Math in Focus

Go through the Math in Focus Grade 8 Workbook Answer Key Cumulative Review Chapters 1-2 to finish your assignments.

Math in Focus Grade 8 Course 3 A Cumulative Review Chapters 1-2 Answer Key

Concepts and Skills

Write the prime factorization of each number in exponential notation. (Lesson 1.1)

Question 1.
16,807
Answer:
Prime factorization of 16,807 is 7×7×7×7×7×7
The number in the exponential notation is 7⁶.

Question 2.
25,920
Answer:
Prime factorization of 25,920 is 2×2×2×2×2×2×3×3×3×3×5.
The number in the exponential form is 2⁶ × 3⁴ × 5.

Simplify each expression. Write your answer in exponential notation. (Lessons 1.2, 1.3)

Question 3.
\(\frac{\left[\left(\frac{3}{5}\right) \cdot\left(\frac{3}{5}\right)^{3}\right]^{4}}{\left[\left(\frac{3}{5}\right)^{2}\right]^{2}}\)
Answer:
((3/5) • (3/5)³)⁴/((3/5)²)²
⅗ • ( ⅗)³)⁴ • ((⅗)²)²
⅗ • (⅗)¹² • (⅗)⁴
Bases are equal powers should be added.
⅗ • (⅗)¹⁶
(⅗)¹⁷

Question 4.
(a⁶ • a⁷)³ ÷ (4a³)²
Answer:
Given (a6 • a7)3 ÷ (4a3)2
(a⁶)³ • (a⁷)³ ÷ (4a)⁶
a¹⁸ • a²¹ ÷ 4a⁶
Bases are equal powers should be added.
a³⁹/4a⁶
¼ × a³⁹/a⁶
¼ × a³³

Simplify each expression. Write your answer using a positive exponent. (Lessons 1.2, 1.3, 1.4, 1.5)

Question 5.
\(\frac{6^{3} \cdot 15^{3}}{\left(7^{0}\right)^{3}}\)
Answer:
6³ • 15³/(7⁰)³
= 90³/7⁰
= 90³/1
= 90³

Question 6.
\(\frac{2^{8} \cdot(-3)^{8} \cdot 3^{0}}{5^{-8}}\)
Answer:
Given 2⁸ • (-3)⁸ • 3⁰/5^-8
= -6⁸ • 1/5^-8
= -6⁸/5^-8

Question 7.
[12² • 3²]³ ÷ 3⁶
Answer:
Given [122 • 32]3 ÷ 36
12⁶ • 3⁶ ÷ 3⁶
36⁶/3⁶
12⁶/1⁶

Question 8.
(16⁷ ÷ 16⁴) • \(\frac{\left(5^{0}\right)^{3}}{2^{3} \cdot 4^{3}}\)
Answer:
Given (167 ÷ 164) • (5⁰)³/2³ • 4³
16⁷/16⁴ • 5⁰/2³ • 4³
16³ • 1/2³ • 4³
64³ • (½)³

Question 9.
8^-2 • \(\frac{3^{0} \cdot 8^{-3}}{4^{-5}}\) .
Answer:
Given 8-2 • 3⁰ • 8-3/4^-5
8-2 • 1 • 8-3/4^-5
8-2 • 8-3/4^-5
8-5/4^-5
2-5/1^-5

Question 10.
6^-4 • (5⁰)^-4 • (\(\frac{1}{3}\))^-4 ÷ 3^-4
Answer:
6^-4 (5⁰)^-4 ∙ (\(\frac{1}{3}\))^-4 ÷ 3^-4 We are given the expression:
6^-4 ∙ (5⁰)^-4 ∙ (\(\frac{1}{3}\))^-4 ÷ 3^-4 Simplify:
= 6^-4 ∙ \(\frac{1}{3^{-4}} \cdot \frac{1}{3^{-4}}\)
= \(\frac{6^{-4}}{3^{-4}} \cdot \frac{1}{\frac{1}{3^{4}}}\)
= \(\left(\frac{6}{3}\right)^{-4} \cdot 3^{4}\)
= 2^-4 ∙ 3⁴
= \(\frac{3^{4}}{2^{4}}\)
= \(\left(\frac{3}{2}\right)^{4}\)

