Math in Focus

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 11 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 11 Review Test Answer Key

Chapter Review/Test

Concepts and Skills

Find the circumference and area of each circle. Use \(\frac{22}{7}\) as an approximation for π.

Question 1.

Answer:
Circumfrence: 307.72 cm,
Area: 7,539.14 square centimeters,

Explanation:
Given diameter of the circle as 98 cm so radius is 98 cm/2 = 49 cm, Circumfrence is 2πr = 2 X 3.14 X 49 cm = 307.72 cm, Area of the circle is πr²  = 3.14 X 49 cm X 49 cm = 7,539.14 square centimeters.

Question 2.

Answer:
Circumfrence: 351.68 cm,
Area: 9,847.04 square centimeters,

Explanation:
Given diameter of the circle as 98 cm so radius is 112 cm/2 = 56 cm, Circumfrence is 2πr = 2 X 3.14 X 56 cm = 351.68 cm, Area of the circle is πr²  = 3.14 X 56 cm X 56 cm = 9,847.04 square centimeters.

Find the distance around each semicircle. Use \(\frac{22}{7}\) as an approximation for π.

Question 3.

Answer:
The distance around the semicircle is 21.98 ft,

Explanation:
Given diameter of the circle as 14 ft so radius is 14 ft/2 = 7ft, Distance around the semicircle is \(\frac{1}{2}\)  X 2πr = 3.14 X 7 ft = 21.98 ft.

Question 4.

Answer:
The distance around the semicircle is 98.91 in,

Explanation:
Given diameter of the circle as 63 in so radius is 63 in/2 = 31.5 in, Distance around the semicircle is \(\frac{1}{2}\)  X 2πr = 3.14 X 31.5 in = 98.91 in.

Find the distance around each quadrant. Round your answer to the nearest tenth. Use 3.14 as an approximation for π.

Question 5.

Answer:
The distance around the quadrant is 7.85 m,

Explanation:
Given radius of the circle is 5 m, Distance around the quadrant is \(\frac{1}{4}\) X 2πr = \(\frac{1}{2}\) X πr  = \(\frac{1}{2}\) X 3.14 X 5 m = 7.85 m.

Question 6.

Answer:
The distance around the quadrant is 23.55 ft,

Explanation:
Given radius of the circle is 15 ft, Distance around the quadrant is \(\frac{1}{4}\) X 2πr = \(\frac{1}{2}\) X πr  = \(\frac{1}{2}\) X 3.14 X 15 ft = 23.55 ft.

Solve. Show your work.

Question 7.
The diameter of a flying disc is 10 inches. Find the circumference and area of the disc. Use 3.14 as an approximation for π.
Answer:
Circumfrence of the flying disc is 31 inches and area of the flying disc is 78.5 square inches,

Explanation:
Given diameter of the flying disc as 10 inches so radius is 10 in/2 = 5 in, Circumfrence is 2πr = 2 X 3.14 X 5 in = 31.4 in, Area of the circle is πr² = 3.14 X 5 in X 5 in = 78.5 square inches.

Question 8.
The area of a compact disc is 452\(\frac{4}{7}\) square centimeters. What is the diameter of the compact disc? Use \(\frac{22}{7}\) as an approximation for π.
Answer:
The diameter of the compact disc is 26.22 centimeters,

Explanation:
Given the area of a compact disc is 452\(\frac{4}{7}\) square centimeters. Let d be the diameter of the compact disc so 452\(\frac{4}{7}\) square centimeters = πr² = π X (d/2)² ,
3.14 X d² = 4 X 452\(\frac{4}{7}\) square centimeters,
d² = 4 X 3,168 sq cms/7 X 3.14,
d² = 12,672 sq cms /21.98 = 576.524 therefore d = square root of 576.524 sq cms = 26.22 centimeters.

Question 9.
The circumference of a circular table is 816.4 centimeters. Find the radius of the table. Use 3.14 as an approximation for π.
Answer:
The radius of the table is 130 cntimeters,

Explanation:
Given the circumference of a circular table is 816.4 centimeters means it is 2πr, So radius = 816.4 cms/2 X 3.14= 816.4 cms/6.28 = 130 centimeters

Problem Solving

Solve. Show your work.

Question 10.
A water fountain shoots up a jet of water. The water falls back down onto the ground in the shape of a circle. Michelle wants the circle of water on the ground to be 0.7 meter wider on each side. She gradually increases the strength of the water jet. The area of the circle of water increases at 0.2 square meter per second. Use \(\frac{22}{7}\) as an approximation for n.
a) Find the area of the original circle of water.
Answer:
The area of the original circle of water is 13.8474 square meters,

Explanation:
The radius of the original circle of water  as diameter 4.2 m is 4.2 m/2 = 2.1 m, So area is  3.14 X 2.1 m X 2.1 m = 13.8474 square meters.

b) Find the area of the larger circle of water.
Answer:
The area of the larger circle of water is 24.6176 square meters,

Explanation:
The radius of the original circle of water  as diameter 4.2 m + 0.7 m + 0.7 m = 5.6 m is 5.6 m/2 = 2.8 m, So area is 2 X 3.14 X 2.8 m X 2.8 m =  24.6176 square meters.

c) How long does it take for the original circle of water to become the larger circle of water? Round your answer to the nearest second.

Answer:
It take for the original circle of water to become the larger circle of water to the nearest second is 54 seconds,

Explanation:
The area increased from the orignal circle to larger circle is 24.6176 square meters – 13.8474 square meters = 10.7702 square meters, if the area of the circle of water increases at 0.2 square meter per second long does it take for the original circle of water to become the larger circle of water is 10.7702/ 0.2 = 53.851 seconds nearest second is 54 seconds.

Question 11.
A machine in an assembly line stamps pieces of metal. The stamping plate on the machine travels in a path shaped like the arc of a quadrant as the stamping plate opens and closes. It takes the machine 5 seconds to open and close the stamping plate one time. Use \(\frac{22}{7}\) as an approximation for π.
a) Find the total distance the outside edge of the stamping plate travels when the machine opens and closes one time.
Answer:
The total distance the outside edge of the stamping plate travels when the machine opens and closes one time is 60.442 cms,

Explanation:
Given a machine in an assembly line stamps pieces of metal. The stamping plate on the machine travels in a path shaped like the arc of a quadrant as the stamping plate opens and closes, So the total distance the outside edge of the stamping plate travels when the machine opens and closes one time is distance around the quadrant is \(\frac{1}{4}\) X 2πr we have radius as 38.5 cm so the distance is \(\frac{1}{4}\) X 2 X 3.14 X 38.5 cm = 60.445 cms.

b) Find the speed of the stamping plate’s outside edge in centimeters per second.
Answer:
The speed of the stamping plate’s outside edge in centimeters per second is 12.089 centimeters per second,

Explanation:
Given it takes the machine 5 seconds to open and close the stamping plate one time, So the speed of the stamping plate’s outside edge in centimeters per second is 60.445 cms/5 seconds = 12.089 centimeters per second.

c) Assume the machine starts and ends in an open position. How many seconds will it take the machine to stamp 500 pieces of metal?

Answer:

Answer:
Seconds will it take the machine to stamp 500 pieces of metal is 6,044.5 seconds,

Explanation:
Assuming the machine starts and ends in an open position. So seconds will it take the machine to stamp 500 pieces of metal is 12.089 centimeters per second X 500 = 6,044.5 seconds.

Question 12.
The figure shows four identical quadrants enclosed in a square. The side length of the square is 20 inches. Find the area of the blue part. Use 3.14 as an approximation for π.

Answer:
The area of the blue part is 321.5 square inches,

Explanation:
The figure shows four identical quadrants enclosed in a square. The side length of the square is 20 inches, The area of the squar is 20 in X 20 in = 400 sq in. As diameter is 20 in radius is 20 in/2 = 10 in so area of four quadrants is \(\frac{1}{4}\) X πr² = \(\frac{1}{4}\) X 3.14 X 10 in X 10 in = 78.5 square inches. Now area of the blue part is 400 square inches – 78.5 square inches = 321.5 square inches.

Question 13.
The figure shows 3 identical circles. X, Y, and Z are the centers of the circles, and the radius of each circle is 15 feet. \(\frac{1}{6}\) of each circle is shaded. What is the total area of the shaded portion? Round your answer to the nearest tenth of a foot. Use 3.14 as an approximation for π.

Answer:
The total area of the shaded portion is 350 square feet,

Explanation:
Given the figure shows 3 identical circles. X, Y, and Z are the centers of the circles, and the radius of each circle is 15 feet. \(\frac{1}{6}\) of each circle is shaded. So the total area of the shaded portion first we calculate area of the circle with radius 15 feet is πr² = 3.14 X 15 feet X 15 feet = 706.5 square feet so the area of  shaded circle is \(\frac{1}{6}\) X 706.5 square feet = 117.75 square feet so we have 3 identical circles shaded so its total area is 3 X 117.75 square feet is 353.25 nearest to 350 square feet.

Question 14.
The figure is made up of one semicircle and two quadrants. The distance around the figure is 97.29 inches. Find the value of k. Use 3.14 as an approximation for π.

Answer:
The value of k is 15.492 inches,

Explanation:
Given the figure is made up of one semicircle and two quadrants. The distance around the figure is 97.29 inches and the diameter is k in. first the distance of semi circle is \(\frac{1}{2}\) X 2πr = 3.14 X k in. = 3.14k in. Now the area of 2 quadrants is 2 X \(\frac{1}{4}\) X 2πr =3.14k in. Given the distance of fiqure as 97.29 inches = 3.14k inches + 3.14 k inches = 6.28k inches, Therefore k = 97.29/6.28 = 15.492 inches.

This handy Math in Focus Grade 6 Workbook Answer Key Cumulative Review Chapters 8-11 detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Cumulative Review Chapters 8-11 Answer Key

Concepts and Skills

Represent the solution set of each Inequality on a number line. (Lesson 8.3)

Question 1.
p ≤ 35
Answer:

 

 

Explanation:
Shown p ≤ 35 inequality with red line on the number line above the point starts from 35 and moves towards left side values less than or equal to 35.

Question 2.
q ≥ 12.6
Answer:

 

 

 

Explanation:
Shown q ≥ 12.6 inequality with green line on the number line above, the points starts from 12.6 on the number lines and moves towards right side greater than or equal to 12.6.

Question 3.
r < \(\frac{4}{5}\)
Answer:

Explanation:
Shown r < \(\frac{4}{5}\) = 0.8 inequality with green line on the number line above the point starts from 0.8 and moves towards left side values less than r < \(\frac{4}{5}\) = 0.8.

Question 4.
s > 13\(\frac{1}{2}\)
Answer:

 

 

 

Explanation:
Shown s > 13\(\frac{1}{2}\) = 13.5 inequality with green line on the number line above, the points starts from 13.5 on the number lines and moves towards right side greater than 13.5.

Use graph paper. Plot the points on a coordinate plane and answer the question. (Lesson 9.1)

Question 5.
a) Plot points A (-2, -2) and B (-10, -2) on a coordinate plane. Connect the two points to form a line segment.
Answer:
Explanation:
Plotted points A (-2, -2) and B (-10, -2) on a coordinate plane and connected the two points to form a line segment as shown above.

b) Point C lies above \(\overline{A B}\), and is 5 units away from the x-axis. If triangle ABC is an isosceles triangle with base \(\overline{A B}\), find the coordinates of point C.
Answer:
If Point C lies above \(\overline{A B}\), and is 5 units away from the x-axis. If triangle ABC is an isosceles triangle with base \(\overline{A B}\) and points A (-2, -2) and B (-10, -2)  as AC = BC, Let Point be (x,y) so Square root of (x-(-2))²  +(y-(-2))²  = square root of (x -(-10))²  + (y-(-2))² ,
(x + 2)²  + (y + 2)²  = (x + 10)²  + (y + 2)² ,
4x + 4  = 20x + 100,
16x = 4 – 100 = -96,
x = -96/4 = -24 and y = 0

c) Points D and E lie below \(\overline{A B}\) such that ABDE is a rectangle. If BD is 7 units, find the coordinates of points D and E.

Answer:

Find the area of each figure. (Lesson 10.2)

Question 6.

Answer:
The area is 42 square meters,

Explanation:
Given figure is trapezoid and area of trapezoid is \(\frac{1}{2}\)(a +b) X h where a and b are bases of parallel sides so a = 9 m, b = 5 m and h = 6 m, therefore area = \(\frac{1}{2}\)(9 m + 5 m) X 6 m = 7 m X 6 m = 42 square meters.

Question 2.

Answer:
The area is 120 square feet,

Explanation:
Given figure is trapezoid and area of trapezoid is \(\frac{1}{2}\)(a +b) X h where a and b are bases of parallel sides so a = 16 ft, b = 8 ft and h = 10 ft, therefore area = \(\frac{1}{2}\)(16 ft + 8 ft) X 10 ft = 12 ft X 10 ft = 120 square feet.

Solve. Show your work. (Lessons 8.2, 11.1, 11.2)

Question 8.
The cost of a shirt is p dollars. The cost of a pair of pants is twice the cost of the shirt. If the cost of the pair of pants is t dollars, express t in terms of p.
Answer:
t = 2p dollars,

Explanation:
Given the cost of a shirt is p dollars. The cost of a pair of pants is twice the cost of the shirt. If the cost of the pair of pants is t dollars, So t in terms of p is t = 2p dollars.

Question 9.
A can has a circular base of diameter 8 centimeters. Find the area of this base. Use 3.14 as an approximation for π.
Answer:
The area of the base is 50.24 square centimeters,

Explanation:
Given A can has a circular base of diameter 8 centimeters. As the area is πr² where radius = 8 cms/2 = 4 cms, So area = 3.14 X 4 cms X 4 cms = 50.24 square centimeters.

Question 10.
The cross section of a bowl is in the shape of a semicircle. The area of the semicircle is 77 square centimeters. Find its radius. Use \(\frac{22}{7}\) as an approximation for π.
Answer:
Radius is approximately 7 cms,

Explanation:
Given the cross section of a bowl is in the shape of a semicircle. The area of the semicircle is 77 square centimeters.
As area of semicircle is \(\frac{1}{2}\) X πr² so 77 sq cms = \(\frac{1}{2}\) X 3.14 X r² ,
r² = 77 sq cms X 2/3.14 = 154 sq cms/3.14 = 49.04 sq cms, therefore r = square root of 49.04 approximately 7 cms.

Question 11.
The circumference of a platinum ring is 44 millimeters. Find its radius. Use \(\frac{22}{7}\) as an approximation for π.
Answer:
The radius of the platinum ring is 7.006 millimeters,

Explanation:
Given the circumference of a platinum ring is 44 millimeters as circumference is 2πr so r = 44 millimeter/2π, r = 44 mm/6.28 = 7.006 millimeters.

Problem Solving

Solve. Show your work.

Question 12.
Emily weighs x pounds. Jonathan weighs 3 times as much as Emily. If Jonathan weighs 81 pounds, write an equation in terms of x and solve it. (chapter 8)
Answer:
Equation is 3x = 81, Emily weighs 27 pounds,

Explanation:
Given Emily weighs x pounds. Jonathan weighs 3 times as much as Emily. If Jonathan weighs 81 pounds, So equation is 3x = 81 pounds so x = 81/3 = 27 pounds.

Question 13.
Kim has a bag of yo-yos. Some of the yo-yos are red. The rest are yellow. The ratio of the number of red yo-yos to the number of yellow yo-yos is 5 : 9. If Kim has a total of b yo-yos, how many more yellow yo-yos than red yo-yos are there? (Chapter 8)
Answer:
Many more yellow yo-yos than red yo-yos are r = (b -9y)/5,

Explanation:
Given Kim has a bag of yo-yos. Some of the yo-yos are red. The rest are yellow. The ratio of the number of red yo-yos to the number of yellow yo-yos is 5 : 9. If Kim has a total of b yo-yos, let r be red yo-yos and y be yellow yo-yos,
b = 5r + 9y, So many more yellow yo-yos than red yo-yos are r = (b -9y)/5.

Question 14.
The length of a rectangle is 4n centimeters, and it is twice as long as the width. The perimeter of the rectangle is twice as long as that of an equilateral triangle. Find the side length of the triangle in terms of n. (Chapter 10)
Answer:
Side length of the triangle in terms of n is 2n,

Explanation:
Given the length of a rectangle is 4n centimeters and it is twice as long as the width. The perimeter of the rectangle is twice as long as that of an equilateral triangle, length of rectangle = 4n and width is 4n/2 = 2n, So Perimeter p = 2(4n + 2n) = 12n, let a be the side length of the triangle so 12n : 2 = 6n, P = 3a, a = 6n : 3, a = 2n, So Side length of the triangle in terms of n is 2n.

Question 15.
The length of the minute hand of a clock is 6.5 centimeters. How far does the tip of the minute hand travel in an hour? Use 3.14 as an approximation for π. (Chapter 11)
Answer:
The minute hand travel in an hour is 40.82 centimeters,

Explanation:
Given the length of the minute hand of a clock is 6.5 centimeters. The tip of the minute hand travel in an hour is the circumference with radius as 6.5 centimeters, so it is 2π r = 2 X 3.14 X 6.5 centimeters = 40.82 centimeters.

Question 16.
The area of trapezoid ABCE is 36 square meters. Find the height of the trapezoid. (Chapter 10)

Answer:
The height of the trapezoid is 6 meters,

Explanation:
Area of trapezoid is \(\frac{1}{2}\)(a +b) X h where a and b are bases of parallel sides given a = 7 m, b = 5 m and let height be h m, therefore 36 square meters = \(\frac{1}{2}\)(7 m + 5 m) X h m, height = 2 X 36 sq mts/12 mts = 6 meters.

Find the area of the shaded region. (Chapter 10)

Question 17.

Answer:
The area of the shaded region is 56 square feet,

Explanation:
Given area consists of 2 rectangles one bigger one with length 12 ft and width 8 ft so area is 12 ft X 8 ft = 96 square feet and another small rectangle inside the bigger rectangle with a length 8 ft and width 5 ft so area is 8 ft X 5 ft = 40 square feet now the area of the shaded region is 96 square feet – 40 square feet = 56 square feet.

Question 18.

Answer:
The area of the shaded region is 172.5 square inches,

Explanation:
The shape is square inside with 2 non shaded regions are trapezoid and triangle, First area of square with side length 15 in is 15 inches X 15 inches = 225 square inches, Now area of non shaded region first trapezoid we have a as 2.5 inches, b as 7.5 inches, height 7.5 inches, so area is (2.5 inches + 7.5 inches)/2 X 7.5 inches = 37.5 square inches, Area of the non shaded triangle with base 4 inches and height 7.5 inches is 1/2 X 4 inches X 7.5 inches = 15 square inches,
Total area of non shaded region is 37.5 square inches + 15 square inches = 52.5 square inches, Now the area of the shaded region is 225 square inches – 52.5 square inches = 172.5 square inches.

