
1- f
2- c
3- d
4- e
5- b
6- a

Name the solid that can be formed from each net.

Question 7.

Answer: The closed figure formed is a cylinder

Explanation:

It forms a cylinder

Question 8.

Answer: The closed figure formed is a cylinder

Explanation:

It is a cylinder.

Question 9.

Answer: The closed figure formed is a Square pyramid

Explanation:

It is a square pyramid

Question 10.

Answer: The closed figure formed is a Triangular prism

Explanation:

It is a triangular prism.

Solve. Show your work.

Question 11.
Tell what cross-section is formed when a plane slices a square pyramid as described.
a) Perpendicular to its base and passes through its vertex.
Answer: The figure Perpendicular to its base and passes through its vertex is a triangle

Explanation:
Thee square pyramid is


So, The figure Perpendicular to its base and passes through its vertex is a triangle

b) Parallel to its base.
Answer: The figure parallel to its base is a square

Explanation:

The square pyramid is


So, The figure parallel to its base is a square.

Question 12.
The diagram shows a cone and its net.

a) Copy the net of the cone and label these dimensions on the net.
Answer:

b) How is the circumference of the base of the cone related to the curve XV?
Answer: The  closed figure  of the net forms a cone and its is a circle

Explanation:

Given,, the radius of a circle is 3 cm
From that the circumference of a circle is 2πr
C = 2 (3.14)(3) =  18.84 cm
So,  The xy becomes the circumference of a circle .

Question 13.
The diagram shows the net of a cylinder. Which sides of the rectangle have the same length as the circumference of the circular base?

Answer: The sides of the rectangle have the same length as the circumference of the circular base are  AB and DC

Explanation:
The closed figure of this  net becomes a cylinder

So, The sides of the rectangle have the same length as the circumference of the circular base are  AB and DC.

Question 14.
A base of each of the following prism is shaded. Name the shape of the cross section formed when each prism is sliced by a plane parallel to each base. Copy each prism. Sketch the cross sections and label them with the dimensions.
a)

Answer:

b)

Answer:

Question 15.
A cross-section that is parallel to one of the bases of a rectangular prism is 3 inches wide and 6 inches long. A cross-section that is perpendicular to its bases and parallel to two other faces is 4 inches wide and 6 inches long. What are the dimensions of the rectangular prism?
Answer: The dimensions of a rectangular prism are

Question 16.
The area of the base of a square pyramid is 64 square centimeters. Several planes slice through the pyramid parallel to the base to form square cross-sections.
a) Besides the cross-section formed by a plane slicing the base, how many cross-sections parallel to the base can be formed with areas that are perfect squares?
Answer: There can be 7 perfect squares after the multiple cross section.

Explanation:
The area of the base of a square pyramid is 64 square centimeters
That is a = 8 , Area a² = 8 × 8 = 64
So, cross-sections parallel to the base can be formed with areas that are perfect squares are
If a = 7 ,  a² = 7 × 7 = 49,
If a = 6 ,  a² = 6 × 6 = 36,
If a = 5 ,  a² = 5 × 5 = 25,
If a = 4 ,  a² = 4 × 4 = 16,
If a = 3 ,  a² = 3 × 3 = 9,
If a = 2 ,  a² = 2 × 2 = 4,
If a =  ,  a² = 1 × 1 = 1,
There can be 7 perfect squares after the multiple cross section.

b) Find the sum of the area of the base and the areas of the cross-sections found in a).
Answer: The sum of the area of the base is 28 cm²
The sum of the area of the cross-sections is  140 cm²