Evaluate the square roots of each number. Round your answer to the nearest tenth when you can. (Lesson 1.6)

Question 11.
576
Answer:
576 = (24)²
Square of 576 is 24
24 rounded to the nearest tenth is 20

Question 12.
1,003.4
Answer:
1,003.4 = (31.6764897046)²
Square of 1,003.4 is 31.6764897046
31.6764897046 rounded to the nearest tenth is 30

Evaluate the cube root of each number. Round your answer to the nearest tenth when you can. (Lesson 1.6)

Question 13.
\(\frac{27}{216}\)
Answer:
27/216
0.125
0.125 = (0.5)³
Cubic root of 27/216 is 0.5

Question 14.
-629.5
Answer:
-629.5 = (-8.57035039349)³
Cubic root of -629.5 is -8.57035039349

Evaluate each expression and write your answer in scientific notation. Identify the greater number. (Lessons 2.1, 2.2, 2.3)

Question 15.
3.27 • 10¹¹ + 3.13 • 10¹¹ and 9.28 • 10¹¹ – 4.15 • 10¹¹
Answer:
3.27 • 1011 + 3.13 • 1011
3.27 + 3.13 • 10¹¹
6.4 • 10¹¹
And 9.28 • 1011 – 4.15 • 1011
9.28 – 4.15 • 10¹¹
5.13 • 10¹¹
6.4 • 10¹¹ is the greater number

Question 16.
9.1 • 10^-5 – 8.2 • 10^-6 and 1.2 • 10^-6 – 5.5 • 10^-7
Answer:
9.1 • 10^-5 – 8.2 • 10^-6
9.1 • 10^-5 – 0.82 • 10^-5
9.1 – 0.82 • 10^-5
8.28 • 10^-5
And 1.2 • 10-6 – 5.5 • 10^-7
1.2 • 10-6 – 0.55 • 10^-6
1.2 – 0.55 • 10^-6
0.65 • 10^-6
8.28 • 10^-5 is the greater number.

Question 17.
8.4 • 10⁵ • 2 • 10⁵ and 3.2 • 10^-7 • 2 • 10^-5
Answer:
8.4 • 10⁵ • 2 • 10⁵
8.4 • 10⁵ • 2 • 10⁵
16.8 • 10¹⁰
And 3.2 • 10^-7 • 2 • 10^-5
6.4 • 10^-12
16.8 • 10¹⁰ is the greater number.

Question 18.
9.1 • 10³ ÷ (7 • 10⁵) and 7.2 • 10^-4 ÷ (1.2 • 10^-4)
Answer:
9.1 • 10³ ÷ (7 • 10⁵)
9.1/7 • 10³/10⁵
1.3 • 1/10²
1.3 • 10-2
And 7.2 • 10^-4 ÷ (1.2 • 10^-4)
7.2/1.2 • 10^-4/10^-4
6 • 1
6 = 0.6 • 10¹
0.6 • 10¹ is the greater number.

Write each measurement in the appropriate unit in prefix form. (Lesson 2.2)

Question 19.
0.000020 meter
Answer:
0.02 • 10³ meter
0.02 millimeter

Question 20.
0.070 gram
Answer:
Given 0.070 gram
0.07 •10^-3 gram

Question 21.
35,000,000 bytes
Answer:
Given 35000000 bytes
35 • 10⁶ bytes
0.035 kilobytes

Question 22.
42,000 volts
Answer:
Given 42,000 volts
42 • 10³ volts
42 millivolts.

Problem Solving

Solve. Show your work.