Find the area of the shaded region. Use 3.14 as an approximation for π.

Question 19.

Answer:
The area of the shaded region is 113.64 square centimeters,

Explanation:
The area of the shaded region is area of triangle – area of the circle so first area of the triangle is \(\frac{1}{2}\) X base X height here base is 25 cm and height is 21.4 cm so area is \(\frac{1}{2}\) X 25 cm X 21.4 cm = 267.5 square centimeters, Area of circle is πr² we have diameter as 14 cm so radius = 14 cm/2 = 7 cm so area is 3.14 X 7 cm X 7 cm = 153.86 sq cms now area of the shaded region is 267.5 sq cms – 153.86 sq cms = 113.64 sq cms.

Question 20.

Answer:
The area of the shaded region is 75.11 square meters,

Explanation:
Given a circle and square in it and in square we have 2 small shaded triangles, First the area of the square is 10 m X 10 m = 100 square meters, The area of the circle with radius 7 m is πr² = 3.14 X 7 m X 7 m = 153.86 square meters,
Area of shaded region without square is 153.86 square meters – 100 square meters = 53.86 square meters, We have 2 triangles shaded regions one traingle area with base side 5 m and height 5 m so its area is 1/2 X 5 m X 5 m = 12.5 square meters another area of square is 1/2 X 5 m X 3.5 m = 8.75 square meters so area of 2 small shaded triangles is 12.5 sq mts + 8.75 sq mts = 21.25 square meters, Now area of the only shaded region is 53.86 square meters + 21.25 square meters = 75.11 square meters.
Solve.

Question 21.
The diagram shows the plan of a square garden. The side length of each grid square is 2 meters. (Chapters 9, 10)

a) A triangular region ABP is surrounded with a wooden fence. The shortest possible distance from point P to \(\overline{A B}\) is 8 meters, and triangle ABP is an isosceles triangle with base \(\overline{A B}\). Find the coordinates of point P.
Answer:
We have A(-8,8) and B(-8,-8) and triangle ABP is an isosceles triangle with base AB means AP = BP,
Square root of  (x – (-8))² + (y – 8)² = Square root of (x  – (-8))² + (y -(-8))² ,
-16y + 64 = 16y + 64,
32y = 0,

b) Find the area and perimeter of the garden.
Answer:
Area is 64 square meters, Perimeter of the garden is 32 meters,

Explanation:
Given the plan of a square garden with 8 mts length so area is 8 mts X 8 mts = 64 square meters and perimeter of the garden is 4 X 8 meters = 32 meters.

c) Find the area of the garden that lies outside triangle ASP in square meters.
Answer:

Solve. Use graph paper to answer the question.

Question 22.
An aspen tree is 300 centimeters tall. It grows 15 centimeters taller each month. The height of the tree, h centimeters, over t months is given by h = 300 + 15t. Copy and complete the table, Graph the relationship between t and h. Use 1 unit on the horizontal axis to represent 1 month and 1 unit on the vertical axis to represent 15 centimeters. Start your vertical axis at 300 centimeters. (Chapters 8, 9)
a)

Answer:

Explanation:
Updated the given table as shown above.

b) What is the height of the tree after 3 months?
Answer:
The height of the tree after 3 months is 345 centimeters,

Explanation:
If an aspen tree is 300 centimeters tall. It grows 15 centimeters taller each month. So after 3 months it will grow
300 centimeters + 15 centimeters + 15 centimeters + 15 centimeters = 345 centimeters.

c) Assuming the growth of the tree is constant for the next year, what is the height of the tree after 10 months?
Answer:
The height of the tree after 10 months is 450 centimeters,
Explanation:
Assuming the growth of the tree is constant for the next year, So the height of the tree after 10 months will be 300 centimeters + 10 X 15 centimeters = 300 centimeters + 150 centimeters = 450 centimeters.

d) If the tree is at least 360 centimeters tall, how many months have passed? Express your answer in the form of an inequality in terms of t, where t stands for the number of months that have passed.
Answer:
360 = 300 + 15t,

Explanation:
Given if the tree is at least 360 centimeters tall, how many months have passed so expressing my answer in the form of an inequality in terms of t, where t stands for the number of months that have passed is 360 = 300 + 15t.

e) Name the dependent and independent variables.
Answer:
Dependent is time and height and independent is 15 centimeters taller each month,

Explanation:
Given an aspen tree is 300 centimeters tall. It grows 15 centimeters taller each month. The height of the tree, h centimeters, over t months is given by h = 300 + 15t. Here dependent is time and height and independent is 15 centimeters taller each month.

Solve. Show your work.

Question 23.
A circular garden is surrounded by a cement path that is 1.5 meters wide. Find the area of the path. Use 3.14 as an approximation for n. (Chapter 11)

Answer:
The area of the path 40.035 square meters,

Explanation:
Given the radius of the circle with circular path is 5 m so its area is πr² = 3.14 X 5 m X 5 m = 78.5 square meter, now area of the circle without the path as radius is 5 m – 1.5 m = 3.5 m is 3.14 X 3.5 m X 3.5 m = 38.465 square meter, Now the area of the path only is 78.5 square meter – 38.465 square meter = 40.035 square meter.

Question 24.
Figure PRSV is a parallelogram. The length of \(\overline{V U}\) and \(\overline{U T}\) are equal, The area of parallelogram QRSW is 30 square inches. The length of \(\overline{R S}\) is 6 inches. (Chapter 10)

a) Find the height of parallelogram QRSW.
Answer:
The height of parallelogram QRSW is 5 inches,

Explanation:
Given the area of parallelogram QRSW is 30 square inches. The length of \(\overline{R S}\) is 6 inches.
As area of parallelogram is base X height we have base as 6 in. so height of the parallelogram QRSW is 30 square inches/6 inches = 5 inches.

b) Find the area of triangle PRV.
Answer:
The area of triangle PRV is 30 square inches,

Explanation:
As given the length of \(\overline{V U}\) and \(\overline{U T}\) are equal, So length of \(\overline{V U}\) = 5 inches so \(\overline{U T}\) = 5 inches the total length of VT is length of \(\overline{V U}\) + length of \(\overline{U T}\)  = 5 inches + 5 inches = 10 inches and PR = VT = 10 inches, PV = RS = 6 inches so the area of triangle PRV is 1/2 X RS X PR = 1/2 X 6 inches X 10 inches = 30 square inches.

c) If you did not know the length of \(\overline{R S}\), explain how you could find the area of triangle PRV.
Answer:
Through Parallelogram,

Explanation:
As we know the area of a triangle is half the product of any of its sides and the corresponding altitude. If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half of the area of the parallelogram.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 11 Lesson 11.3 Real-World Problems: Circles detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 11 Lesson 11.3 Answer Key Real-World Problems: Circles

Math in Focus Grade 6 Chapter 11 Lesson 11.3 Guided Practice Answer Key

Complete. Use 3.14 as an approximation for π.

Question 1.
The circumference of the moon is the approximate distance around a circle with radius 1,736 kilometers. Find the circumference of the moon.
a) Round your answer to the nearest 10 kilometers.
Circumference of moon = 2πr
≈ • •
= km
The circumference of the moon to the nearest 10 kilometers is kilometers.
Answer:
The circumference of the moon to the nearest 10 kilometers is 10,900 kilometers,

Explanation:
Given the radius of circle as 1,736 kilometers, So circumference of the moon is 2πr = 2 X 3.14 X 1,736 kilometers = 10,902.08 kilometer nearest 10 is 10,900 kilometers.

b) Round your answer to the nearest 1,000 kilometers.
The circumference of the moon to the nearest 1,000 kilometers is kilometers.
Answer:
The circumference of the moon to the nearest 1,000 kilometers is 11,000 kilometers,

Explanation:
We got the circumference of the moon as 10,902.08 to the nearest 1,000 kilometers rounding we get 11,000 kilometers.

Question 2.
A greeting card is made up of three semicircles. O is the center of the large semicircle. Sarah wants to decorate the distance around the card with a ribbon. How much ribbon does Sarah need? Round your answer to the nearest inch.

Length of semicircular arc AB = \(\frac{1}{2}\) • 2πr
≈ • • •
= 1 • •
= in.
Semicircular arcs AO and OB have the same length.
Total length of semicircular arcs AO and OB
= 2 • \(\frac{1}{2}\) • πd
≈ • • •
= 1 • •
= in.
Distance around the card
= length of semicircular arc AB + total length of semicircular arcs AO and OB
= +
= in.
≈ in.
Sarah needs approximately inches of ribbon
Answer:
Sarah needs approximately 49 inches of ribbon,

Explanation:
A greeting card is made up of three semicircles. O is the center of the large semicircle. Sarah wants to decorate the distance around the card with a ribbon. Ribbon does Sarah need length of semicircular arc AB = \(\frac{1}{2}\) • 2πr = πr, radius is 5.2 in. so length of semicircular arc AB = 3.14 X 5.2 in. = 16.328 inches, Total length of semicircular arcs AO and OB = 2 X \(\frac{1}{2}\) X πd as diameter is 5.2 in. + 5.2 in. = 10.4 inches, So total length of semicircular arcs AO and OB = 3.14 X 10.4 inches = 32.656 inches. Now distance around the card is length of semicircular arc AB + total length of semicircular arcs AO and OB = 16.328 inches + 32.656 inches = 48.984 inches approximately 49 inches.

Question 3.
As part of her artwork, Sally bends a length of wire into the shape shown. The shape is made up of a semicircle and a quadrant. Find the length of the wire.

Distance around the shape
= length of semicircular arc PQ + length of arc RO + RP + OQ
= + + +
= cm
The length of the wire is approximately centimeters.
Answer:
The length of the wire is approximately 81 centimeters,

Explanation:
As part of her artwork, Sally bends a length of wire into the shape shown. The shape is made up of a semicircle and a quadrant. To find the length of the wire first we calculate length of semicircular arc \(\frac{1}{2}\) X 2πr = πr as r = 12 cm so it is 3.14 X 12 cm = 37.68 cm, Length of arc RO = \(\frac{1}{4}\) X 2πr = \(\frac{1}{2}\) X πr = \(\frac{1}{2}\) X 3.14 X 12 cm = 18.84 cm, Now Distance around the shape
= length of semicircular arc PQ + length of arc RO + RP + OQ = 37.68 cm + 18.84 cm + 12 cm + 12 cm = 80.52 cm approximately 81 centimeters.

Complete. Use \(\frac{22}{7}\) as an approximation for π.

Question 4.
Judy baked a pizza and had part of it for lunch. After the meal, the shape of the remaining pizza is made up of a semicircle and a quadrant. Find the area of the remaining pizza.

Area of remaining pizza = area of quadrant + area of semicircle
= +
= cm²
The area of the remaining pizza is approximately square centimeters.
Answer:
The area of the remaining pizza is approximately 2,885 square centimeters,

Explanation:
Given Judy baked a pizza and had part of it for lunch. After the meal, the shape of the remaining pizza is made up of a semicircle and a quadrant. So to find the area of the remaining pizza first we calculate area of quadrant \(\frac{1}{4}\) X πr²  as radius is 35 cm it is \(\frac{1}{4}\) X 3.14 X 35 cm X 35 cm = 961.625 square centimeters, Second we calculate area of semicircle = \(\frac{1}{2}\) X πr² = \(\frac{1}{2}\) X 3.14 X 35 cm X 35 cm = 1,923.25 square centimeters, So area of remaining pizza = area of quadrant + area of semicircle =
961.625 square centimeters + 1,923.25 square centimeters = 2,884.875 square centimeters approximately 2,885 square centimeters.

Question 5.
A rug is made up of a quadrant and two semicircles. Find the area of the rug.

Total area of two semicircles = 2 • \(\frac{1}{2}\) • πr²
≈ • • •
= 1 • • •
= in.²
Area of figure = area of quadrant + total area of two semicircles
= +
= in.²
The area of the rug is approximately square inches.
Answer:
The area of the rug is approximately 2,770 square inches,

Explanation:
A rug is made up of a quadrant and two semicircles. To calculate the area of the rug first we calculate area of quadrant \(\frac{1}{4}\) X πr²  as radius is 42 in. \(\frac{1}{4}\) X 3.14 X 42 in. X 42 in. = 1,384.74 square inches, radius of semicircle is half the diameter 42 in/2 = 21 in. now total area of two semicircles = 2 X \(\frac{1}{2}\) X πr² = 3.14 X 21 in. X 21 in. = 1,384.74 square inches therefore area of figure = area of quadrant + total area of two semicircles = 1,384.74 square inches + 1,384.74 square inches = 2,769.48 square inches
approximately 2,770 square inches.

Question 6.
The diameter of a bicycle wheel is 60 centimeters. How far does the wheel travel when it makes 35 revolutions? Give your answer in meters.
Circumference of wheel = πd
≈ •
= cm
Distance traveled = circumference of wheel • number of revolutions
= •
= cm
= m Divide by 100 to convert to meters.
The wheel travels approximately meters.
Answer:
The wheel travels approximately 66 meters,

Explanation:
Given the diameter of a bicycle wheel is 60 centimeters. Circumference of wheel = πd = 3.14 X 60 centimeters = 188.4 centimeters, Distance traveled = circumference of wheel X number of revolutions = 188.4 centimeters X 35 revolutions = 6,594 now we convert to meters by dividing 100 so we get 6,594/100 = 65.94 meters approximately
66 meters.

Question 7.
A park is shaped like the diagram below. It is a rectangle with semicircles at the two ends. There is a running track around the park.

a) The total length of the track is 220 yards. Find the length of \(\overline{P S}\).
The track is made up of semicircular arcs PQ and SR, and sides PS and QR. Semicircular arcs PQ and SR are equal.
Total length of semicircular arcs PQ and SR
= 2 • \(\frac{1}{2}\) • πd
≈ 1 • •
= yd
The length of \(\overline{P S}\) and \(\overline{Q R}\) are equal.
Total length of track = 220
PS + QR + total length of semicircular arcs PQ and SR = 220
PS + QR + = 220
PS + QR = 220
2 • PS =
PS =
= ys
The length of \(\overline{P S}\) is approximately yards.
Answer:
The length of \(\overline{P S}\) is approximately 45 yards.

Explanation:
The total length of the track is 220 yards. The length of \(\overline{P S}\), The track is made up of semicircular arcs PQ and SR, and sides PS and QR. Semicircular arcs PQ and SR are equal. Total length of semicircular arcs PQ and SR ith diameter 35 yd is 2 X \(\frac{1}{2}\) X πd = 3.14 X 35 yd = 109.9 yards,
The length of \(\overline{P S}\) and \(\overline{Q R}\) are equal.
Total length of track = 220 yards,
PS + QR + total length of semicircular arcs PQ and SR = 220 yards,
PS + QR + 109.9 yards = 220 yards,
PS + QR = 220 yards – 109.9 yards = 90.1 yards, (PS = QR),
2 PS = 90. 1 yards,
PS = 90.1 yards/2 = 45.05 yards approximately 45 yards.

b) A jogger runs once around the track in 125 seconds. What is his average speed in yards per second?
The jogger runs yards in 125 seconds.
125 seconds →
1 second →
= yd
The average speed of the jogger is yards per second.
Answer:
The average speed of the jogger is 1.76 yards per second,

Explanation:
Given a jogger runs once around the track in 125 seconds, The jogger runs 220 yards in 125 seconds.
125 seconds = 220 yards, 1 second = 220/125 yards = 1.76 yards, So the average speed of the jogger is 1.76 yards per second.

c) A gardener is hired to water the grass in the park. Using a machine, he waters 4 square yards per second. How many minutes will he take to water the entire park? Round your answer to the nearest minute.
Radius = diameter ÷ 2
= ÷ 2
= yd
The areas of the two semicircles are equal.
Total area of two semicircles = 2 • \(\frac{1}{2}\) • πr²
= 2 • \(\frac{1}{2}\) • •
= 1 • • •
= yd²
Area of rectangle PQRS = lw
= •
= yd²
Area of park = area of rectangle PQRS + total area of two semicircles
= +
= yd²
Time taken = area of park ÷ rate of watering park
= ÷
= s
= min
≈ min Round to the nearest minute.

The gardener will take approximately minutes to water the entire park.
Answer:
The gardener will take approximately 11 minutes to water the entire park,

Explanation:
Given A gardener is hired to water the grass in the park. Using a machine, he waters 4 square yards per second. So many minutes will he take to water the entire park, Radius = diameter ÷ 2 = 35 yd ÷ 2 = 17.5 yd, The areas of the two semicircles are equal. Total area of two semicircles = 2 X \(\frac{1}{2}\) X πr² = 3.14 X 17.5 yd X 17.5 yd = 961.625 yd²,
Area of rectangle PQRS = length X width,
= 45  yd X 35 yd,
= 1,575 yd²,
Area of park = area of rectangle PQRS + total area of two semicircles
= 1,575 yd² + 961.625 yd² = 2,536.625 yd²,
Time taken = area of park ÷ rate of watering park
= 2,536.625 yd² X seconds ÷ 4 yd² ,
=  634.15625 s
= 634.15625 / 60 minutes = 10.5692 min
≈ 11 min round to the nearest minute.

Math in Focus Course 1B Practice 11.3 Answer Key

Solve. Show your work.

Question 1.
The radius of a circular pond is 8 meters. Find its area and circumference. Use 3.14 as an approximation for π.
Answer:
Area = 200.96 square meters,
Circumference = 50.24 meters,

Explanation:
Given the radius of a circular pond is 8 meters so area is πr² = 3.14 X 8 mts X 8 mts = 200.96 square meters,
Circumference = 2πr = 2 X 3.14 X 8 mts = 50.24 meters.

Question 2.
The diameter of a metal disc is 26 centimeters. Find its area and circumference. Use 3.14 as an approximation for π.
Answer:
Area = 530.66 square centimeters,
Circumference = 81.64 centimeters,

Explanation:
Given the diameter of a metal disc is 26 centimeters so radius = diameter ÷ 2 = 26 cms ÷ 2 = 13 cms, so area is πr² = 3.14 X 13 cms X 13 cms = 530.66 square centimeters, Circumference = 2πr = 2 X 3.14 X 13 cms = 81.64 centimeters.

Question 3.
The shape of a carpet is a semicircle. Use \(\frac{22}{7}\) as an approximation for π.
a) Find its area.
Answer:
Area of semicircle carpet is 76.93 square feet,

Explanation:
Area of the carpet which is semi-circle with diameter 14 ft first radius = 14 ft/2 = 7 ft and area of carpet is \(\frac{1}{2}\) X πr² = \(\frac{1}{2}\) X 3.14 X 7 ft X 7 ft = 76.93 square feet.

b) Janice wants to put a fringed border on all sides of the carpet. How many feet of fringe are needed?

Answer:
21.98 feet of fringe is needed,

Explanation:
Given Janice wants to put a fringed border on all sides of the carpet. So feet of fringe are needed \(\frac{1}{2}\) X 2πr = πr and we have diameter as 14 ft radius = diameter/2 = 14 ft/2 = 7 ft now 3.14 X 7 ft = 21.98 feet.

Question 4.
The circumference of the rim of a wheel is 301.44 centimeters. Find the diameter of the rim. Use 3.14 as an approximation for π.
Answer:
The diameter of the rim is 96 centimeters,

Explanation:
Given the circumference of the rim of a wheel is 301.44 centimeters. So it is 301.44 cms = 2πr ,
r = 301.44 cms/2 X 3.14 = 310.44 cms/6.28 = 48 cms, Now diameter = 2r = 2 X 48 cms = 96 centimeters.

Question 5.
A Japanese fan is made out of wood and cloth. The shape of the fan is made up of two overlapping quadrants. What is the area of the portion that is made of cloth? Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The area of the portion that is made of cloth is 1,846.32 square centimeters,

Explanation:
Given a Japanese fan is made out of wood and cloth. The shape of the fan is made up of two overlapping quadrants.
First we calculate area of cloth and wood with radius 28 cm as we know area πr² = 3.14 X 28 cm X 28 cm = 2,461.76 square cms, Area of wood with radius 14 cms is 3.14 X 14 cms X 14 cms = 615.44 square centimeters, Therefore area of the cloth is area of cloth and wood – area of the wood = 2,461.76 square centimeters – 615.44 square centimeters = 1,846.32 square centimeters.

Question 6.
A pancake restaurant serves small silver-dollar pancakes and regular-size pancakes. Use 3.14 as an approximation for π.

a) What is the area of a small silver dollar-pancake? Round your answer to the nearest tenth of an inch.
Answer:
The area of a small silver dollar-pancake to the nearest tenth of an inch is 10 square inches,

Explanation:
The area of small silver dollar-pan cake with diameter 3.5 in. as radius = 3.5 in./2 = 1.75 in. now area = πr² = 3.14 X 1.75 in. X 1.75 in. = 9.61625 square inches to the nearest tenth is 10 square inches.

b) What is the area of a regular-size pancake? Round your answer to the nearest tenth of a square inch.
Answer:
The area of a regular-size pancake to the  nearest tenth of a square inch is 28 square inches,

Explanation:
The area of a regular-sizepan cake with diameter 6 in. as radius = 6 in./2 = 3 in. now area = πr² = 3.14 X 3 in. X 3 in. = 28.26 square inches to the nearest tenth is 28 square inches.

c) If the total price of 6 small silver-dollar pancakes is the same as the total price of 3 regular-size pancakes, which is a better deal?
Answer:
3 regular-size pancakes is a better deal,

Explanation:
6 X area of small silver dollar-pan cake = 6 X 9.61625 square inches = 57.6975 sqaure inches and 3 X area of a regular-sizepan cake = 3 X 28.26 square inches = 84.78 sqaure inches if the total price of 6 small silver-dollar pancakes is the same as the total price of 3 regular-size pancakes then the better deal is 3 regular-size pancakes as it has bigger area.

Question 7.
A park is shaped like a rectangle with a semicircle on one end, and another semicircle cut out of one side.
a) Find the distance around the park.
Answer:
The distance around the park is 480 m,

Explanation:
The semicircle on one end is equal to another semicircle which is cut out of one side, so the distance around the park is the rectangle so it is 2 X (170 m + 70 m) = 2 X 240 m = 480 m.

b) Find the area of the park. Use \(\frac{22}{7}\) as an approximation for π.

Answer: 11,900 square meters

Explanation:
As the semicircle on one end is equal to another semicircle which is cut out of one side, so it the complete rectangle so the area of rectangle is length X breadth = 170 m X 70 m = 11,900 square meters.

Question 8.
The diameter of a circular fountain in a city park is 28 feet. A sidewalk that is 3.5 feet will be built around the fountain. Use \(\frac{22}{7}\) as an approximation for π.
a) Find the area of the sidewalk.
Answer:
The area of the sidewalk is 346.185 square feet,

Explanation:
Given the diameter of a circular fountain in a city park is 28 feet. A sidewalk that is 3.5 feet will be built around the fountain so first area of the circular fountain with diameter 28 ft, radius = 28 ft/2 = 14 ft so area of the cirular fountain is πr² = 3.14 X 14 ft X 14 ft = 615.44 square feet now the area of circular fountain with the side walk is
radius = 28 ft + 7 ft /2= 35 ft/2 = 17.5 ft so area is πr² = 3.14 X 17.5 ft X 17.5 ft =
If the sidewalk is 3.5 feet means it is radius so the area = πr² = 3.14 X 3.5 feet X 3.5 feet = 961.625 square feet, So only the side walk area = area of circular fountain with side walk – area of circular fountain = 961.625 square feet – 615.44 square feet = 346.185 square feet.

b) 0.8 bag of concrete will be needed for every square foot of the new sidewalk. What is the minimum number of bags needed?

Answer:
The minimum number of bags needed are 277 bags,

Explanation:
The area of the sidewalk is 346.185 square feet and 0.8 bag of concrete will be needed for every square foot of the new sidewalk. So the minimum number of bags needed are 346.185 X 0.8 = 276.948 bags approximately 277 bags.

Question 9.
The diagram shows an athletic field with a track around it, The track is 4 feet wide. The field is a rectangle with semicircles at the two ends. Find the area of the track. Use 3.14 as an approximation for π.

Answer:
The area of the track is 1,110.4 square feet,

Explanation:
The field is a rectangle with semicircles at the two ends. First we calculate the area with out the track so we have
rectangle area as length = 84 ft – 8 ft = 76 ft and width is 36 ft so area is 76 ft X 36 ft = 2,736 square feet,
Now area of two semicircles with diameter 36 ft so radius is 36 ft/2 = 18 ft, so area is 2 X \(\frac{1}{2}\) X πr² = 3.14 X 18 ft X 18 ft = 1,017.36 square feet, Field area is 2,736 sq ft + 1,017.36 sq ft = 3,753.36 square feet,
Now field area with track area of rectangle is length 76 ft and width is 36 ft + 8 ft = 44 ft is 76 ft X 44 ft = 3,344 square feet, area of two semicircles with diameter 44 ft so radius is 44 ft/2 = 22 ft, so area is 2 X \(\frac{1}{2}\) X πr² = 3.14 X 22 ft X 22 ft = 1,519.76 square feet, Field area with track is 3,344 square feet + 1,519.76 square feet = 4,863.76 square feet. Now the area of the track is field area with track – field area = 4,863.76 square feet – 3,753.36 square feet = 1,110.4 square feet.

Question 10.
The petal of a paper flower is created by cutting along the outlines of two overlapping quadrants within a square. Use 3.14 as an approximation for π.
a) Find the distance around the shaded part.
Answer:
The distance around the shaded part is 15.7 cm,

Explanation:
Distance around the shaded part is of the 2 quadrants = 2 X circumference ÷ 4 and C = πd or 2πr as given diameter = 10 cm, So distance around the shaded part  =  2 X πd ÷ 4 = 2 X 3.14 X 10 cm ÷ 4 = 62.8 cm ÷ 4 = 15.7 cm.

b) Find the area of the shaded part.

Answer:
39.25 square centimeters,

Explanation:
The area of the 2 quardants is 2 X \(\frac{1}{4}\) X πr² as diameter is 10 cm radius is 10 cm/2 = 5 cm,
Now area is 2 X \(\frac{1}{4}\) X 3.14 X 5 cm X 5 cm = 39.25 square centimeters.

Question 11.
Wheels A and B are placed side by side on a straight road. The diameter of wheel A is 56 inches. The diameter of wheel B is 35 inches. Suppose each wheel makes 15 revolutions. Find the distance between the wheels after they have made these 15 revolutions.
Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The distance between the wheels after they have made these 15 revolutions is 494.55 inches,

Explanation:
Given wheels A and B are placed side by side on a straight road. The diameter of wheel A is 56 inches. The diameter of wheel B is 35 inches. Radius of wheel A is diameter of wheel A/2 = 56 inches/2 = 28 inches, Radius of wheel B is diameter of wheel B/2 = 35 inches/2 = 17.5 inches, Circumference of wheel A = πr = 3.14 X 28 inches = 87.92 inches,
Circumference of wheel B = πr = 3.14 X 17.5 inches = 54.95 inches, If Wheel A makes 15 revolutions the distance is
87.92 inches X 15 = 1,318.8 inches and If wheel B makes 15 revolutions the distance is 54.95 inches X 15 = 824.25 inches so the distance between the wheels after they have made these 15 revolutions is distance of 15 revolutions of wheel A – distance of 15 revolutions of wheel B = 1,318.8 inches – 824.25 inches = 494.55 inches.

Question 12.
Nine identical circles are cut from a square sheet of paper whose sides are 36 centimeters long. If the circles are as large as possible, what is the area of the paper that is left after all the circles are cut out? Use 3.14 as an approximation for π.
Answer:
The area of the paper that is left after all the circles are cut out is 278.64 square centimeters,

Explanation:
Given nine identical circles are cut from a square sheet of paper whose sides are 36 centimeters long. If the circles are as large as possible, So the area of the paper that is left after all the circles are cut out is area of the square – area of 9 circles, The area of the square is 36 cms X 36 cms = 1,296 square cms, diameter = 38 cms/3 approximately 12 cms, radius is 12 cms / 2 cms = 6 cms, Area of 9 circles is 9 X πr² = 9 X 3.14 X 6 cms X 6 cms = 1,017.36 square centimeters, therefore the area of the paper that is left after all the circles are cut out = 1,296 sq cms – 1,017.36 sq cms = 278.64 square centimeters.

Question 13.
A designer drew an icon as shown below. O is the center of the circle, and AB is a diameter. Two semicircles are drawn in the circle. If \(\overline{A B}\) is 28 millimeters, find the area of the shaded part. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The area of the shaded part is 307.72 square millimeters,

Explanation:
Given a designer drew an icon as shown below. O is the center of the circle, and AB is a diameter. Two semicircles are drawn in the circle. If \(\overline{A B}\) is 28 millimeters, if we see the design the shaded region is half of the circle so the area of the shaded part as diameter is 28 mm its radius is 28 mm/2 = 14 mm, Now area of shaded region = \(\frac{1}{2}\) X πr² = \(\frac{1}{2}\) X 3.14 X 14 mm X 14 mm = 307.72 square millimeters.

Use graph paper. Solve.

Question 14.
Mary wants to draw the plan of a circular park on graph paper. The coordinates of the center of the park are A (3, 4). The circle has a radius of 3 units.
a) Use a compass and draw the plan of the circular park on graph paper.
Answer:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Explanation:
Used a compass and drawn the plan of the circular park on graph paper as shown above.

b) Assume that the y-axis points north and south. A barbecue pit is located at the northernmost part of the park. Plot and label the location of the barbecue pit as point B. Give the coordinates of point B.
Answer:

c) Connect points A, B, and the origin to form a triangle. Find the area of the triangle.
Answer:

Question 15.
A wire is bent to make the shape below. The shape is made up of four identical circles. Each circle intersects two other circles. The four circles meet at a common point T, which is the center of square PQRS. Use \(\frac{22}{7}\) as an approximation for π.
a) Find the length of the wire.
Answer:
The length of the wire is 112 cm,

Explanation:
Given to find the length of the wire it is perimeter of the square and length of square is 28 cm therfore length of the wire is 4 X 28 cm = 112 cms.

b) Find the area of the whole shape.

Answer:
The area of the whole space is 2,014.88 square centimeters,

Explanation:
Given a wire is bent to make the shape below. The shape is made up of four identical circles. Each circle intersects two other circles. The four circles meet at a common point T, which is the center of square PQRS. So the area of the whole shape is (4 X \(\frac{1}{2}\) the area of circle) + area of the square, radius = diameter/2 = 28 cm/2 = 14 cm, So [4 X (\(\frac{1}{2}\) X 3.14 X 14 cm X 14 cm)] + (28 cm X 28 cm) = (2 X 615.44 square centimeter) +  784 square centimeters = 1,230.88n square centimeters + 784 square centimeters = 2,014.88 square centimeters.

Brain @ Work

Question 1.
The figure shows two identical overlapping quadrants. Find the distance around the shaded part. Use 3.14 as an approximation for π. Round your answer to the nearest tenth of a centimeter.

Answer:
The distance around the shaded part is 47.1 cm,

Explanation:
Distance around the shaded part is of the 2 quadrants = 2 X circumference ÷ 4 and C = πd or 2πr as given radius = 15 cm, So distance around the shaded part  =  2 X 2 X 3.14 X 15  cms ÷ 4 = 3.14 X 15 cm = 47.1 cm.

Question 2.
A cushion cover design is created from a circle of radius 7 inches, and 4 quadrants. Find the total area of the shaded parts of the design. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The total area of the shaded parts of the design is 42.14 square inches,

Explanation:
Given a cushion cover design is created from a circle of radius 7 inches, and 4 quadrants. The total area of the shaded parts of the design is area of square – 4 X area of the quadrant = 14 in X 14 in – 4 X \(\frac{1}{4}\) X 3.14 X 7 in X 7 in = 196 square inches – 153.86 square inches = 42.14 square inches.

Question 3.
Two identical wheels are placed along a straight path so that their centers are 9.31 meters apart. The radius of each wheel is 3.5 centimeters. They are pushed towards each other at the same time, each making one revolution per second. How long does it take for them to knock into each other? Use \(\frac{22}{7}\) as an approximation for π.

Answer:
Long does it will take for them to knock into each other is 4.0926 meters,

Explanation:
Given two identical wheels are placed along a straight path so that their centers are 9.31 meters apart. The radius of each wheel is 3.5 centimeters. Circumfrence of the one wheel is 2 X 3.14 X 3.5 cms = 21.98 cms converting into meters 21.98 X 0.01 m = 0.2198 meters so 2 wheels means 0.2198 m X 2 =  0.4396 m their centers are apart of 9.31 meters so the distance for the 2 wheels to knock into each other is 9.31 meters X 0.4396 = 4.0926 meters.

Question 4.
A stage prop is made up of a semicircle and a quadrant. Its area is 924 square inches, Find the value of x. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The value of x is 9.903 inches,

Explanation:
Given a stage prop is made up of a semicircle and a quadrant. Its area is 924 square inches with radius 2x in. So 924 sqaure inches = area of the quadrant + area of the semicircle,
924 sq in = \(\frac{1}{4}\) X 3.14 X 2x in X 2x in + \(\frac{1}{2}\) X 3.14 X 2x in X 2x in,
924 sq in = 3.14x² sq in + 6.28x² sq in,
924 sq in =9.42x² sq in,
x² = 924 sq in/9.42 sq in = 98.089 sq in, therefore x = square root of 98.089 sq in = 9.903 inches.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 11 Lesson 11.2 Area of a Circle detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 11 Lesson 11.2 Answer Key Area of a Circle

Math in Focus Grade 6 Chapter 11 Lesson 11.2 Guided Practice Answer Key

Complete. Use 3.14 as an approximation for π.

Question 1.
Find the area of a circle that has a radius of 18 centimeters.
Area = πr²
≈ •
= •
= cm²
The area of the circle is approximately square centimeters
Answer:
The area of the circle is approximately 1,017 square centimeters,

Explanation:
Given radius as 18 centimeters as the area of the circle is πr², So area = 3.14 X 18 cms X 18 cms = 1,017.36 cm² approximately 1,017 square centimeters.

Question 2.
Find the area of a circle that has a radius of 15 inches.
Area = πr²
≈ •
= •
= in.²
The area of the circle is approximately square inches.
Answer:
The area of the circle is approximately 707 square inches,

Explanation:
Given radius as 15 inches as the area of the circle is πr², So area = 3.14 X 15 in X 15 in =  706.5 in² approximately 707 square inches.

Question 3.
Find the area of a circle that has a diameter of 26 centimeters.
Radius = diameter ÷ 2
= ÷
= cm
Area = πr²
≈ • •
= •
= cm²
The area of the circle is approximately square centimeters.
Answer:
The area of the circle is approximately 531 square centimeters,

Explanation:
Given diameter as 26 cms so radius = diameter ÷ 2 = 26 cms ÷ 2 = 13 cms as the area of the circle is πr², So area = 3.14 X 13 cms X 13 cms = 530.66 cm² approximately 531 square centimeters.

Complete. Use \(\frac{22}{7}\) as an approximation for π.

Question 4.
Find the area of a quadrant.

The area of a quadrant is approximately square feet.
Answer:
The area of the quadrant is approximately 154 square feet,

Explanation:
Given radius is 14 ft as area of quardant is \(\frac{1}{4}\) X area of circle = \(\frac{1}{4}\) X πr² = \(\frac{1}{4}\) X 3.14 X 14 feet X 14 feet = 153.86 square feet approximately 154 square feet.

Question 5.
The diameter of a circle is 42 inches. Find the area of a quadrant.

Radius = diameter ÷ 2
= ÷ 2
= in.
Area of quadrant = • area of circle
= • πr²
≈ • •
= • • •
= in.²
The area of a quadrant is approximately square inches.
Answer:
The area of the quadrant is approximately 346 square inches,

Explanation:
Given diameter as 42 inches so radius = 42 inches/2 = 21 inches as area of quardant is \(\frac{1}{4}\) X area of circle = \(\frac{1}{4}\) X πr² = \(\frac{1}{4}\) X 3.14 X 21 inches X 21 inches = 346.185 square inches approximately 346 square inches.

Math in Focus Course 1B Practice 11.2 Answer Key

Find the area of each circle. Use 3.14 as an approximation for π.

Question 1.

Answer:
The area of the circle is 314 square centimeters,
Explanation:

Explanation:
Given radius as 10 cm as the area of the circle is πr², So area = 3.14 X 10 cm X 10 cm = 314 cm² .

Question 2.

Answer:
The area of the circle is approximately 1,963 square inches,

Explanation:
Given diameter as 50 inches so radius = diameter ÷ 2 = 50 in ÷ 2 = 25 in as the area of the circle is πr², So area = 3.14 X 25 in X 25 in =1962.5 inches² approximately 1,963 square inches.

Find the area of each semicircle. Use \(\frac{22}{7}\) as an approximation for π.

Question 3.

Answer:
The area of semicircle is 307.72 square feet,

Explanation:
Given diameter as 28 ft  so radius = diameter ÷ 2 = 28 ft ÷ 2 = 14 ft as the area of the circle is πr²,  So area = 3.14 X 14 ft X 14 ft = 615.44 square feet, The area of semicircle is half the area of the circle so it is \(\frac{1}{2}\) X
615.44 square feet = 307.72 square feet.