Question 23.
The total surface area of a cube is 4,704 square inches. What is the length of each side? (Chapter 1)

Answer:
From the given question
Total surface area = 4,704 in².
We know that
The total surface area of a cube is 6a²
Where a = side of the cube
6a² = 4704
a² = 4704/6
a² = 784
a = √784
a = 28 in

Question 24.
The volume of a spherical balloon is 12.348π cubic feet. (Chapter 1)
a) Find its radius. Round to the nearest tenth.
Answer:
Given that the volume of a spherical balloon is 12.348π cubic feet.
We know that
The volume of the spherical balloon is 4/3 × πr³
4/3 × πr³ = 12.348π cubic feet.
r³ = 12.348π/4/3× π
r³ = 12.348/ 1.33
r³ = 9.261
r = 2.1
2.1 rounded to the nearest tenth is 2.1

b) Air is pumped into the balloon, so that its radius doubles every 10 seconds. Using 3.14 as an approximation for n, find its surface area after 30 seconds. Round to the nearest tenth.
Answer:
Let r be the radius of the balloon
The radius of the balloon doubles for every 10 seconds.
For 10 seconds radius = r²
For 20 seconds radius = r³
For 30 seconds radius = r⁴
Therefore r⁴ = 30
r = ± 2.340
2.340 rounded to the nearest tenth is 2.3

Question 25.
An oxygen atom has a total of 8 protons. If the mass of one proton is 1.67 • 10^-24 gram, find the total mass of the protons in the oxygen atom. Write your answer in scientific notation. Round the coefficient to 3 significant digits. (Chapters 1, 2)
Answer:
Given that the oxygen atom has a total of 8 protons
Mass of one proton = 1.67 • 10-24
Mass of 8 protons = 8
Total mass of a protons = 8 × 1.67 • 10-24
= 13.36• 10-24
13.36• 10-24 Round the coefficient to 3 significant digits is 13.4

Question 26.
The table lists the energy in Calories contained in 100 grams of fruits. (Chapter 2)

a) Calculate the total energy of the three fruits. Write your answer in scientific notation.
Answer:
Given that the energy in calories contained in 100 grams of apples = 4.9 • 10⁴
The energy in calories contained in 100 grams of oranges = 6.2 • 10⁴
The energy in calories contained in 100 grams of pear = 3.5 • 10⁴
Total energy in all the three fruits = 4.9 • 10⁴ + 6.2 • 10⁴ + 3.5 • 10⁴
= 4.9 + 6.2 + 3.5 • 10⁴
= 14.6 • 10⁴

b) Find the difference in energy contained between 100 grams of apple and 100 grams of pear.
Answer:
Given that the energy in calories contained in 100 grams of apples = 4.9 • 10⁴
The energy in calories contained in 100 grams of pear = 3.5 • 10⁴
Difference = 4.9 • 10⁴ – 3.5 • 10⁴
= 1.4 • 10⁴
The difference in the energy contained between 100 grams of apple and 100 grams of pear is 1.4 • 10⁴

c) How many times more energy does 100 grams .of orange have compared to 100 grams of apple? Round to the nearest tenth.
Answer:
Given that the energy in calories contained in 100 grams of oranges = 6.2 • 10⁴
The energy in calories contained in 100 grams of apples = 4.9 • 10⁴
6.2 • 10⁴ – 4.9 • 10⁴
6.2 – 4.9 • 10⁴
1.3 • 10⁴
The energy in calories contained in 100 grams of oranges is 1.3 • 10⁴ times more than the 100 grams of apple

Question 27.
Jim deposits $2,000 in a bank, which gives 6% interest, compounded yearly. Use the formula A = P (1 + r )ⁿ to find the amount of money in his account after 15 years. A represents the final amount of investment, P is the original principal, r is the interest rate, and n is the number of years it was invested. (Chapter 1)
Answer:
Given that
Jim deposits $2,000 in a bank
It gives 6% interest
Using the formula A = P (1 + r )n
P is the original principal
r is the interest rate
n is the number of years
A = 2000(1+6)¹⁵
A = 2000(7)¹⁵
A = 2000 • 7¹⁵

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Lesson 8.3 Rotations to score better marks in the exam.