Question 4.

Answer:
The area of semicircle is 76.93 square meter,

Explanation:
Given radius as 7 m the area of the circle is πr²,  So area = 3.14 X 7 m X 7 m = 153.86 square meter, The area of semicircle is half the area of the circle so it is \(\frac{1}{2}\) X
153.86 square meter = 76.93 square meter.

Find the area of each quadrant to the nearest tenth. Use 3.14 as an approximation for π.

Question 5.

Answer:
The area of quardant is 113.04 square inches,

Explanation:
Given radius as 12 inches the area of the circle is πr²,  So area = 3.14 X 12 in. X 12 in. = 452.16 square inches, The area of quardant is 1/4 of the area of the circle so it is \(\frac{1}{4}\) X 452.16 square inches = 113.04 square inches.

Question 6.

Answer:
The area of quardant is 283.385 square meters,

Explanation:
Given radius as 19 meters the area of the circle is πr²,  So area = 3.14 X 19 m X 19 m = 1,133.54 square meters, The area of quardant is 1/4 of the area of the circle so it is \(\frac{1}{4}\) X 1,133.54 square meters = 283.385 square meters.

Solve. Show your work.

Question 7.
A circular pendant has a diameter of 7 centimeters. Find its area. Use \(\frac{22}{7}\) as an approximation for π.
Answer:
Area = 38.465 square centimeters,

Explanation:
Given a circular pendant has a diameter of 7 centimeters. So radius = diameter ÷ 2 = 7 cms ÷ 2 = 3.5 cms, As area of circle = πr², So area = 3.14 X 3.5 cms X 3.5 cms = 38.465 square centimeters.

Question 8.
The shape of the stage of a lecture theater is a semicircle. Find the area of the stage. Use 3.14 as an approximation for π.

Answer:
The area of the stage is 226.08 square meter,

Explanation:
Given shape of the stage of a lecture theater is a semicircle with radius 12 m, First area of circle is πr², So area = 3.14 X 12 m X 12 m = 452.16 square meter, Now area of semicircle is \(\frac{1}{2}\) X 452.16 sq mt = 226.08 square meter.

Question 9.
The shape of a balcony floor is a quadrant. Find the area of the balcony floor. Use 3.14 as an approximation for π.

Answer:
The area of the balcony floor is 113.04 square feet,

Explanation:
Given shape of a balcony floor is a quadrant with radius 12 ft, First area of circle is πr², So area = 3.14 X 12 ft X 12 ft = 452.16 square feet, Now area of quadrant is \(\frac{1}{4}\) X 452.16 square feet = 113.04 square feet.

Question 10.
The cost of an 8-inch pizza is $4. The cost of a 16-inch pizza is $13. Use 3.14 as an approximation for π.
a) How much greater is the area of the 16-inch pizza than the area of the 8-inch pizza?
Answer:
Greater is the area of the 16-inch pizza than the area of the 8-inch pizza is 150.72 square inches,

Explanation:
Area of 8 inch pizza radius is 8 inch ÷ 2 = 4 inch, Area of 8 inch pizza = 3.14 X 4 inch X 4 inch = 50.24 square inch,
Area of 16 inch pizza is 16 inch ÷ 2 = 8 inch, Area of 16 inch pizza = 3.14 X 8 inch X 8 inch = 200.96 square inch,
Now the area of the 16-inch pizza than the area of the 8-inch pizza is 200.96 square inch – 50.24 square inch = 150.72 square inches.

b) Which is a better deal? Explain your reasoning.
Answer:
8 inch pizza,

Explanation:
Cost of 16 inch pizza is $13, So cost of 1 inch pizza is $13/16 = $0.8125,
Cost of 8 inch pizza is $4, So cost of 1 inch pizza is $4/8 = $0.5 therefore it is better to buy 8 inch pizza as in 1 inch it cost $0.5 than 16 inch pizza in which 1 inch is $0.8125.

Question 11.
Four identical drinking glasses each have a radius of 5 centimeters. The glasses are arranged so that they touch each other as shown in the figure below. Find the area of the green portion. Use 3.14 as an approximation for π.

Answer:
The area of the green portion is 21.5 square centimeters,

Explanation:
Given to find the area of green portion so it is the area of square minus area of 4 quadrants so ((the area of square – 4(\(\frac{1}{4}\) X πr²)) = 10 cms x 10 cms – 4(\(\frac{1}{4}\) X 3.14 X 5 cms X 5 cms) = 100 square cms – 3.14 X 25 square cms = 100 square cms – 78.5 square cms =  21.5 square centimeters.

Question 12.
The figure is made up of trapezoid ABCD and a semicircle. The height of trapezoid ABCD is \(\frac{5}{6}\) the length of \(\overline{B C}\). Find the area of the figure. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
Total area of trapezoid ABCD and a semicircle is 18,235.7925 square meters,

Explanation:
The area of trapezoid is 1/2 X (base1 + base2) X height, where base1 and base2 are lengths of parallel sides, we have height as \(\frac{5}{6}\) X 124.8 m = 104 m, Now area of trapezoid is 1/2 X (124.8 m + 119 m) X 104 m = 1/2 X 243.8 m X 104 m =12,677.6 square meters. First we calculate area of circle as diameter is 119 m so radius = 119 m/2 = 59.5 m, Area = 3.14 X 59.5 m x 59.5 m = 11,116.385 square meter, Now area of semicircle is half the area of circle = 11,116.385 square mt/2 = 5,558.1925 square meter, So total area of trapezoid ABCD and a semicircle is
area trapezoid + area of semicircle = 12,677.6 square meters + 5,558.1925 square meters = 18,235.7925 square meters.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 11 Lesson 11.1 Radius, Diameter, and Circumference of a Circle to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 11 Lesson 11.1 Answer Key Radius, Diameter, and Circumference of a Circle

Math in Focus Grade 6 Chapter 11 Lesson 11.1 Guided Practice Answer Key

Question 1.
In the figure, 0 ¡s the center of the circle with \(\overline{A B}\), \(\overline{C D}\), and \(\overline{E D}\) as shown.

a) Name all the diameters that are drawn in the circle.
Answer:
AB, CD,

Explanation:
A diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circle, all the diameters that are drawn in the circle are AB, CD.

b) Which line segment that joins two points on the circle ¡s not a diameter? Explain why it ¡s not a diameter.
Answer:
DE,

Explanation:
A line segment that crosses the circle by passing through its center is called a diameter. As DE doesnot cross through the circle.

Question 2.
The radius of a circle is 6 centimeters. What is the length of its diameter?
Diameter = 2 • radius
=
= cm
The diameter of the circle is centimeters.
Answer:
12 centimeters,

Explanation:
Given the radius of a circle is 6 centimeters, So the length of its diameter as diameter = 2 X radius,
Diameter = 2 X 6 centimeters = 12 centimeters.

Question 3.
The diameter of a circle is 15 inches. What is the length of its radius?
Radius = diameter ÷ 2
= ²
= in.
The radius of the circle is inches.
Answer:
7.5 inches,

Explanation:
Given the diameter of a circle is 15 inches, So the length of it radius is as radius = diameter ÷ 2 = 15 inches ÷ 2 = 7.5 inches.

Hands-On Activity

Materials

• compass

DRAWING CIRCLES USING A COMPASS

Step 1: Measure 5 centimeters on a ruler with a compass.
Example

Step 2: Then draw a circle. Draw and label the center O and four radii: \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\) and \(\overline{O S}\).
Example

Step 3: Measure the radii \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\), and \(\overline{O S}\). What can you say about the lengths of \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\), and \(\overline{O S}\)?
Answer:
\(\overline{O P}\) = 5 centimeters, \(\overline{O Q}\) = 5 centimeters, \(\overline{O R}\), = 5 centimeters and \(\overline{O S}\) = 5 centimeters,

Explanation:
The lengths of \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\), and \(\overline{O S}\) are 5 centimeters as we measured 5 centimeters on a ruler with a compass and then drawa a circle and labelled the center O and four radii \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\) and \(\overline{O S}\) so the lengtha are 5 cm each.

Hands-On Activity

INVESTIGATING THE RELATIONSHIP BETWEEN THE CIRCUMFERENCE AND DIAMETER OF A CIRCLE

Step 1: Lisa uses a string to measure the circumference of each circle to the nearest tenth of a centimeter and records it in a table.

Copy the table. Divide the circumference of each circle by its diameter. Round your answers to the nearest tenth. Record your results.

What do you notice about the quotients in the last column?
The circumference of any circle divided by its diameter always gives the same value.

The Greek letter π is used to represent this value.

Step 2: To see the value of π up to 9 decimal places, press π Enter on your calculator. Round the value of π to
a) the nearest tenth.
b) the nearest hundredth.
c) the nearest thousandth.

Step 3: In Step 1, you learned that the circumference of any circle divided by its diameter is equal to n. You can use this fact to write a formula for the circumference of a circle.
Complete the following statement.
Since circumference ÷ diameter = π,
Circumference = π •
You also know that the diameter of a circle is 2 times its radius. You can use this fact to write a related formula for the circumference of a circle. Complete the following statement.
Circumference = π • diameter
= π • 2 •
= 2 • π •
Using C for circumference, d for diameter, and r for radius, you can write these formulas as
C = πd
C = 2πr
Math Note
πd means π • d and 2πr means 2 • π • r.
Answer:

Explanation:
The quotients in the last column is the circumference of any circle divided by its diameter always gives the same value 3.14.

Copy and complete the table. Use \(\frac{22}{7}\) as an approximation for π.

Question 4.

Answer:

Explanation:
Given A has diameterr 14 cm so as circumfrence = πd so diameter is circumfrence/ π, 14 = circumfrence/ π,
so circumfrence = 14 X π = 14 X 3.14 = 43.96 cms, Now radius = circumfrence/ 2π = 43.96/ 2 X 3.14 = 43.96/6.28 = 7 cm. Circle B has radius 21 cm so diameter =  2r = 2 X 21 cm = 42 cms, circumfrence = 2πr = 2 X 3.14 X 21 cm = 131.88 cm and Circle C has radius 10.5 cm so diameter =  2r = 2 X 10.5 cm =  21 cms, circumfrence = 2πr = 2 X 3.14 X 10.5 cm = 65.94 cm. Completed the table as shown above.

Copy and complete the table. Use 3.14 as an approximation for π.

Question 5.

Answer:

Explanation:
Given D has diameterr 25 cm so as circumfrence = πd so diameter is circumfrence/ π, 25 = circumfrence/ π,
so circumfrence = 25 X π = 25 X 3.14 = 78.5 cms, Now radius = circumfrence/ 2π = 78.5/ 2 X 3.14 =78.5/6.28 = 12.5 cm. Circle E has radius 16 cm so diameter =  2r = 2 X 16 cm = 32 cms, circumfrence = 2πr = 2 X 3.14 X 16 cm = 100.48 cm and Circle F has radius 8.25 cm so diameter =  2r = 2 X 8.25 cm = 16.5 cms, circumfrence = 2πr = 2 X 3.14 X 8.25 cm = 51.81 cm. Completed the table as shown above.

Complete.

Question 6.
A circular hoop is cut into two equal parts. Its diameter is 35 inches. Find the length of each semicircular arc. Use \(\frac{22}{7}\) as an approximation for π.
Circumference of hoop = πd
≈
= in.
Length of each semicircular arc = \(\frac{1}{2}\) • circumference of hoop
=
= in.
The length of each semicircular arc is approximately inches.
Answer:
The length of each semicircular arc is approximately 55 inches,

Explanation:
Given a circular hoop is cut into two equal parts. Its diameter is 35 inches. So the length of each semicircular arc. Using \(\frac{22}{7}\) as an approximation for π. Circumference of hoop = πd = 3.14 X 35 inches = 109.9 inches, Now Length of each semicircular arc = \(\frac{1}{2}\) • circumference of hoop = \(\frac{1}{2}\) X 109.9 inches = 54.95 approximately 55 inches.

Question 7.
A quadrant is cut from a square. The side of the square is 10 centimeters. Find the length of the arc of the quadrant. Use 3.14 as an approximation for π.

Circumference of circle = 2πr
≈ 2 • •
= cm
Length of the arc of the quadrant = circumference ÷ 4
= ÷
= cm
The length of the arc of the quadrant is approximately centimeters.
Answer:
The length of the arc of the quadrant is approximately 15.7 centimeters.,

Explanation:
Given quadrant is cut from a square. The side of the square is 10 centimeters. So the length of the arc of the quadrant by using 3.14 as an approximation for π, Circumference of circle = 2πr,  2 X 3.14 X 10 cm =62.8 cms,
Length of the arc of the quadrant = circumference ÷ 4 = 62.8 ÷ 4 = 15.7 cm, therefore the length of the arc of the quadrant is approximately 15.7 centimeters.

Hands-On Activity

DRAWING SEMICIRCLES AND QUADRANTS

Materials

• compass
• drawing triangles
• ruler
• protractor

Step 1: Use a compass to draw a circle of radius 2 inches. Label the center O.

Step 2: Draw and label a diameter of the circle \(\overline{P Q}\). What do you notice?
A diameter of a circle divides it into semicircles.
This figure is one of the semicircles.

Find the distance around the semicircle. Use 3.14 as an approximation for π.

Step 3: In your circle, draw a second diameter perpendicular to \(\overline{P Q}\) using a ruler and protractor, or a drawing triangle. Label it \(\overline{R S}\). What do you notice?
a) Two perpendicular diameters of a circle divide it into quadrants.
This figure is one of the quadrants.

Find the distance around the quadrant. Use 3.14 as an approximation for π.
Answer:
A diameter of a circle divides it into 2 semicircles,
The distance around the quadrant is 3.14 inches,

b) This figure is made up of the semicircular arc, the arc of a quadrant, and the radii \(\overline{O R}\) and \(\overline{O Q}\). Find the distance around the figure. Use 3.14 as an approximation for π.

Answer:
The distance around the fiqure is 9.42 inches,

Explanation:
Step 1: Using a compass to draw a circle of radius 2 inches. Label the center O.
Step 2: Drawn and labelled a diameter of the circle \(\overline{P Q}\).
A diameter of a circle divides it into 2 semicircles.
Step 3: In my circle, drawn a second diameter perpendicular to \(\overline{P Q}\) using a ruler and protractor, or a drawing triangle. Labeledit \(\overline{R S}\).
Length of the arc of the quadrant = circumference ÷ 4, As circumfrence = 2πr,  2 X 3.14 X 2 in =12.56 inches,
= 12.56 inches/4 = 3.14 inches, The distance around the fiqure is 3 X 3.14 inches =  9.42 inches.

Math in Focus Course 1B Practice 11.1 Answer Key

Use the figure to complete. In the figure, O is the center of the circle and \(\overline{X Y}\) is a straight line.

Question 1.
\(\overline{O X}\), \(\overline{O Y}\), and \(\overline{O Z}\) are of the circle.
Answer:
\(\overline{O X}\), \(\overline{O Y}\), and \(\overline{O Z}\) are radius of the circle,

Explanation:
Radius of a circle is the distance from the center of the circle to any point on it’s circumference. So \(\overline{O X}\), \(\overline{O Y}\), and \(\overline{O Z}\) are radius of the circle.

Question 2.
XY is a of the circle.
Answer:
Diameter,

Explanation:
The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circumference of the circle. So XY ia a diameter of the circle.

Question 3.
OX = =
Answer:
OX = OY = OZ,

Explanation:
As OX is radius of the circle and the distance of OY and OZ is same so OX = OY = OZ.

Question 4.
XY = • OZ
Answer:
2 X OZ,

Explanation:
As XY id diameter so XY = 2 X OZ.

Question 5.
Circumference of the circle = π •
Answer:
Circumference of the circle = 2 X π X OX or 2 X π X OY or 2 X π X OZ,

Explanation:
As radius of circle is OX, OY, OZ so Circumference of the circle = 2 X π X OX or 2 X π X OY or 2 X π X OZ.

Find the circumference of each circle. Use \(\frac{22}{7}\) as an approximation for π.

Question 6.

Answer:
The circumfrence is 43.96 cms,

Explanation:
As we know circumference of circle is C = 2πr given radius r = 7 cm so circumfrence = 2 X 3.14 X 7 cm = 43.96 cms.

Question 7.

Answer:
The circumfrence is 65.94 inches,

Explanation:
As we know circumference of circle is C = πd given diameter d = 21 inches so  circumfrence = 3.14 X 21 inches = 65.94 inches.

Question 8.

Answer:
The circumfrence is 153.86 m,

Explanation:
As we know circumference of circle is C = πd given diameter d = 49 m so circumfrence = 3.14 X 49 m = 153.86 m.

Question 9.

Answer:
The circumfrence is 17.584 cms,

Explanation:
As we know circumference of circle is C = 2πr given radius r = 2.8 ft so circumfrence = 2 X 3.14 X 2.8 ft = 17.584 cms.

Find the length of each arc. Use \(\frac{22}{7}\) as an approximation for π.

Question 10.

Answer:
Length of the arc = 12.089 cm,

Explanation:
Length of arc = \(\frac{1}{2}\) X circumference, Given diameter d = 7.7 cm as we know C = πd so length of
arc is \(\frac{1}{2}\) X 3.14 X 7.7 cm = 12.089 cm.

Question 11.

Answer:
Length of the arc = 15.386 ft,

Explanation:
Length of arc = \(\frac{1}{2}\) X circumference, Given diameter d = 9.8 ft as we know C = πd so length of
arc is \(\frac{1}{2}\) X 3.14 X 9.8 ft = 15.386 ft.

Question 12.

Answer:
Length of the arc = 3.297 m,

Explanation:
Length of the arc of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 2.1 m so circumfrence = 2 X 3.14 X 2.1 m = 13.188 m, So length of the arc =  13.188 m ÷ 4 = 3.297 m.

Question 13.

Answer:
Length of the arc = 17.584 cm,

Explanation:
Length of the arc of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 11.2 cm so circumfrence = 2 X 3.14 X 11.2 cm = 70.366 cm, So length of the arc =  70.366 cm ÷ 4 = 17.584 cm.

Find the distance around each semicircle. Use 3.14 as an approximation for π.

Question 14.

Answer:
Length of the semicircle = 31.4 in,

Explanation:
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 20 in as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 20 in = 31.4 in.

Question 15.

Answer:
Length of the semicircle = 31.4 in,

Explanation:
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 10 cm as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 10 cm = 31.4 in.

Question 16.

Answer:
Length of semicircle = 78.5 m,

Explanation:
Length of the semicircle = \(\frac{1}{2}\) X circumference and C = 2πr as given radius = 25 m so length of the semicircle = \(\frac{1}{2}\) X 2 X 3.14 X 25 m = 78.5 m.

Question 17.

Answer:
Length of semicircle = 23.55 ft,

Explanation:
Length of the semicircle = \(\frac{1}{2}\) X circumference and C = 2πr as given radius = 7.5 ft so length of the semicircle = \(\frac{1}{2}\) X 2 X 3.14 X 7.5 ft = 23.55 ft.

Find the distance around each quadrant. Use \(\frac{22}{7}\) as an approximation for π.

Question 18.

Answer:
Distance around quadrant = 5.495 in,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 3.5 in. So circumfrence = 2 X 3.14 X 3.5 in = 21.98 in , So length of the quadrant =  21.98 in ÷ 4 = 5.495 in.

Question 19.

Answer:
Distance around quadrant = 16.485 cm,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 10.5 cm so circumfrence = 2 X 3.14 X 10.5 cm = 65.94 cm, So length of the quadrant =  65.94 cm ÷ 4 = 16.485 cm.

Question 20.

Answer:
Distance around quadrant = 27.475 m,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 17.5 m so circumfrence = 2 X 3.14 X 17.5 m = 109.9 m , So length of the quadrant =  109.9 m ÷ 4 = 27.475 m.

Question 21.

Answer:
Distance around quadrant = 43.96 ft,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 28 ft so circumfrence = 2 X 3.14 X 28 ft = 175.84 ft, So length of the quadrant =  175.84 ft ÷ 4 = 43.96 ft.

Solve. Show your work. Use 3.14 as an approximation for π.

Question 22.
A circular garden has a diameter of 120 feet. Find its circumference.
Answer:
The circumfrence is 376.8 feet,

Explanation:
Given a circular garden has a diameter of 120 feet as we know circumference of circle is C = πd  given diameter = 120 feet so circumfrence = 3.14 X 120 ft = 376.8 feet.

Question 23.
A circular coaster has a radius of 4 centimeters. Find its circumference.
Answer:
The circumfrence is 25.12 cms,

Explanation:
Given a circular coaster has a radius of 4 centimeters as we know circumference of circle is C = 2πr given radius r = 4 cm so circumfrence = 2 X 3.14 X 4 cms =  25.12 cms.

Question 24.
The diameter of a roll of tape is 5\(\frac{1}{2}\) centimeters. Find its circumference.
Answer:
The circumfrence is 17.27 centimeters,

Explanation:
Given a diameter of a roll of tape 5\(\frac{1}{2}\) centimeters as we know circumference of circle is C = πd  given diameter = 5\(\frac{1}{2}\) centimeters so circumfrence = 3.14 X  5\(\frac{1}{2}\) centimeters = 3.14 X \(\frac{11}{2}\) centimeters = 17.27 centimeters.

Question 25.
The shape of a floor mat is a semicircle. Find the distance around the mat.

Answer:
Diatance around the mat = 119.32 cm,

Explanation:
Distance of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 38 cm + 38 cm = 76 cm as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 76 cm = 119.32 cm.

Question 26.
A small playground is shaped like a quadrant, as shown. Find the distance around the playground.

Answer:
The distance around the playground is 37.68 ft,

Explanation:
Given a small playground is shaped like a quadrant so distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 24 ft so circumfrence = 2 X 3.14 X 24 ft =150.72 ft, So length of the arc = 150.72 ft ÷ 4 = 37.68 ft.

Question 27.
If the radius of a wheel is 14 inches, what is the distance traveled when the wheel turns around 100 times?

Answer:
The distance traveled when the wheel turns around 100 times is 8,792 inches,

Explanation:
Given the radius of a wheel is 14 inches, the distance traveled is circumference so it is C = πd or 2πr as given radius = 14 in so circumfrence = 2 X 3.14 X 14 inchs = 87.92 inches so the distance traveled when the wheel turns around 100 times is 100 X 87.92 inches = 8,792 inches.

Find the distance around each figure. Use \(\frac{22}{7}\) as an approximation for π.

Question 28.
The figure is made up of a rectangle and a semicircle.

Answer:
Length of a rectangle is 80 cm,
Length of the semicircle = 43.96 cm,

Explanation:
Length or Perimeter of rectangle = 2(length + width) = 2(28 cm + 12 cm) = 2(40 cm) = 80 cm,
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 28 cm as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 28 cm = 43.96 cm.

Question 29.
The figure is made up of an equilateral triangle and a semicircle.

Answer:
Length of equilateral triangle =
Length of the semicircle = 65.94 cm,

Explanation:
Length of equilateral triangle = 42 cm + 42 cm + 42 cm = 126 cm,
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 42 cm as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 42 cm = 65.94 cm.

Find the distance around each figure. Use 3.14 as an approximation for π.

Question 30.
The figure is made up of a rectangle and two identical semicircles.

Answer:
Length of a rectangle is 172 ft,
The length of each identical semicircles is 50.24 ft,

Explanation:
Length or Perimeter of rectangle = 2(length + width) = 2(54 ft + 32 ft) = 2 X 86 ft = 172 ft,
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 32 ft as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 32 ft = 50.24 ft.

Question 31.
The figure is made up of two identical quadrants, a semicircle, and an equilateral triangle.

Answer:
The length of each quardrants is 36.11 ft,
The length of a semicircle is 72.22 ft,
Length of equilateral triangle is 69 ft,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 23 ft so circumfrence = 2 X 3.14 X 23 ft = 144.44 ft, So length of the each quadrant = 144.44 ft ÷ 4 = 36.11 ft.
Length of semicircle = \(\frac{1}{2}\) X circumference, Given radius = 23 ft as we know C = 2πr so length of semicircle is \(\frac{1}{2}\) X 3.14 X 2 X 23 ft = 72.22 ft. Length of equilateral triangle is 23 ft + 23 ft + 23 ft = 69 ft.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 11 Circumference and Area of a Circle to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 11 Answer Key Circumference and Area of a Circle

Math in Focus Grade 6 Chapter 11 Ouick Check Answer Key

Add.

Question 1.
0.451 + 3.12
Answer:
3.571,

Explanation:
Adding 3.12 to 0.451 we will get
0.451
+3.120
3.571

Question 2.
0.861 + 6.95
Answer:
7.811,

Explanation:
Adding 6.95 to 0.861 we will get
0.861
+6.950
7.811

Question 3.
13.74 + 3.791
Answer:
17.531,

Explanation:
Adding 3.791 to 13.74 we will get
13.740
+3.791
17.531

Subtract.

Question 4.
5.45 – 1.78
Answer:
3.67,

Explanation:
Subtacting 1.78 from 5.45 we will get
5.45
-1.78
3.67

Question 5.
12.795 – 0.816
Answer:
11.979,

Explanation:
Subtracting 0.816 from 12.795 we will get
12.795
-0.816
11.979

Question 6.
42.781 – 36.19
Answer:
6.591

Explanation:
Subtracting 36.19 from 42.781 we will get
42.781
-36.190
6.591

Multiply

Question 7.
1.34 × 9
Answer:
12.06,

Explanation:
Given to multiply 1.34 with 9 we will get
3,3
1.34
X 9
12.06

Question 8.
4.246 × 2
Answer:
8.492,

Explanation:
Given to multiply 4.246 by 2 we will get
1
4.246
X   2
8.492

Question 9.
7.487 × 8
Answer:
59.896,

Explanation:
Given to multiply 7.487 by 8 we will get
3,6,5
7.487
X  8
59.896

Divide

Question 10.
2.56 ÷ 5
Answer:
0.512,

Explanation:
5)2.56(0.512
   2.5
0.06
0.05
0.01
0.01
0

Question 11.
2.429 ÷ 7
Answer:
0.347,

Explanation:
7)2.429(0.347
   2.1
    0.32
0.28
0.049
0.049
0

Question 12.
1.143 ÷ 9
Answer:
0.127,

Explanation:
9)1.143(0.127
   0.9
   0.24
    0.18
   0.063
    0.063
      0

Question 13.
4.671 ÷ 9
Answer:
0.519,

Explanation:
9)4.671(0.519
   4.5
   0.17
   0.09
  0.081
   0.081
  0

Question 14.
0.656 ÷ 8
Answer:
0.082,

Explanation:
8)0.656(0.082
    0.64
    0.016
     0.016
0

Question 15.
0.867 ÷ 3
Answer:
0.289,

Explanation:
3)0.867(0.289
    0.6
0.26
0.24
0.027
0.027
0

Round to the nearest whole number.

Question 16.
4.56
Answer:
5,

Explanation:
Asked to round 4.56 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 4.56 is 5.

Question 17.
12.05
Answer:
12,

Explanation:
Asked to round 12.05 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 12.05 is 12.

Question 18.
6.48
Answer:
6,

Explanation:
Asked to round 6.48 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 6.48 is 6.

Question 19.
6.50
Answer:
7,

Explanation:
Asked to round 6.50 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 6.50 is 7.

Question 20.
14.15
Answer:
14,

Explanation:
Asked to round 14.15 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 14.15 is 14.

Question 21.
46.59
Answer:
47,

Explanation:
Asked to round 46.59 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 46.59 is 47.

Round to the nearest tenth.

Question 22.
7.68
Answer:
7.7,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 7.68 is 7.7.

Question 23.
3.05
Answer:
3.00,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 3.05 is 3.00.

Question 24.
19.92
Answer:
20.00,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 19.92 is 20.00.

Question 25.
5.55
Answer:
5.60,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 5.55 is 5.60.

Question 26.
8.17
Answer:
8.0,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 8.17 is 8.0.

Question 27.
2.44
Answer:
2.4,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 2.44 is 2.4.

Question 28.
43.65
Answer:
43.7,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 43.65 is 43.7.

Question 29.
23.73
Answer:
23.8,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 23.73 is 23.8.

Question 30.
17.51
Answer:
17.5,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 17.51 is 17.5.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 The Coordinate Plane to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Answer Key The Coordinate Plane

Math in Focus Grade 6 Chapter 9 Quick Check Answer Key

Use the coordinate plane below.

Question 1.
Give the coordinates of points B, C, and D.
Answer:
Point B:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is 3.
So, the point P will be (3,3).
Point C:
The first coordinate represents distance along x axis which is 1 and the second coordinate represents distance along y axis which is 4.
So, the point C will be (1,4).
Point D:
The first coordinate represents distance along x axis which is 5 and the second coordinate represents distance along y axis which is 2.
So, the point D will be (5,2).

Use graph paper. Plot the points on a coordinate plane.

Question 2.
P (3, 2), Q (2, 3), and R (0, 4)
Answer:
To plot point P (3, 2):
Here, the x-coordinate is 3 and the y-coordinate is 2. Start at the Origin. As the x coordinate is positive, move 3 units along the positive x-axis and 2 units along positive y -axis.
Thus, the required point P (3, 2) is marked.
To plot point Q (2, 3):
Here, the x-coordinate is 2 and the y-coordinate is 3. Start at the Origin. As the x coordinate is positive, move 2 units along the positive x-axis and 3 units along positive y -axis.
Thus, the required point P (2, 3) is marked.
To plot point R (0, 4):
Here, the x-coordinate is 0 and the y-coordinate is 4.  As the y coordinate is positive, move 5 units along the positive y-axis and mark it.
Thus, the required point(0, 4) is marked

Identify the number that each indicated point represents.

Question 3.

Answer:
To the left of the origin, the x coordinate values will be negative.
Start counting from the origin and note down the values.
The first value will be -1, next will be -4, -6 and -8.

Draw a horizontal number line to represent each set of numbers.

Question 4.
-3, 0, 1, 5, 8
Answer:
-3 is the negative value; therefore mark it to the left of the origin.
0 is the origin. 1,5,8 are the positive values, which are to be marked to the right of the origin.

Question 5.
-15, -11, -9, -7, -2
Answer:
Here, all are negative values. Hence all the values are to be marked to the left of the origin.

Use the symbol || to write the absolute values of the following numbers.

Question 6.
11
Answer:
The distance of an integer from ‘

$0$

‘ on the number line irrespective of its direction is called the absolute value of that integer.
Two vertical bars ‘| |’ are used to denote the absolute value.
The absolute value of any positive number is the number itself.
Therefore, the absolute value of 11 will be 11.

Question 7.
-16
Answer:
The absolute value of negative number is the positive of it.
Therefore, the absolute value of -16 or |-16| will be 16.

Question 8.
-21
Answer:
The absolute value of negative number is the positive of it.
Therefore, the absolute value of -21 or |-21| will be 21.

Find the perimeter of each polygon.

Question 9.
Figure ABC is an isosceles triangle.

Answer:
Given triangle ABC has two equal sides, thus it will form an isosceles triangle.
The triangle will measure 8 in, 8 in and 5 in.
Perimeter of the triangle will be the sum of all sides, 8+8+5 = 21 in

Question 10.
Figure DEFis an equilateral triangle.

Answer:
Given DEF is an equilateral triangle. Equilateral trinagle will have all equal sides.
All sides of the triangle will measure 4 in.
Perimeter of the triangle will be the sum of all sides, 4+4+4 = 12 in

Question 11.
Figure PQRS is a trapezoid.

Answer:
Given figure PQRS is a trapezoid. One of the pair of opposite sides are equal in length.
The trapezoid measures 16cm,10cm and 9cm.
Perimeter of the trapezoid will be the sum of all sides, 16+10+9+9 = 44cm

Question 12.
Figure WXYZ is a parallelogram.

Answer:
Given figure WXYZ is a parallelogram. The opposite sides are equal in length.
The parallelogram measures 8cm and 7cm.
Perimeter of the parallelogram will be the sum of all sides, 8+7+8+7 = 30cm

Question 13.
Figure JKLM is a rhombus.

Answer:
Given figure JKLM is a rhombus. All the sides are equal in length.
Each side will measure 6m in length.
Perimeter of the rhombus will be the sum of all sides, 6+6+6+6 = 24cm

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 Lesson 9.1 Points on the Coordinate Plane to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Lesson 9.1 Answer Key Points on the Coordinate Plane

Math in Focus Grade 6 Chapter 9 Lesson 9.1 Guided Practice Answer Key

Use the coordinate plane below.

Question 1.
Give the coordinates of points P, Q, R, S, T, U, and V. In which quadrant is each point located?

Answer:
Point P:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is 0.
So, the point P will be (3,0).
The point having y-coordinate as 0 lie on the X-axis and thus is not located in any quadrant.

Point Q:
The first coordinate represents distance along x axis which is -3 and the second coordinate represents distance along y axis which is 3.
So, the point Q will be (-3,3).
If the x axis coordinate is negative and y axis coordinate is positive, the point will lie in quadrant II.

Point R:
The first coordinate represents distance along x axis which is -5 and the second coordinate represents distance along y axis which is 1.
So, the point R will be (-5,1).
If the x axis coordinate is negative and y axis coordinate is positive, the point will lie in quadrant II.

Point S:
The first coordinate represents distance along x axis which is -6 and the second coordinate represents distance along y axis which is -4.
So, the point S will be (-6,-4).
If the both the coordinates are negative then, the point will lie in quadrant III.

Point T:
The first coordinate represents distance along x axis which is -4 and the second coordinate represents distance along y axis which is -7.
So, the point T will be (-4,-7).
If the both the coordinates are negative then, the point will lie in quadrant III.

Point U:
The first coordinate represents distance along x axis which is 4 and the second coordinate represents distance along y axis which is -5.
So, the point U will be (4,-5).
If the x axis coordinate is positive and y axis coordinate is negative , the point will lie in quadrant IV.

Point V:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is -2.
So, the point V will be (3,-2).
If the x axis coordinate is positive and y axis coordinate is negative , the point will lie in quadrant IV.

Use graph paper.

Question 2.
Plot points A (-4, 3), 6(3, -4), C (5, 0), D (0, -5), E(-2, -1), and F(2, -1)on a coordinate plane.
Answer:
To plot point A(-4,3):
Here, the x-coordinate of A is -4 and the y-coordinate is 3. Start at the Origin. As the x coordinate is negative, move 4 units to the left along the x-axis and 3 units to the top along y -axis.
Thus, the required point A(-4,3) is marked.

To plot point B(3,-4):
Here, the x-coordinate of B is 3 and the y-coordinate is -4. Start at the Origin. As the x coordinate is positive, move 3 units to the right along the x-axis. As y cooridnate is negative, move 4 units to the down along y -axis.
Thus, the required point B(-3,4) is marked.

To plot point C(5,0):
Here, the x-coordinate of C is 5 and the y-coordinate is 0. As the x coordinate is positive, move 5 units to the right along the x-axis and mark it. Thus, the required point(5,0) is marked.

To plot point D(0,-5):
Here, the x-coordinate of D is 0 and the y-coordinate is -5.  As the y coordinate is negative, move 5 units to the down along the y-axis and mark it. Thus, the required point(0,-5) is marked.

To plot point E(-2, -1):
Here, the x-coordinate of E is -2 and the y-coordinate is -1.  As the x coordinate is negative, move 2 units to the left along the x-axis. As the y coordinate is negative, move 1 units to the down along the y-axis and mark it.
Thus, the required point (-2,-1) is marked.

To plot point F(2, -1):
Here, the x-coordinate of E is 2 and the y-coordinate is -1. As the x coordinate is positive, move 2 units to the right along the x-axis. As the y coordinate is negative, move 1 units to the down along the y-axis and mark it.
Thus, the required point (2,-1) is marked.

Question 3.
Points P and Q are reflections of each other about the x-axis. Give the coordinates of point Q if the coordinates of point P are the following:
a) (-6, 2)
Answer:
Given that the points P and Q are reflections of each other about the x-axis.
Given that Q is reflection of point P in the x-axis and point P is (-6,2).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be -6 and y coordinate will be -2.
Therefore, point Q will be (-6,-2).

b) (-2, -4)
Answer:
Given that Q is reflection of point P in the x-axis and point P is (-2, -4).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be -2 and y coordinate will be 4.
Therefore, point Q will be (-2,4).

c) (4, 5)
Answer:
Given that Q is reflection of point P in the x-axis and point P is (4, 5).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be 4 and y coordinate will be -5.
Therefore, point Q will be (4,-5).

d) (7, -3)
Answer:
Given that Q is reflection of point P in the x-axis and point P is (7, -3).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be 7 and y coordinate will be 3.
Therefore, point Q will be (7,3).

Question 4.
Points R and S are reflections of each other about the y-axis. Give the coordinates of point S if the coordinates of point R are the following:

a) (-6,2)
Answer:
Given that the points R and S are reflections of each other about the y-axis.
Given that R is reflection of point S in the y-axis and point R is (-6,2).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 6 and y coordinate will be 2.
Therefore, point S will be (6,2).

b) (-2, -4)
Answer:
Given that R is reflection of point S in the y-axis and point R is (-2, -4).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 2 and y coordinate will be -4.
Therefore, point S will be (2,-4).

c) (4, 5)
Answer:
Given that R is reflection of point S in the y-axis and point R is (4, 5).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -4 and y coordinate will be 5.
Therefore, point S will be (-4,5).

d) (7, -3)
Answer:
Given that R is reflection of point S in the y-axis and point R is (7, -3).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -7 and y coordinate will be -3.
Therefore, point S will be (-7,-3).