Math in Focus Grade 7 Course 3 B Chapter 8 Lesson 8.3 Answer Key Rotations

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

Solve. Show your work.

Question 1.
P is rotated counterclockwise to P’ about the origin. State the angle of rotation.

a)

Answer:

b)

Answer:

Question 2.
The tip of a fan blade for a ceiling fan rotates from position W to W. The angle of rotation is 180°.

Answer:

A 180° rotation is also called a half turn. You do not have to include the direction of rotation for half-turn, because 180° clockwise is the same as 180° counterclockwise.

a) On a copy of the graph, mark and label the center of rotation as T.
Answer:

T = (x1+x2 / 2, y1+y2 / 2)
T = (-2 + 2/2, 1+(-3) / 2)
T = (0/2, -0/2)
T = (0, -1)
Therefore, the centre of rotation T has a coordinate (0, -1)

b) W’ is rotated 90° clockwise to W” about the center T. Label W” on the graph in a).
Answer:

Complete.

The hour hand of a clock turns through an angle from 12 noon to 4 P.M. State the following.

a) The center of rotation
Answer: The hands of the clock are placed on the center while moving from 12 to 4. While undergoing a rotation, this point where the other end of the hands does not move. Thus the center of the rotation is the center of the clock.

b) The angle and direction of rotation
Answer: The complete rotation of the hands of the whole clock is 360°. The clock is divided into 12 representing the 12 hours. By dividing 360° to 12, then for every hour the hand of the clock rotates at an angle of 30°. Thus, turning from 12 to 4, then the angle of rotation of the hand of the clock is 120°. Also, the hand moves from left to right direction then the rotation is in a clockwise direction.

Technology Activity

Explore The Properties Of Rotations With Geometry Software

Materials:

• geometry software

Step 1.
Draw a line segment AB using a geometry software program.

Step 2.
Select the Rotate function, within the Transform menu. Enter 90° to rotate \(\overline{A B}\) counterclockwise about the origin. Call the image \(\overline{A^{\prime} B^{\prime}}\).

Step 3.
Make a table of coordinates for the segment endpoints and their images. How are the lengths of the two segments related?

Step 4.
Rotate \(\overline{\mathrm{AB}}\) 90° clockwise by changing the angle of rotation to -90°. Call this image \(\overline{A^{\prime \prime} B^{\prime \prime}}\). Repeat the activity in Step 3.

Step 5.
Choose a different angle of rotation and rotate \(\overline{\mathrm{AB}}\) once again. Repeat . How do the lengths of \(\overline{\mathrm{AB}}\) and \(\overline{A^{\prime} B^{\prime}}\) compare?

Step 6.
Repeat Step 2 to Step 5 using a rectangle as the original figure. Observe how the image is related to the original rectangle.

Math Journal Which of these properties does a rotation seem to preserve: length, shape, parallel lines, or perpendicular lines? Explain.

Draw images after rotations about the origin.

In the activity, you may have observed the following:

Rotations preserve shape and size. They also preserve parallelism and perpendicularity.

You have seen how a line segment can be rotated. A rotation can be described as a transformation that rotates all points on a line or figure clockwise or counterclockwise angle about the center of rotation. You will see in the next example how to find the image of a figure after a rotation.

Copy and complete on graph paper.

Question 4.
A rotation of ∆ABC 90° clockwise about the origin, O, produces the image ∆A’B’C’. Draw and label the image ∆A’B’C’.

Answer:

Copy and complete on graph paper.

Question 5.
An animation artist draws a fish on the coordinate plane and marks the points A, B, C, and D. Then the artist rotates the fish 180° about the origin, O. Complete the table of coordinates to show the coordinates of the image points A’, B’, C’, and D’.

Answer:

Question 6.
DEFG is rotated 90° counterclockwise about O.
a) Draw and label the image D’E’FG’.
Answer:

b) Complete the table of coordinates for DEFG and its image D’E’F’G’.

Answer:

Math in Focus Course 3B Practice 8.3 Answer Key

Solve. Show your work.

Question 1.
A rotation of point P clockwise about O maps onto P’. State the angle of rotation.

a)

Answer:

b)

Answer:

Question 2.
\(\overline{O N}\) is rotated about the origin, O to form the image \(\overline{O N^{\prime}}\). State the angle and direction of each rotation.

a)

Answer:
The point N is on (1,3), N’ is on (-1,-3) and the point of rotation is point O. The line created from N to O and O to N’ is a straight line and has an angle of 180°. To ensure the angle of rotation is 180°, the coordinates should have changed from (x,y) to (-x,-y). Since N is on (1,3) and its image N’ is on (-1,-3) then the angle of rotation is indeed 180°.

b)

Answer:
The point N is on (3, -2), N’ is on (2, 3) and the point of rotation is point O. The lines created from P to O and O to P’ makes a right angle. To ensure that the angle of rotation is 90° counterclockwise, the coordinate should have changed from (x, y) to (-y, x). Since N is on (3, -2) and its image N’ is on (2, 3) then the angle of rotation is indeed 90° in a counterclockwise direction.

Solve. Show your work.

Question 3.
At an amusement park, Olivia is riding the carousel at point P. She is rotated from P by each of the following rotations. Mark and label her position after each rotation from P on a copy of the graph.

a) A: 90° counterclockwise about the origin
Answer: P is at (3, 5). The rotation at 90° counterclockwise changes the coordinates from (x, y) to (-y, x). Therefore, Olivia at (3, 5) will be on point A(-5, 3) when the carousel rotates at this degree and at this direction.

b) B : 90° clockwise about the origin
Answer: The rotation at 90° clockwise shifts the coordinates from (x, y) to (y, -x). So Olivia at (3, 5) will be at B (5, -3) after the rotation at this angle and in this direction.

c) C: 270° counterclockwise about the origin
Answer: A 270° rotation counterclockwise is basically just a 180° plus 90° counterclockwise direction. A 180° angle of rotation moves the coordinates from (x, y) to (-x, -y). Thus, P(3, 5) will be on (-3,-5). Then rotating it again 90°counterclockwise, the new point should be at the form (-y, x). Therefore, Olivia will be on C(5, -3) after this rotation at this direction.

d) D: Half turn about the origin
Answer:

Answer:

Question 4.
A cam of an automobile rotates about a shaft at the origin, O. Point P on the cam rotates to point Q.
a) Describe the rotation.
Answer: The point is on (2, 2) and Q is on (-2, 2). The coordinate (x, y) shifted to (-y, x). Therefore, the angle of rotation is 90° counterclockwise direction because the coordinates changed from (x, y) to (-y, x).

b) A point (-5, 4) undergoes the same rotation. Find the coordinates of the image.

Answer: (-5,4) is rotated counterclockwise at 90° then the coordinates (x, y) should shift to (-y, x). Therefore, the image of (-5, 4) will have the coordinates (-4, -5).

Question 5.
The hinges on a door are at (0, 0), looking down from above. Its keyhole is at position (2, 4) when the door is closed. The door swings open. Find the position of the keyhole under each rotation below.

a) 90° clockwise
Answer: The point is on (2,4) should changed from (x, y) to (-y, x) for 90° clockwise rotation. Therefore, the keyhole at (2, 4) will be on point (4,-2) when the door is open 90° clockwise.

b) 90°counterclockwise
Answer: (2, 4) is rotated counterclockwise at 90° then the coordinates (x, y) should shift to (-y, x). Therefore, the keyhole at (2, 4) will have the coordinates (4, -2) when the door is open 90° clockwise.

c) 180°
Answer: Point (2, 4) has a 180° angle of rotation then the coordinates should have changed from (x, y) to (-y, x). Therefore, the keyhole at (2, 4) will have the coordinates from (-2, -4) when the door is open rotates 180°.