Use graph paper. For each exercise, plot the given points on a coordinate plane. Then join the points in order with line segments to form a closed figure. Name each figure formed.

Question 5.
A (3, 4), B (-6, -3), and C (2, -4)

Answer:

Given points are, A (3, 4), B (-6, -3), and C (2, -4).
To get a closed figure, plot the above points and join them in order.
It is forming a three-sided polygon, which is a triangle.
Thus, ABC will form a triangle as shown in the graph above.

Question 6.
D(1, 1), E(0, 0), and F (-4, 4)
Answer:

Given points are, D(1, 1), E(0, 0), and F (-4, 4).
To get a closed figure, plot the above points and join them in order.
It is forming a straight line.
Thus, DEF is a straight line as shown in the graph above.

Question 7.
J (-3, 0), K (0, 5), and L (3, 0)
Answer:

Given points are, J (-3, 0), K (0, 5), and L (3, 0).
To get a closed figure, plot the above points and join them in order.
It is forming a three-sided polygon, which is a triangle.
Thus, JKL will form a triangle as shown in the graph above.

Question 8.
P (3, 2), Q (-1, 2), R (-1, -2), and S (3, -2)
Answer:

Given points are, P (3, 2), Q (-1, 2), R (-1, -2), and S (3, -2).
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the sides are equal in length. A square have all equal sides.
Thus, PQRS will form a square as shown in the graph above.

Question 9.
W(-3, 2), X (1, -2), Y(3, 0), and Z(-1, 4)
Answer:

Given points are, W(-3, 2), X (1, -2), Y(3, 0), and Z(-1, 4)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. The opposite sides are equal in length. A rectangle will have opposite sides equal.
Thus, WXYZ will form a rectangle as shown in the graph above.

Question 10.
A (-5, 2), B (-5, -1), C(-1, -1), and D(1, 2)
Answer:

Given points are, A (-5, 2), B (-5, -1), C(-1, -1), and D(1, 2)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are of different length, and a pair of opposite sides seems to be parallel which can be a trapezuim.
Thus, ABCD will form a trapezium as shown in the graph above.

Question 11.
E (-2, -2), F (-5, -5), G (-2, -5), and H (1, -2)
Answer:

Given points are E (-2, -2), F (-5, -5), G (-2, -5), and H (1, -2)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. The opposite sides are parallel to each other. A parallelogram will have opposite sides in parallel.
Thus, EFGH will form a parallelogram as shown in the graph above.

Question 12.
J (-4, 1), K (-3, -1), L (0, -1), and M (2, 1)
Answer:

Given points are, J (-4, 1), K (-3, -1), L (0, -1), and M (2, 1)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are of different length, and a pair of opposite sides seems to be parallel which can be a trapezuim.
Thus, JKLM will form a trapezium as shown in the graph above.

Question 13.
P (4, 0), Q (0, 4), P (-4, 0), and S (0, -4)
Answer:

Given points are, P (4, 0), Q (0, 4), R (-4, 0), and S (0, -4)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are equal in length and diagonals are bisecting each other.
Thus, PQRS will form a rhombus.

Question 14.
W (-2, 0), X (-3, -3), Y(1, 1), and Z(2, 4)
Answer:

Given points are W (-2, 0), X (-3, -3), Y(1, 1), and Z(2, 4)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. The opposite sides are parallel to each other. A parallelogram will have opposite sides in parallel.
Thus, WXYZ will form a parallelogram as shown in the graph above.

Hands-On Activity

IDENTIFYING QUADRILATERALS DRAWN ON A COORDINATE PLANE

Work in pairs.

Step 1.
Plot four points on a coordinate plane and connect them to form a special quadrilateral such as a parallelogram, a rectangle, or a rhombus. Do not let your partner see your quadrilateral.

Step 2.
Tell your partner the coordinates of three out of the four coordinates of the points you plotted in Step 1. Also tell your partner the type of quadrilateral you plotted, and in which quadrant the fourth point is located. Have your partner guess the coordinates of the fourth point.
Example

Points A (1, 5), B (-2, 1), C(1, -3), and D can be joined to form a rhombus. If point D is in Quadrant I, what are the coordinates of point D?

Step 3.
Switch roles with your partner and repeat the activity with other quadrilaterals.

Math in Focus Course 1B Practice 9.1 Answer Key

Use the coordinate plane below.

Question 1.
Give the coordinates of each point. In which quadrant is each point located?

Answer:

Point A:
The first coordinate represents distance along x axis which is 2 and the second coordinate represents distance along y axis which is 6.
So, the point A will be (2,6).
If the x axis coordinate and y axis coordinate are positive, the point will lie in quadrant I.

Point B:
The first coordinate represents distance along x axis which is 6 and the second coordinate represents distance along y axis which is 2.
So, the point B will be (6,2).
If the x axis coordinate and y axis coordinate are positive, the point will lie in quadrant I.

Point C:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is 0.
So, the point C will be (3,0).
The point having y-coordinate as 0 lie on the X-axis and thus is not located in any quadrant.

Point D:
The first coordinate represents distance along x axis which is 4 and the second coordinate represents distance along y axis which is 9.
So, the point D will be (4,-9).
If the x axis coordinate is positive and y axis coordinate is negative, then the point will lie in quadrant IV.

Point E:
The first coordinate represents distance along x axis which is 0 and the second coordinate represents distance along y axis which is 2.
So, the point E will be (0,-2).
The point having x-coordinate as 0 lie on the y-axis and thus is not located in any quadrant.

Point F:
The first coordinate represents distance along x axis which is 2 and the second coordinate represents distance along y axis which is 4.
So, the point F will be (-2,-4).
If the x axis coordinate and y axis coordinate are positive, the point will lie in quadrant III.

Point G:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is 0.
So, the point G will be (-3,0).
The point having y-coordinate as 0 lie on the x-axis and thus is not located in any quadrant.

Point H:
The first coordinate represents distance along x axis which is 1 and the second coordinate represents distance along y axis which is 9.
So, the point F will be (-1,9).
If the x axis coordinate is negative and y axis coordinate is positive, than the point will lie in quadrant II.

Use graph paper. Plot the points on a coordinate plane. In which quadrant is each point located?

Question 2.
A (3, 7), B (2, 0), C (8, -1), D (0, -6), E(-3, -5), and F(-6, 7)
Answer:

Point A (3, 7) lies in quadrant I, as x-coordinate and y-coordinate are both positive.
Point B (2, 0) is having y-coordinate as 0 lie on the x-axis does not lie in any quadrant.
Point C (8, -1) lies in quadrant IV, as x axis is positive and y-axis is negative.
Point D (0, -6) is having x-coordinate as 0 lie on the y-axis does not lie in any quadrant.
Point E (-3, -5) lies in quadrant III, as x-coordinate and y-coordinate are both negative.
Point F (-6, 7) lies in quadrant II, as x-coordinate is negative and y-coordinate is positive.

Use graph paper. Points A and B are reflections of each other about the x-axis. Give the coordinates of point B if the coordinates of point A are the following:

Question 3.
(4, 1)
Answer:
Given that the points A and B are reflections of each other about the x-axis.
Given that B is reflection of point A in the x-axis and point A is (4,1).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be 4 and y coordinate will be -1.
Therefore, point B will be (4,-1).

Question 4.
(-2, 3)
Answer:
Given that B is reflection of point A in the x-axis and point A is (-2,3).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be -2 and y coordinate will be -3.
Therefore, point B will be (-2,-3).

Question 5.
(2, -2)
Answer:
Given that B is reflection of point A in the x-axis and point A is (2,-2).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be 2 and y coordinate will be 2.
Therefore, point B will be (2,2).

Question 6.
(-1, -3)
Answer:
Given that B is reflection of point A in the x-axis and point A is (-1,-3).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be -1 and y coordinate will be 3.
Therefore, point B will be (-1,3).

Use graph paper. Points C and D are reflections of each other about the y-axis. Give the coordinates of point D if the coordinates of point C are the following:

Question 7.
(4, 1)
Answer:
Given that the points A and B are reflections of each other about the y-axis.
Given that B is reflection of point A in the x-axis and point A is (4,1).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -4 and y coordinate will be 1.
Therefore, point S will be (-4,1).

Question 8.
(-2, 3)
Answer:
Given that the points A and B are reflections of each other about the y-axis.
Given that B is reflection of point A in the x-axis and point A is (2,-3).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 2 and y coordinate will be -3.
Therefore, point S will be (2,-3).

Question 9.
(2, -2)
Answer:
Given that the points A and B are reflections of each other about the y-axis.
Given that B is reflection of point A in the x-axis and point A is (2,-2).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -2 and y coordinate will be -2.
Therefore, point S will be (-2,-2).

Question 10.
(-1, -3)
Answer:
Given that the points A and B are reflections of each other about the y-axis.
Given that B is reflection of point A in the x-axis and point A is (-1,-3).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 1 and y coordinate will be -3.
Therefore, point S will be (1,-3).

Use graph paper. For each exercise, plot the given points on a coordinate plane. Then join the points in order with line segments to form a closed figure. Name each figure formed.

Question 11.
H(-5, 1), J(-3, -1), K (-1, 1), and L(-3, 3)
Answer:

Given points are H(-5, 1), J(-3, -1), K (-1, 1), and L(-3, 3)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are equal in length and diagonals are bisecting each other.
Thus, HJKL will form a rhombus as shown in the graph above.

Question 12.
R (2, 1), S (-1, -3), T(4, -3), and U(7, 1)
Answer:

Given points are R (2, 1), S (-1, -3), T(4, -3), and U(7, 1)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. The opposite sides are parallel to each other. A parallelogram will have opposite sides in parallel.
Thus, RSTU will form a parallelogram as shown in the graph above.

Question 13.
W (-5, -2), X (-6, -5), Y (-1, -5), and Z (-3, -2)
Answer:

Given points are, W (-5, -2), X (-6, -5), Y (-1, -5), and Z (-3, -2)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are of different length, and a pair of opposite sides seems to be parallel which can be a trapezuim.
Thus, WXYZ will form a trapezium as shown in the graph above.

Use graph paper. Plot the points on a coordinate plane and answer each question.

Question 14.
a) Plot points A (-6, 5), C (5, 1), and D (5, 5) on a coordinate plane.
Answer:

Given points are, A (-6, 5), C (5, 1), and D (5, 5)
To get a closed figure, plot the above points and join them in order.
It is forming a three-sided polygon, which is a triangle. Angle CDA is forming a right angle, so it is a right angle triangle.
Thus, ADC will form a right angle triangle as shown in the graph above.

b) Figure ABCD is a rectangle. Plot point B and give its coordinates.
Answer:

Given points are, A (-6, 5), C (5, 1), and D (5, 5)
A rectangle is combination of two right angle triangle. The base length of given trinagle is 11 units. In a rectangle, opposite sides are equal in length. Therefore, if AD is of 11 units, BC will also be of 11 units and the height will also be same which is 4 units.
Move 11 units towards left from point C and 4 units towards down from point A to mark point B.
Thus, point B will be (-6,1)
Plot point B to form a rectangle.

c) Figure ACDE is a parallelogram. Plot point E above \(\overline{\mathrm{AD}}\) and give its coordinates.
Answer:
Given that figure ACDE is a parallelogram. Point E is above \(\overline{\mathrm{AD}}\).
In parallelogram, opposite sides are parallelo to each other. Move 4 units towards upwards from point A nd mark as E.
Join the point EC to form the required parallelogram ACDE.
Thus, Point E will be (-6,9)

Question 15.
a) Plot points A (-3, 2) and B (-3, -2) on a coordinate plane.
Answer:

To plot point A(-3,2):
Here, the x-coordinate of A is -3 and the y-coordinate is 2. Start at the Origin. As the x coordinate is negative, move 3 units to the left along the x-axis and 2 units to the top along y -axis.
Thus, the required point A(-3,2) is marked.
To plot point B(-3,-2):
Here, the x-coordinate of A is -3 and the y-coordinate is 2. Start at the Origin. As the x coordinate is negative, move 3 units to the left along the x-axis and 2 units to the down along y -axis as it also is negative.
Thus, the required point A(-3,-2) is marked.

b) Join points A and B with a line segment.
Answer:

Join A and B points to form a straight line.

c) \(\overline{A B}\) is a side of square ABCD. Name two possible sets of coordinates that could be the coordinates of points C and D.
Answer:

Given that AB is a side of square ABCD. In square all sides are equal. Length of AB is 4 units.
Set 1:
So, first we will start moving towards right from B along with the x-axis and mark the point as C.
Move 4 units towards up from C or towards right from A, mark the point D.
Join ABCD to form a square.
The first possible set of coordinates for C and D will be (1,-2) and (1,2) respectively.
Set 2:
Next, start moving 4 units towards the left from point B and mark it as F.
Move 4 uniits towards up from F or towards left from A, mark the point as E.
Join ABEF to form a square.
The second possible set of coordinates for C and D will be (-7,-2) and (-7,2) respectively.

Question 16.
Plot points A (2, 5) and 8 (2, -3) on a coordinate plane. Figure ABC is a right isosceles triangle. If point C is in Quadrant III, give the coordinates of point C.
Answer:
To plot point A(2,5):
Here, the x-coordinate of A is 2 and the y-coordinate is 5. Start at the Origin. As the x coordinate is positive, move 2 units to the right along the x-axis and 5 units to the top along y -axis.
Thus, the required point A(2,5) is marked.
To plot point B(2,-3):
Here, the x-coordinate of A is 2 and the y-coordinate is -3. Start at the Origin. As the x coordinate is positive, move 2 units to the right along the x-axis and 3 units to the top along y -axis.
Thus, the required point B(2,-3) is marked.
Point C is in Quadrant III and it should form a right angle triangle.
The side AB is of length 8 units, so BC will also remain the same.
Figure ABC is a right isosceles triangle.

Question 17.
Plot points A (0, 4), B (-4, 0), and C (0, -4) on a coordinate plane.

Given points A (0, 4), B (-4, 0), and C (0, -4) are to be plotted on a coordinate plane.
Joining them will form a triangle.

a) What kind of triangle is triangle ABC?
Answer:
Since the point B is at equidistant from A and C.
The lengths of BC and AB will be same.
Thus, ABC will form an isosceles triangle.

b) Figure ABCD is a square. Plot point D on the coordinate plane and give its coordinates.
Answer:

In square, all sides are equal. The length of AB and BC equals 4 units.
D is reflection of point B in the x-axis and point B is (-4,0).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Therefore, point D will be (4,0)
Thus, the required square ABCD is formed.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 Lesson 9.2 Length of Line Segments to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Lesson 9.2 Answer Key Length of Line Segments

Math in Focus Grade 6 Chapter 9 Lesson 9.2 Guided Practice Answer Key

Use graph paper. Plot each pair of points on a coordinate plane. Connect the points to form a line segment and find its length.

Question 1.
E(-6, 0) and D(8, 0)
Answer:
Point E(-6, 0)
Here, the x-coordinate of E is -6 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is negative, move 6 units along the negative x-axis and mark it. Thus, the required point(-6,0) is marked.
Point D(8,0)
Here, the x-coordinate of D is 8 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is positive, move 8 units along the positive x-axis and mark it. Thus, the required point(8,0) is marked.

After joining the E and D, we will get a line segment of 18 units.

Question 2.
E(-6, 0) and F(-2, 0)
Answer:
Point E(-6, 0)
Here, the x-coordinate of E is -6 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is negative, move 6 units along the negative x-axis and mark it. Thus, the required point(-6,0) is marked.
Point F(-2,0)
Here, the x-coordinate of D is -2 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is negative, move 2 units along the negative x-axis and mark it. Thus, the required point(-2,0) is marked.

After joining the E and F, we will get a line segment of 4 units.

Question 3.
G(-7, 0) and H(1, 0)
Answer:
Point E(-7, 0)
Here, the x-coordinate of E is -7 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is negative, move 7 units along the negative x-axis and mark it. Thus, the required point(-7,0) is marked.
Point H(1, 0)
Here, the x-coordinate of H is 1 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is positive, move 1 units along the positive x-axis and mark it. Thus, the required point(1,0) is marked.

After joining the G and H, we will get a line segment of 8 units.

Question 4.
J (0, 5) and K (0, 2)
Answer:
To plot J(0,5):
Here, the x-coordinate is 0 and the y-coordinate is 5. As the x-coordinate is zero and y coordinate is positive, move 5 units along positive y-axis and mark it. Thus, the required point(0,5) is marked.
To point K (0, 2):
Here, the x-coordinate is 0 and the y-coordinate is 2. As the x-coordinate is zero and y coordinate is positive, move 2 units along positive y-axis and mark it. Thus, the required point(0,2) is marked.

After joining the J and K, we will get a line segment of 3 units.

Question 5.
M(0, -6)and N(0, -3)
Answer:
To plot M(0, -6):
Here, the x-coordinate is 0 and the y-coordinate is -6. As the x-coordinate is zero and y coordinate is negative, move 6 units along negative y-axis and mark it. Thus, the required point (0,-6) is marked.
To point N(0, -3):
Here, the x-coordinate is 0 and the y-coordinate is -3. As the x-coordinate is zero and y coordinate is negative, move 3 units along negative y-axis and mark it. Thus, the required point(0,-3) is marked.

After joining the M and N, we will get a line segment of 3 units.

Question 6.
P(0, -3) and Q(0, 5)
Answer:
To plot P(0, -3):
Here, the x-coordinate is 0 and the y-coordinate is -3. As the x-coordinate is zero and y coordinate is negative, move 3 units along negative y-axis and mark it. Thus, the required point (0,-3) is marked.
To point Q(0, 5):
Here, the x-coordinate is 0 and the y-coordinate is 5. As the x-coordinate is zero and y coordinate is positive, move 5 units along positive y-axis and mark it. Thus, the required point(0,5) is marked.

After joining the P and Q, we will get a line segment of 8 units.

Use graph paper. Plot each pair of points on a coordinate plane. Connect the points to form a line segment and find its length.

Question 7.
A(1, -2) and B(6, -2)
Answer:
To plot A(1, -2):
Here, the x-coordinate is 1 and the y-coordinate is -2. Start at the Origin. As the x coordinate is positive, move 1 units to the right along the x-axis and 2 units to the bottom along y -axis.
Thus, the required point (1,-2) is marked.
To point B(6, -2):
Here, the x-coordinate is 6 and the y-coordinate is -2. Start at the Origin. As the x coordinate is positive, move 6 units to the right along the x-axis and 2 units to the bottom along y -axis.
Thus, the required point (6, -2) is marked.