Question 6.
Pentagon ABCDE is drawn on the coordinate plane.

a) ABCDE is rotated 90° clockwise about the origin, O. Draw and label the image A’B’C’D’E’.
Answer:

b) State the coordinates of A’, B’, C’, D’, and E’.
Answer: The rotation changes the coordinates from (x,y) to (y,-x). Thus, the coordinates of points A’, B’, C’, D’ and E’ are (-1, -2), (-2, -3), (-3, -3), (-3, -1) and (-2, -1) respectively.

c) ABCDE is rotated 90° counterclockwise about the origin, O. Draw the image A”B”C”D”E”. State the coordinates for A”B”C”D”E”.
Answer: A 90° rotation counterclockwise direction changes the coordinates (x,y) to (-y,x). Thus, A at (2,-1) will have an image A” at (1, 2), B(3, -2) will be mapped onto B”(2, 3), C at (3, -3) will have an image C” at (3, -3) D(1, -3) will be mapped onto D”(3, 1) and E at (1, -2) will have an image E” at (2, 1). So, plotting the points of the images, A”, B”, C”, D”, E”.

d) How are A’B’C’D’E’ and A”B”C”D”E” related?
Answer: The coordinates of points A’B’C’D’E’ are (-1, -2), (-2, -3), (-3, -3), (-3,-1) and (-2, -1) respectively while the coordinates of points A”B”C”D”E” (1, 2), (2, 3), (3, 3), (3, 1) and (2, 1) respectively. On the other hand, the 180° rotation changes the coordinates from (x, y) to (-x, -y) which is the case if pentagon A’B’C’D’E’ will be the original figure and the image will be A”B”C”D”E”. Therefore A”B”C”D”E” is the image of A’B’C”D’E’ if it is rotated at an angle of 180°.

Question 7.
A regular hexagon ABCDEF is rotated about its center, O, so that its appearance stays the same, but the vertices are rotated to different positions. For example in one rotation, A moves to B, B to C, and so on. Which clockwise rotations will cause this effect? Which points are invariant under a rotation? Explain.

Answer: A whole hexagon has a complete angle of 720°. The figure has 6 equal angles, dividing 720° to 6 than a angle to angle rotation has 120° which is the case in this problem. Therefore, the angle of rotation is 120°.

Question 8.
Math Journal Which points are invariant under a rotation? Explain.
Answer: The points that are invariant are those that did not change even after the rotation. If a point or line or figure is rotated at any angle at any direction about (0, 0) then the invariant is (0, 0) since this point will not change at any rotation the point or line or figure will undergo. If it is rotated at about (1, 2) then the invariant is (1, 2). Therefore, the centers of rotation are the invariant points.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Lesson 8.4 Dilations to score better marks in the exam.

Math in Focus Grade 7 Course 3 B Chapter 8 Lesson 8.4 Answer Key Dilations

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

Solve.

Question 1.
Which triangles are dilations of one another? Explain.

a)

Answer:
Δ STU and Δ PTR. They have a center of dilation T and the sides of Δ PTR are twice as long as the sides of Δ STU.

b)

Answer:
Δ PQS and Δ PUT. They have a center of dilation, P, and the sides of Δ PUT are 1 1/2 times as long as the sides of Δ PQS.

Copy and complete.

Question 2.
A rectangle has coordinates A (5, 1), B (3, 1), C (3, 4), and D (5, 4).

a) Find the length and width of ABCD.
The length of ABCD is units. Its width is units.
Answer:

The length of ABCD is 3 units. Its width is 2 units.

b) Find the length and width of the image of ABCD when dilated with scale factor 2.
Length of image: • = units
Width of image: • = units
Answer:
scale factor = 2
Length of image: 3 • 2 = 6 units
Width of image: 2 • 2 = 4 units

c) Find the length and width of the image of ABCD when dilated with scale factor \(\frac{1}{2}\).
Length of image: • = units
Width of image: • = units
Answer:
Scale factor \(\frac{1}{2}\).
Length of image: 3 • \(\frac{1}{2}\) = 3\(\frac{1}{2}\). units
Width of image: 2 • \(\frac{1}{2}\). = 1 units

d)
What are the coordinates of the image rectangle under each dilation if the center of dilation is at the origin?