After joining the A and B, we will get a line segment of 5 units.

Question 8.
C(-1, 3) and D(5, 3)
Answer:
Plot C(-1, 3):
Here, the x-coordinate is -1 and the y-coordinate is 3. Start at the Origin. As the x coordinate is negative, move 1 units to the left along the x-axis and 3 units to the above along positive y -axis and mark it.
Thus, the required point (-1, 3) is marked.
Point D(5, 3):
Here, the x-coordinate is 5 and the y-coordinate is 3. Start at the Origin. As the x-coordinate and y coordinate is positive, move 5 unit along positive x-axis and 3 units to the above along positive y -axis and mark it.
Thus, the required point (5, 3) is marked.

After joining the C and D, we will get a line segment of 6 units.

Question 9.
E (-3, 4) and F (1, 4)
Answer:
Plot E (-3, 4):
Here, the x-coordinate is -3 and the y-coordinate is 4. Start at the Origin. As the x coordinate is negative, move 3 units to the left along the x-axis and 4 units to the above along positive y -axis and mark it.
Thus, the required point (-3, 4) is marked.
Point F (1, 4):
Here, the x-coordinate is 5 and the y-coordinate is 3. Start at the Origin. As the x-coordinate and y coordinate is positive, move 1 unit along positive x-axis and 4 units to the above along positive y -axis and mark it.
Thus, the required point (1, 4) is marked.

After joining the E and F, we will get a line segment of 4 units.

Question 10.
G (-3, 2) and H (-3, 6)
Answer:
Plot G (-3, 2):
Here, the x-coordinate is -3 and the y-coordinate is 2. Start at the Origin. As the x coordinate is negative, move 3 units to the left along the x-axis and 2 units to the above along positive y -axis and then mark it.
Thus, the required point (-3, 2) is marked.
Point H (-3, 6):
Here, the x-coordinate is -3 and the y-coordinate is 6. Start at the Origin. As the x-coordinate is negative, move 3 units to the left along the x-axis and 6 units along positive y-axis and then mark it.
Thus, the required point (-3, 6) is marked.

After joining the G and H, we will get a line segment of 4 units.

Question 11.
J(-1, -6) and K(-1, 4)
Answer:
Plot J(-1, -6):
Here, the x-coordinate is -1 and the y-coordinate is -6. Start at the Origin. As the x coordinate is negative, move 1 unit to the left along the x-axis and 6 units along negative y -axis and then mark it.
Thus, the required point (-1, -6) is marked.
Point K(-1, 4):
Here, the x-coordinate is -1 and the y-coordinate is 4. Start at the Origin. As the x-coordinate is negative, move 1 units to the left along the x-axis and 4 units along positive y-axis and then mark it.
Thus, the required point (-1, 4) is marked.

After joining the J and K, we will get a line segment of 10 units.

Question 12.
L(5, 6) and M(5, 1)
Answer:
Point L(5, 6):
Here, the x-coordinate is 5 and the y-coordinate is 6. Start at the Origin. As the x-coordinate and y coordinate is positive, move 5 units along positive x-axis and 6 units to the above along positive y -axis and mark it.
Thus, the required point (5, 6) is marked.
Point M(5, 1):
Here, the x-coordinate is 5 and the y-coordinate is 1. Start at the Origin. As the x-coordinate and y coordinate is positive, move 5 units along positive x-axis and 1 unit to the above along positive y -axis and mark it.
Thus, the required point (5, 1) is marked.

After joining the L and M, we will get a line segment of 5 units.

In the diagram, triangle ABC represents a plot of land. The side length of each grid square ¡s 5 meters. Use the diagram to answer questions 13 to 15.

Question 13.
Give the coordinates of points A, B, and C.
Answer:
Point A:
The first coordinate represents distance along x axis which is 10 units and the second coordinate represents distance along y axis which is 20 units.
So, the point A will be (10,20).
Point B:
The first coordinate represents distance along x axis which is 10 units and the second coordinate represents distance along y axis which is 5 units.
So, the point B will be (10,5).
Point C:
The first coordinate represents distance along x axis which is 30 units and the second coordinate represents distance along y axis which is 20 units.
So, the point C will be (30,20).

Question 14.
Mr. Manning wants to build a fence around the plot of land. If BC is 25 meters, how many meters of fencing does he need?
AB = –
= m
AC = –
Perimeter of triangle ABC = AB + BC + AC
= + +
= m
Mr. Manning needs meters of fencing.
Answer:
Here, 1 grid square represents 5 meters. From the figure,
AB equals 3 grid square so it will measure 3×5=15m
AC equals 4 grid square so it will measure 4×5=20m
Given BC equals 25m.
Perimeter of triangle ABC = AB + BC + AC
= 15+25+20
= 60 m

Question 15.
A pole is located at point D on the plot of land at a distance of 10 meters from \(\overline{A B}\) and 5 meters from \(\overline{A C}\). Give the coordinates of point D.
1 grid square represents 5 meters.
10 m = ÷ 5
= grid squares
Answer:
Given that point D is located a distance of 10 meters from \(\overline{A B}\) and 5 meters from \(\overline{A C}\).
10 m = 10 ÷ 5
= 2 grid squares.

For point D to be on the plot of land, the x-coordinate has to be grid squares to the right of \(\overline{\mathrm{AB}}\). So, point D is + = grid squares to the right of the y-axis. The x-coordinate of point D is × = .
Answer:
For point D to be on the plot of land, the x-coordinate has to be 2 grid squares to the right of
\(\overline{\mathrm{AB}}\). So, point D is 2+2 = 4 grid squares to the right of the y-axis.
The x-coordinate of point D is 4×5  = 20.

For point D to be on the plot of land, the y-coordinate has to be grid square below \(\overline{\mathrm{AC}}\). So, point D is – = grid squares above the x-axis. The y-coordinate of point D is × = .
The coordinates of D are (, )
Answer:
For point D to be on the plot of land, the y-coordinate has to be 1 grid square below \(\overline{\mathrm{AC}}\). So, point D is 4-1 = 3 grid squares above the x-axis. The y-coordinate of point D is 3×5  = 15.
The coordinates of D are (20,15)

In the diagram, rectangle PQRS represents a parking lot of a supermarket. The side length of each grid square is 4 meters. Use the diagram to answer questions 16 to 18.

Question 16.
Give the coordinates of points P, Q, R, and S.
Answer:
Point P:
The first coordinate represents distance along x axis which is 16 units and the second coordinate represents distance along y axis which is 44 units.
So, the point P will be (16,44).
Point Q:
The first coordinate represents distance along x axis which is 16 units and the second coordinate represents distance along y axis which is 4 units.
So, the point Q will be (16,4).
Point R:
The first coordinate represents distance along x axis which is 36 units and the second coordinate represents distance along y axis which is 4 units.
So, the point QRwill be (36,4).
Point S:
The first coordinate represents distance along x axis which is 36 units and the second coordinate represents distance along y axis which is 44 units.
So, the point S will be (36,44).

Question 17.
The manager of the supermarket wants to build a concrete wall around the parking lot. What is the perimeter of the wall?
Answer:
Here, 1 grid square represents 4 meters.
Length of PQ = 44-4 = 40 m
Length of QR = 36-16 = 20m
Since PQRS is a rectangle, opposite lengths will be equal.
Perimeter of wall will be = 40+20+40+20 = 120m

Question 18.
The entrance of the supermarket is at point T. It lies on \(\overline{P Q}\), and point T is 8 meters from point P. Give the coordinates of point T.
Answer:
Point T lies on \(\overline{P Q}\), and it is 8 meters from point P.
Thus, the entrance will be on PQ.
8 m means 8÷4=2 square grid
Move 2 square grid below point P on \(\overline{P Q}\)
Thus, point T will be (16,36)

Math in Focus Course 1B Practice 9.2 Answer Key

Use graph paper. Plot each pair of points on a coordinate plane. Connect the points to form a line segment and find its length.

Question 1.
A(5, 0) and B(8, 0)
Answer:
Point A(5, 0)
Here, the x-coordinate is 5 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is positive, move 5 units along the positive x-axis and mark it. Thus, the required point(5,0) is marked.
Point B(8,0)
Here, the x-coordinate is 8 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is positive, move 8 units along the positive x-axis and mark it. Thus, the required point(8,0) is marked.

After joining the A and B, a line segment of 3 units is formed.

Question 2.
C(-3, 4)and D(3, 4)
Answer:
Plot C(-3, 4):
Here, the x-coordinate is -3 and the y-coordinate is 4. Start at the Origin. As the x coordinate is negative, move 3 units along the negative x-axis and 4 units to the above along positive y -axis and mark it.
Thus, the required point (-3, 4) is marked.
Plot D(3, 4):
Here, the x-coordinate is 3 and the y-coordinate is 4. Start at the Origin. As the x coordinate is positive, move 3 units along the positive x-axis and 4 units to the above along positive y -axis and mark it.
Thus, the required point (3, 4) is marked.

After joining the C and D, a line segment of 6 units is formed.

Question 3.
E (-5, -2) and F (8, -2)
Answer:
Plot point E(-5,-2)
Here, the x-coordinate is -5 and the y-coordinate is -2. Start at the Origin. As the x coordinate is negative, move 5 units along the negative x-axis and 2 units along negative y -axis and mark it.
Thus, the required point (-5, -2) is marked.
Plot point F(8, -2)
Here, the x-coordinate is 8 and the y-coordinate is -2. Start at the Origin. As the x coordinate is positive, move 8 units along the positive x-axis and 2 units along negative y -axis and mark it.
Thus, the required point (8, -2) is marked.

After joining the E and F, a line segment of 13 units is formed.

Question 4.
G(0, -5) and H(0, 2)
Answer:
Plot point G(0, -5)
Here, the x-coordinate is 0 and the y-coordinate is -5. As the x-coordinate is zero and y coordinate is negative, move 5 units along negative y-axis and mark it. Thus, the required point(0,-5) is marked.
Plot point H(0, 2)
Here, the x-coordinate is 0 and the y-coordinate is 2. As the x-coordinate is zero and y coordinate is positive, move 2 units along positive y-axis and mark it. Thus, the required point(0,2) is marked.

After joining the G and H, a line segment of 7 units is formed.

Question 5.
J(-5, -3) and K(-5, -8)
Answer:
Plot point J(-5, -3)
Here, the x-coordinate is -5 and the y-coordinate is -3. Start at the Origin. As the x coordinate is negative, move 5 units along the negative x-axis and 3 units along negative y -axis and mark it.
Thus, the required point (-5, -3) is marked.
Plot point K(-5,-8):
Here, the x-coordinate is -5 and the y-coordinate is -8. Start at the Origin. As the x coordinate is negative, move 5 units along the negative x-axis and 8 units along negative y -axis and mark it.
Thus, the required point (-5, -8) is marked.

After joining the J and K, a line segment of 5 units is formed.

Question 6.
M(1, 7) and N(1, -8)
Answer:
Plot point M(1, 7)
Here, the x-coordinate is 1 and the y-coordinate is 7. Start at the Origin. As the x coordinate is positive, move 1 unit along the positive x-axis and 7 units along positive y -axis and mark it.
Thus, the required point (1, 7) is marked.
Plot point N(1, -8)
Here, the x-coordinate is 1 and the y-coordinate is -8. Start at the Origin. As the x coordinate is positive, move 1 unit along the positive x-axis and 8 units along negative y -axis and mark it.
Thus, the required point (1, -8) is marked.

After joining the M and N, a line segment of 15 units is formed.

Use graph paper. Find the coordinates.

Question 7.
Rectangle PQRS is plotted on a coordinate plane. The coordinates of P are (-1, -3) and the coordinates of Q are (-1, 2). Each unit on the coordinate plane represents 1 centimeter, and the perimeter of rectangle PQRS is 20 centimeters. Find the coordinates of points R and S given these conditions:
a) Points R and S are to the left of points P and Q.
Answer:
Given that PQRS is a rectangle. The coordinates of P are (-1, -3) and the coordinates of Q are (-1, 2).
The perimeter of rectangle PQRS is 20 centimeters.
Length of PQ is 5 units or 5 cm which is equal to length of RS. Length of RQ is equal to length of SP.
Perimeter of rectangle = Sum of four sides
20 = PQ+QR+RS+SP
20 = 5+QR+5+SP
20 = 5+SP+5+SP
20 = 10+2×SP
20-10 = 10+2×SP-10
10 = 2×SP
10÷2 = (2×SP)÷2
5 = SP
Length of SP equals to 5 units or 5 cm.
Start from point P and move 5 square grids towards left and point R. Similarly, point R from Q and then join the points to form the rectangle.

b) Points R and S are to the right of points P and Q. The coordinates of P are (-1, -3) and the coordinates of Q are (-1, 2).
Answer:
We know the measurements of the PQRS, where length of all sides are equal.
Start from point P and move 5 square grids towards right and point R. Similarly, point R1 from Q and then join the points to form the rectangle.

Question 8.
Rectangle ABCD is plotted on a coordinate plane. The coordinates of A are (2, 3) and the coordinates of B are (-2, 3). Each unit on the coordinate plane represents 3 centimeters, and the perimeter of rectangle ABCD is 48 centimeters. Find the coordinates of points C and D given these conditions:
a) Points C and D are below points A and B.
Answer:
Given that ABCD is a rectangle. The coordinates of A are (2, 3) and the coordinates of B are (-2, 3).
The perimeter of rectangle ABCD is 48 centimeters.
Length of AB is 4 units or 4×3 = 12cm which is equal to length of CD. Length of AC is equal to length of BD.
Perimeter of rectangle = Sum of four sides
48 = AB+BD+DC+CA
48 = 12+3BD+12+3BD (BD will be equal to 3BD as length of each unit on the coordinate plane represents 3 cm)
48 = 24+2×6×BD
48-24 = 24+12×BD-24
24 = 12×BD
24÷12 = (12×BD)÷12
2= BD
Length of BD equals to 2 units or 2×3=6 cm.
Start from point A and move 2 square grids below and point C. Similarly, point D from B and then join the points to form the rectangle.

b) Points C and D are above points A and B.
Answer:
We know the measurements of the ABCD, where length of opposite sides are equal.
Start from point A and move 2 square grids upwards and point C1. Similarly, point D1 from B and then join the points to form the rectangle.

Question 9.
Rectangle PQRS is plotted on a coordinate plane. The coordinates of P are (-1, 4) and the coordinates of Q are (-1, -4). Each unit on the coordinate plane represents 1 centimeter, and the area of rectangle PQRS is 64 square centimeters. Find the coordinates of points R and S given these conditions:

a) Points R and S are to the left of points P and Q.
Answer:
Given that PQRS is a rectangle. The coordinates of P are (-1, 4) and the coordinates of Q are (-1, -4).
The area of rectangle PQRS is 64 sq.cm
Length of PQ is 8 units or 8 cm which is equal to length of RS. Length of QS is equal to length of RP.
Area of rectangle = length×width
64 = PQ×QR
64 = 8×QR
64÷8 = (8×QR)÷8
8 = QR
Length of QR equals to 8 units or 8 cm.
Start from point P and move 8 square grids to the left and mark point R. Similarly, mark point S from Q and then join the points to form the rectangle.

b) Points R and S are to the right of points P and Q.
Answer:
We know the measurements of the PQRS, where length of all sides are equal.
Start from point P and move 8 square grids towards right and point it as R. Similarly, point S from Q and then join the points to form the rectangle.

In the diagram, rectangle ABCD represents a shopping plaza. The side length of each grid square Is 10 meters. Use the diagram to answer questions 10 to 14.

Question 10.
Give the coordinates of points A, B, C, and D.
Answer:
Point A:
The first coordinate represents distance along x axis which is 60 units and the second coordinate represents distance along y axis which is 90 units.
So, the point A will be (-60,90).
Point B:
The first coordinate represents distance along x axis which is 60 units and the second coordinate represents distance along y axis which is 50 units.
So, the point A will be (-60,50).
Point C:
The first coordinate represents distance along x axis which is 50 units and the second coordinate represents distance along y axis which is 50 units.
So, the point A will be (50,-50).
Point D:
The first coordinate represents distance along x axis which is 50 units and the second coordinate represents distance along y axis which is 90 units.
So, the point A will be (50,90).

Question 11.
Write down the shortest distance of points A, B, C, and D from the y-axis.
Answer:
The shortest distance of points A from the y-axis is 60 units.
The shortest distance of points B from the y-axis is 60 units.
The shortest distance of points C from the y-axis is 50 units.
The shortest distance of points D from the y-axis is 50 units.

Question 12.
Write down the shortest distance of points A, B, C, and D from the x-axis.
Answer:
The shortest distance of points A from the x-axis is 90 units.
The shortest distance of points B from the x-axis is 50 units.
The shortest distance of points C from the x-axis is 50 units.
The shortest distance of points D from the x-axis is 90 units.

Question 13.
Find the area and perimeter of the shopping plaza.
Answer:
The side length of each grid square Is 10 meters.
Length of AB is 140 square grid, 140×10=1400 m
Length of BC is 110 square grid, 110×10=1100 m
Thus, the length of the shopping plaza is 1400m and width is 1100m
Area of the shopping plaza = length × width
= 1400×1100
= 1540000 sq.m
Perimeter of the shopping plaza = 2×(length+width)
= 2×(1400+1100)
= 2×2500
= 5000 m

Question 14.
A man at the shopping plaza is standing 50 meters from \(\overline{\mathrm{AD}}\), and 40 meters from \(\overline{\mathrm{DC}}\).
a) Find the coordinates of the point representing the man’s location.
Answer:
Given that:
A man at the shopping plaza is standing 50 meters from \(\overline{\mathrm{AD}}\)
AD – 50 = 90-50 = 40
and 40 meters from \(\overline{\mathrm{DC}}\).
DC – 40 = 50-40 = 10

b) Find the shortest distance in meters from the man’s location to the side \(\overline{\mathrm{BC}}\).
Answer:
Count the square grids from the man’s location to the side BC.
There are 9 square grids, 9×10 = 90m
Thus, 90m is the shortest distance in meters from the man’s location to the side BC

In the diagram, triangle PQR represents a triangular garden. The side length of each grid square Is 5 meters. Use the diagram to answer questions 15 to 19.