Answer:
Scale factor 2:
A (5, 1) = 5 × 2 , 1 × 2 = (10, 2)
B (3, 1) = 3 × 2, 1 × 2 = (6, 1)
C (3, 4) = 3 × 2, 4 × 2 = (6, 8)
D (5, 4) = 5 × 2, 4 × 2 = (10, 8)
Scale factor 1/2:
A (5, 1) = 5 × 1/2 , 1 × 1/2 = (2.5, 0.5)
B (3, 1) = 3 × 1/2, 1 × 1/2 = (1.5, 0.5)
C (3, 4) = 3 × 1/2, 4 × 1/2 = (1.5, 2)
D (5, 4) = 5 × 1/2, 4 × 1/2 = (2.5, 2)

You may want to draw the rectangle and its images on the coordinate plane to solve c).

Technology Activity

Materials:

• geometry software

Explore The Properties Of Dilations With Geometry Software

Step 1.
Draw a line segment AB using a geometry software program.

Step 2.
Select the Dilate function, within the Transform menu. Enter the scale factor 2 to dilate the line segment about the origin. Record your results in a table of coordinates.

Step 3.
Describe how the image of \(\overline{A B}\) is related to \(\overline{A B}\).

Step 4.
Repeat Step 1 to Step 3 using a rectangle as the original figure. For Step 2, enter the scale factor \(\frac{1}{2}\).

Step 5.
Repeat Step 1 to Step 3 using a rectangle as the original figure. For Step 2, enter the scale factor -2.

Math Journal Observe any changes in the size or shape of the figure after the dilation. Which of these properties does a dilation preserve: lengths, shape, parallel lines, or perpendicular lines? Explain.

Copy and complete on graph paper.

Question 3.
The management of a swimming pool built a springboard above the pool. The height of the springboard is a dilation of the depth of the pool with center at the origin, O and scale factor –\(\frac{1}{3}\). The depth of the pool is 4.5 meters, represented by \(\overline{S T}\) on the coordinate plane. The floor is represented by the positive x-axis and the surface of the water is represented by the negative x-axis. Draw the location and height of the stand for the springboard, \(\overline{U V}\), on a copy of this vertical cross section of the pool.

Answer:

Use graph paper. Use 1 grid square on both axes to represent 1 unit for the interval from -7 to 4.

Question 4.
The triangles are each mapped onto their images by a dilation. Draw each triangle and its image on a coordinate plane. Then mark and label C as the center of dilation. Find the scale factor for each triangle.
a)

Answer:

C(-1, 1), scale factor = -2

b)

Answer:

C(3, 4) and scale factor = 3

Math in Focus Course 3B Practice 8.4 Answer Key

Tell whether each transformation is a dilation. Explain.

Question 1.

Answer:
Yes, Δ ABC and Δ ADE have center of dilation at A and the sides of Δ ADE are twice as long as the sides of Δ ABC.

Question 2.

Answer: No it is not a dilation

Solve. Show your work.

Question 3.
Nikita wants to make a mosaic for a T-shirt’s design. She makes some dilated copies of a drawing with a photocopier. The drawing is 6 inches by 4 inches. Find the length and width of each copy with the scale factor given in a) to d). State whether each copy is an enlargement or reduction of the drawing.
a) 1.5
Answer:
Scale factor = 1.5
(6, 4) = 6 × 3/2, 4 × 3/2 = 9/2 in, 6 in, Enlargement

b) 2
Answer:
Scale factor = 2
(6, 4) = 6 × 2, 4 × 2 = (12, 8), Enlargement

c) \(\frac{1}{4}\)
Answer:
Scale factor = \(\frac{1}{4}\)
(6, 4) = 6 × \(\frac{1}{4}\), 4 × \(\frac{1}{4}\) = 1.5 in, 1 in, Reduction

d) 140%
Answer:
Scale factor = 140% = 1.4
(6, 4) = 6 × 1.4, 4 × 1.4 = 8.4 in, 5.6 in. Enlargment

Copy and complete on graph paper.