Question 15.
Use graph paper. A rectangular region ABCR in the garden is to be fenced in. Point A lies on \(\overline{P R}\), and is 35 meters away from point P. Point C lies below \(\overline{P R}\), and is 20 meters away from point R. Plot and label points A, B, and C on the coordinate plane. Write the coordinates of points A, B, and C.
Answer:
A rectangular region ABCR in the garden is to be fenced in. Point A lies on \(\overline{P R}\), and is 35 meters away from point P.
The side length of each grid square Is 5 meters.
Since point A is 35m away from P, 35÷5=7 square grids.
Move 7 units away from point P and mark it as A.
Thus, the point A will be (5,15)

Point C lies below \(\overline{P R}\), and is 20 meters away from point R.
Since point C is 20m away from point R, 20÷5=4 square grids.
Move 4 units to the down from point R and mark it as C.
Thus, the point C will be (30, -5)

Since ABCR is a rectangular region, the opposite sides are equal.
Start from point C and move 5 units towards left, and mark the point as B.
Thus, the point B will be (5,-5)

Quadrant 16.
If PQ is 75 meters, what is the perimeter of the triangular garden in meters?
Answer:
Given that PQ is 75 m.
Length of PR = 12 square grids.
Here, one square grid equals 5m. So, 12 square grids = 12×5 = 60m
Length of RQ = 9 square grids.
So, 9 square grids = 9×5 = 45m
Perimeter of the triangular garden = 75+60+45 = 180m

Question 17.
Find the area of the enclosed region ABCR in square meters.
Answer:
Length of AR = 5 square grids, 5×5=25m
Width of AB = 4 square grids, 4×5=20m
Area of rectangle = length×width
=25×20
=500
Area of the enclosed region ABCR will be 500 sq.m

Question 18.
Find the perimeter of the enclosed region ABCR in meters.
Answer:
Length of AR = 25m
Width of AB = 20m
Perimeter = AR+RC+CB+BA
= 25+20+25+20
= 90 m

Question 19.
If PQ is 75 meters, what is the perimeter of the garden that is not enclosed?
Answer:
Given PQ is 75m.
Length of AP = 7 square grids,7×5=35 m
Length of AB = 4 square grids,4×5=20 m
Length of BC = 5 square grids,5×5=25 m
Length of CQ = 5 square grids,5×5=25 m
Perimeter of the garden that is not enclosed = 75+35+20+25+25 = 180 m

The diagram shows the outline of a park. The side length of each grid square is 10 meters. Use the diagram to answer questions 20 to 22.

Question 20.
Find the area of the park in square meters.
Answer:
Let us assume AHGF as a rectangle, AH be the length and HG be the width.
Length of AH is 80m and length of HG is 180m.
Area of rectangle = length×width
= 80×180
= 14400 sq.m
Area of AHGFA will be 14400 sq.m
Area of rectangle BCDE:
Length of BC = 50m
Length of CD = 120m
Area of BCDE will be 50×120 = 6000 sq.m
Area of the park can be calculated by subtracting the area of rectangle BCDE from the rectangle AHGF
= 14400-6000
= 8400 sq.m

Question 21.
Brandon starts at point A and walks all the way around the perimeter of the park. If he walks at 1.5 meters per second, about how many seconds pass before he returns to point A? Round your answer to the nearest second.
Answer:
The side length of each grid square Is 10 m.
Length of HA: 8 square grids = 8×10 = 80 m
Length of AB: 3 square grids = 3×10 = 30 m
Length of BC: 5 square grids = 5×10 = 50 m
Length of CD: 12 square grids = 12×10 = 120 m
Length of DE: 5 square grids = 5×10 = 50 m
Length of EF: 3 square grids = 3×10 = 30 m
Length of FG: 8 square grids = 8×10 = 80 m
Length of GH: 8 square grids = 18×10 = 180 m
Perimeter of the park will be 80+30+50+120+50+30+80+180 = 620m
He walks at 1.5 meters per second.
620÷1.5 = 413.3
Round off for 413.3 equals to 413 sec

Question 22.
A picnic table in the park is 20 meters from \(\overline{B C}\), and is closer to point B than it is to point C. Write down two possible pairs of coordinates for the location of the picnic table.
Answer:
Given that a picnic table in the park is 20 m from \(\overline{B C}\), and is closer to point B than it is to point C.
20m away from B will be 2 square grids towards up which can be (-30,80) and (-40,80)

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 Lesson 9.3 Real-World Problems: Graphing to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Lesson 9.3 Answer Key Real-World Problems: Graphing

Math in Focus Grade 6 Chapter 9 Lesson 9.3 Guided Practice Answer Key

Graph an equation on a coordinate plane.

Angela is driving to the Raccoon River. The distance traveled, d miles, after t hours, is given by d = 40t. Graph the relationship between d and t. Use 2 units on the horizontal axis to represent 1 hour and 2 units on the vertical axis to represent 20 miles.

a) What type of graph is it?
It is a straight line graph.This is also called a linear graph.

b) How far did Angela drive in 3.5 hours?
From the graph, Angela drove 140 miles.

c) What is the speed at which Angela is driving?

= 40 mi/h
Angela is driving at 40 miles per hour.

d) Angela has driven for 4 hours. If she drives for another hour at this • constant speed, how far will she drive in all?
Distance = speed × time
= 40 × 5
= 200 mi
She will drive 200 miles.

e) If Angela wants to drive at least 120 miles, how many hours will she need to drive? Express your answer in the form of an inequality in terms of t, where t stands for the number of hours.
t ≥ 3

f) Name the dependent and independent variables.
d is the dependent variable, and t is the independent variable.

Use graph paper. Solve.

Question 1.
A car uses 1 gallon of gas for every 20 miles traveled. The amount of gas left in the gas tank, x gallons, after traveling y miles is given by y = 240 – 20x. Copy and complete the table. Graph the relationship between x and y.
Use 1 unit on the horizontal axis to represent 1 gallon and 1 unit on the vertical axis to represent 20 miles.
a)

Answer:
Given that, A car uses 1 gallon of gas for every 20 miles traveled. The amount of gas left in the gas tank, x gallons, after traveling y miles is given by y = 240 – 20x.
If y equals to 80 miles then x will be equal to
y = 240 – 20x
80 = 240-20x
80+20x = 240-20x+20x
80+20x = 240
80+20x-80 = 240-80
20x = 160
20x÷20 = 160÷20
x = 8
If y equals to 160 miles then x will be equal to
y = 240 – 20x
160 = 240-20x
160+20x = 240-20x+20x
160+20x = 240
160+20x-160 = 240-160
20x = 80
20x÷20 = 80÷20
x = 4

b) What type of graph is it? It is a graph.
Answer:
It is a straight line graph.This is also called a linear graph.

c) How many gallons of gas will be left in the tank after the car has traveled 60 miles?
From the graph, there will be gallons of gas left,
Answer:
From the graph, there will be 8 gallons of gas left in the tank after the car has traveled 60 miles.

d) How many gallons of gas will be left in the tank after the car has traveled 100 miles?
From the graph, there will be gallons of gas left.
Answer:
From the graph, there will be 4 gallons of gas left in the tank after the car has traveled 100 miles.

e) After the car has traveled 160 miles, how much farther can the car travel before it runs out of gas?
After 160 miles, gallons of gas were left.
The car uses 1 gallon for every miles traveled.
Distance = number of gallons × mileage
= ×
= mi
The car can travel another miles.
Answer:
After 160 miles, 4 gallons of gas were left.
The car uses 1 gallon for every 20 miles traveled.
Distance = 4 × mileage
160 =  4×mileage
160÷4 = (4×mileage)÷4
40 = mileage
The car can travel another 40 miles.

f) If the car travels more than 40 miles, how much gas is left in the tank?
Express your answer in the form of an inequality in terms of x, where x stands for the amount of gas left in the gas tank.
If the distance traveled is more than 40 miles, then .

Answer:
If the car travels more than 40 miles, then x ≥10, where x stands for the amount of gas left in the gas tank.

g) Name the dependent and independent variables.
Answer:
From the above analysis, we can observe that,
x is independent and y is dependent

Question 2.
Sarah plants a seed. After t weeks, the height of the plant, h centimeters, is given by h = 2t. Copy and complete the table. Graph the relationship between t and h. Use 1 unit on the horizontal axis to represent 1 week and 1 unit on the vertical axis to represent 2 centimeters.
a)

Answer:
Given that Sarah plants a seed. After t weeks, the height of the plant, h centimeters, is given by h = 2t
To calculate the height, we can use this relation,  h = 2t
If t=0, h = 2t = 2×0 = 0 cm
If t=4, h = 2t = 2×4 = 8 cm
If t=5, h = 2t = 2×5 = 10 cm

b) What type of graph is it?
Answer:
By joining all the points it forms a straight line graph.
This is also called a linear graph.

c) What is the height of the plant after 3 weeks?
Answer:
From the graph, we can observe that the height of the plant will be 6 cm after 3 weeks.

d) What is the height of the plant after 5 weeks?
Answer:
From the graph, we can observe that the height of the plant will be 10 cm after 5 weeks.

e) What is the height of the plant if less than 4 weeks have passed? Express your answer in the form of an inequality in terms of h, where h stands for the height of the plant in centimeters.
Answer:
The height of the plant if less than 4 weeks have passed will be 8cm.
Therefore, the height in the form of an inequality in terms of h will be h ≥ 8, where h stands for the height of the plant in cm.

f) Name the dependent and independent variables.
Answer:
From the above observation, we can say the dependent variable is h
and the independent variable is t.

Math in Focus Course 1B Practice 9.3 Answer Key

Use graph paper. Solve.

Question 1.
A cyclist took part in a competition. The distance traveled, d meters, after t minutes, is given by d = 700t. Graph the relationship between t and d. Use 2 units on the horizontal axis to represent 1 minute and 1 unit on the vertical axis to represent 350 meters.

a) What type of graph is it?
Answer:
By joining all the points it forms a straight line graph.
This is also called a linear graph.

b) What is the distance traveled in 2.5 minutes?
Answer:
Given that the distance traveled, d meters, after t minutes, is given by d = 700t.
d = 700×2.5
= 1750 m
Thus, 1750 m is the distance traveled in 2.5 min.

c) What is the distance traveled in 3.5 minutes?
Answer:
Given that the distance traveled, d meters, after t minutes, is given by d = 700t.
d = 700×3.5
= 2450 m
Thus, 2450 m is the distance traveled in 3.5 min.

d) What is the average speed of the cyclist?
Answer:
The average speed of the cyclist can be calculated by using the below formula,
Average speed = \(\frac{total distance}{total elapsed time}\)
Total distance = 0+700+1400+2100+2800 = 7000 m
Total time = 0+1+2+3+4 = 10 min
= \(\frac{7000}{10}\)
= 700
The average speed of the cyclist is 700 metre per min.

e) Assuming that the cyclist travels at a constant speed throughout the competition, what distance will he travel in 7 minutes?
Answer:
Let us assume that the cyclist travels at a constant speed of 700m in a min.
So, the distance he travel in 7 mins will be 7×700=4900 m.

f) If the cyclist needs to cycle for at least 2.1 kilometers, how many minutes will he need to cycle? Express your answer in the form of an inequality in terms of t, where t stands for the number of minutes.
Answer:
The cyclist needs to cycle for at least 2.1 kilometers.
1 kilometer equals to 1000 m.
2.1 km will be equal to 2.1×1000 = 2100 m
Given that the distance traveled, d meters, after t minutes, is given by d = 700t.
2100 = 700t
2100÷700 =(700×t)÷700
300 = t
The number of minutes the cyclist will need to cycle 2.1 km will be 300 min.

g) Name the dependent and independent variables.
Answer:
From the above observation, we can say the dependent variable is d
and the independent variable is t.

Question 2.
A bus uses 1 gallon of diesel for every 7 miles traveled. The amount of diesel left in the gas tank, p gallons, after traveling q miles, is given by q = 112 – 7p. Copy and complete the table. Graph the relationship between p and q. Use 1 unit on the horizontal axis to represent 1 gallon and 1 unit on the vertical axis to represent 7 miles.

a)

Answer:
Given that a bus uses 1 gallon of diesel for every 7 miles traveled. The amount of diesel left in the gas tank, p gallons, after traveling q miles, is given by q = 112 – 7p.
If p is 14, then q will be 112 – 7×14 = 112-98 = 14
If p is 8, then q will be 112 – 7×8 = 112-56 = 56

b) How many gallons of diesel were left after the bus has traveled 49 miles?
Answer:
The amount of diesel left in the gas tank, p gallons, after traveling q miles, is given by q = 112 – 7p.
Here q is 49 miles.
49 = 112 – 7p
49 +7p = 112 – 7p +7p
49 +7p = 112
49 +7p – 49 = 112 – 49
7p = 63
7p÷7 = 63÷7
p=9
9 gallons of diesel were left after the bus has traveled 49 miles

c) After the bus has traveled for 56 miles, how much farther can the bus travel before it runs out of diesel?
Answer:
For 56 miles, 8 gallons of diesel will be used
Here p is 8
8 = 112 – 7p
8+7p = 112 – 7p + 7p
8+7p = 112
8+7p-8 = 112-8
7p = 104
7p÷7 = 104÷7
p = 14.8
After the bus has traveled for 56 miles, it will travel 0.8 m before it runs out of diesel.

d) If the bus travels more than 28 miles, how much diesel is left? Express your answer in the form of an inequality in terms of p, where p stands for the amount of diesel left.
Answer:
If the bus travels more than 28 miles, 12 gallons will be utilised.
It can be expressed as p ≥ 12, where p stands for the amount of diesel left.

Question 3.
A kettle of water is heated and the temperature of the water, j°C, after k minutes, is given by j = 5k + 30. Copy and complete the table. Graph the relationship between k and j. Use 1 unit on the horizontal axis to represent 1 minute and 1 unit on the vertical axis to represent 5°C.

a)

Answer:
Given that a kettle of water is heated and the temperature of the water, j°C, after k minutes, is given by j = 5k + 30.
If k = 0 then j will be
j = 5k + 30
j = 5×0 + 30
j = 30
If k = 4 then j will be
j = 5k + 30
j = 5×4 + 30
j = 20+30
j = 50
If j=70
70 = 5k + 30
70-30 = 5k+30-30
40 = 5k
40÷5 = 5k÷5
8 = k

b) What is the temperature of the water after 5 minutes?
Answer:
Given that a kettle of water is heated and the temperature of the water, j°C, after k minutes, is given by j = 5k + 30.
If k = 5 then j will be
j = 5k + 30
j = 5×5 + 30
j = 25+30
j = 55°C

c) What is the average rate of the heating?
Answer:
The average rate of the heating can be calculated by using the below formula,
Average rate = \(\frac{total temp}{total elapsed time}\)
Total temp = 30+40+50+60+70 = 250
Total time = 0+2+4+6+8 = 20 min
= \(\frac{250}{20}\)
= 12.5 rate of heat per min

d) Assuming the temperature of the water rises at a constant rate, what is the temperature of the water after 10 minutes?
Answer:
Given that a kettle of water is heated and the temperature of the water, j°C, after k minutes, is given by j = 5k + 30.
Here, k=10 min
j = 5k + 30
= 5×10 + 30
= 50+30
= 80°C
80°C is the temperature of the water after 10 minutes.

e) The kettle of water needs to be heated till the water boils. For how many minutes does the kettle need to be heated? Express your answer in terms of k, where k stands for the number of minutes. (Hint: Water boils at 100°C.)
Answer:
Given that the kettle of water needs to be heated till the water boils and it will boils at 100°C
j = 5k + 30
100 = 5k + 30
100-30 = 5k + 30 – 30
70 = 5k
70÷5 = 5k÷5
k = 14 min
The kettle needs to be heated for 14 mins.

Brain @ Work

Use graph paper. For each exercise, plot the points on a coordinate plane.

Question 1.
A (-5, 1), B (-3, -3), C (3, 1), and D (-1, 5)
Answer:
Point A (-5, 1):
As the x coordinate is negative, start from origin and move 5 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 1 point towards positive y axis.
Thus, the required point A is plotted.
Point B (-3, -3):
As the x coordinate is negative, start from origin and move 3 points towards negative x axis.
As the y-coordinate is negative, start from origin and move 3 point towards negative y axis.
Thus, the required point B is plotted.
Point C (3,1):
As the x coordinate is positive, start from origin and move 3 points towards positive x axis.
As the y-coordinate is positive, start from origin and move 1 point towards positive y axis.
Thus, the required point C is plotted.
Point D(-1, 5):
As the x coordinate is negative, start from origin and move 1 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 5 point towards positive y axis.
Thus, the required point D is plotted.

Question 2.
J (4, 2), K (-2, 4), L (-4, 0), and M (0, -2)
Answer:
Point J (4, 2):
As the x coordinate is positive, start from origin and move 4 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 2 point towards positive y axis.
Thus, the required point J is plotted.
Point K (-2, 4):
As the x coordinate is negative, start from origin and move 2 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 4 point towards positive y axis.
Thus, the required point K is plotted.
Point L (-4, 0):
As the x coordinate is negative, start from origin and move 4 points towards negative x axis.
As the y-coordinate is zero, mark at origin.
Thus, the required point L is plotted.
Point M (0, -2):
As the x coordinate is zero, mark at origin.
As the y-coordinate is negative, start from origin and move 2 point towards negative y axis.
Thus, the required point M is plotted.

Question 3.
S (-1, 3), T (-3, -1), U (1, -1), and V(5, 3)
Answer:
Point S (-1, 3):
As the x coordinate is negative, start from origin and move 1 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 3 point towards positive y axis.
Thus, the required point A is plotted.
Point T (-3, -1):
As the x coordinate is negative, start from origin and move 3 points towards negative x axis.
As the y-coordinate is negative, start from origin and move 1 point towards negative y axis.
Thus, the required point B is plotted.
Point U (1,-1):
As the x coordinate is positive, start from origin and move 1 points towards positive x axis.
As the y-coordinate is negative, start from origin and move 1 point towards negative y axis.
Thus, the required point C is plotted.
Point V(5, 3):
As the x coordinate is positive , start from origin and move 5 points towards positive x axis.
As the y-coordinate is positive, start from origin and move 3 point towards positive y axis.
Thus, the required point D is plotted.

Question 4.
In questions 1 to 3, what is the figure formed?
Answer:
Join the plotted points A,B,C and D.
It will form a four sided figure or a quadrilateral.

Join the plotted points K,L,M and N.
It will form a four sided figure or a quadrilateral.

Join the plotted points S,T,U and V.
It will form a four sided figure or a quadrilateral.

Question 5.
a) For each figure in questions to mark the middle of each side and connect the points in order.
Answer:
Mark the mid-points of AB, BC, CD and DA line segment.

Mark the mid-points of JK, KL, LM and MJ line segment.

Mark the mid-points of SV, VU, UT and TS line segment.

b) What are the figures formed? Explain your answers.
Answer:
By joining the midpoints on ABCD:
Join the points to form a quadrilateral.
A parallelogram is formed as the opposite sides are parallel to each other.
By joining the midpoints of JLKM:
Join the points to form a quadrilateral.
A rectangle is formed as the opposite sides are equal to each other.
By joining the midpoints of STUV:
Join the points to form a quadrilateral.
A rectangle is formed as the opposite sides are equal to each other.