Question 4.
Timothy uses a lens to view a 2-inch pencil that is represented by \(\overline{\mathrm{AB}}\) on the coordinate plane. \(\overline{\mathrm{AB}}\) is mapped onto \(\overline{A^{\prime} B^{\prime}}\) by a dilation with center at the origin, O. Draw each image for the given scale factor.

a) Scale factor -0.5

Answer:
Observe that the center of dilation is at the origin then the method of multiplying the coordinates of \(\overline{A B}\) to the scale factor can be used. Let j be the coordinates after the dilation of A whose coordinate is (3, 2).
j = A ·  (-0.5)    Formula.
= (3, 2) · (-0.5)    Substitute the Coordinates of A.
= (-1.5, -1)    Multiply.
Thus, point A’ is on (-1.5, -1). Next, Let k be the coordinates after the dilation of B whose coordinate is (3,0).
k = B · (-0.5)    Formula.
= (3, 0) · (-0.5)   Substitute the Coordinates of B.
= (-1.5. 0)    Multtiply.
Thus, point B’ is on (-1.5, 0). Therefore, the points of \(\overline{A^{\prime} B^{\prime}}\) are on (-1.5, -1) and (-1.5, 0). The illustration for the image of AB would be as depicted below.

b) Scale factor 0.5

Answer:
Since the center of dilation is at the origin then the method of multiplying the coordinates of \(\overline{A B}\) to the scale factor can be used. Let l be the coordinates after the dilation of A whose coordinate is (2, 3).
l =  A · 0.5    Formula.
= (2, 3) · 0.5    Substitute the Coordinates of A.
= (1, 1.5) Multiply.
Thus, point A’ is on (1, 1.5). Next, Let m be the coordinates after the dilation of B whose coordinate is (2, 1).
m = B · 0.5    Formula.
= (2. 1) · 0.5    Substitute the Coordinates of B.
= (1, 0.5)    Multiply.
Thus, point B’ is on (1, 0.5). Therefore, the points of \(\overline{A^{\prime} B^{\prime}}\) are on (1, 1.5) and (1, 0.5). The illustration for the image of \(\overline{A B}\) wouLd be as depicted below.

Question 5.
Each figure is each mapped onto its image by a dilation with its center at the origin, O. On a copy of the coordinate plane, draw each image.

a) Triangle LMN is mapped onto triangle L’M’N’ with scale factor –\(\frac{1}{2}\).

Answer:

b) Rectangle PQRS is mapped onto rectangle P’Q’R’S’ with scale factor 3

Answer:

Solve on graph paper. Show your work.

Question 6.
In a room, a flashlight is used to illuminate objects and cast their shadows on a wall. Each shadow is a dilation of the object’s profile.
a) A pen at point (4, 2) is mapped onto its shadow at (6, 3) with the origin as the center of dilation. Find the scale factor.
Answer:
To find the factor of the dilation, solve for the ratio of the dilated points to the originaL points. Take the x-coordinates of the two points. The x- coordinate of the shadow is 6 and x- coordinate of the original object is 4, then
Scale Factor = \(\frac{x-\text { coordinate of Dilated Image }}{x-\text { coordinate of the Original Point }}\) Formula.
= \(\frac{6}{4}\)   Substitution.
= \(\frac{3}{2}\)    Simplify.
Therefore, the scale factor is \(\frac{3}{2}\). If the y-coordinates of the two points is the chosen one to find the scale factor, then the y- coordinate of the shadow is 3 and y- coordinate of the original object is 2, so
Scale Factor = \(\frac{y-\text { coordinate of Dilated Image }}{y-\text { coordinate of the Original Point }}\) Formula.
= \(\frac{3}{2}\)   Substitution.
Therefore, the scale factor is \(\frac{3}{2}\).

b) The shadow of a circular disc has its center at (6, 2) and radius 3 units. The circular disc has center at (2, 2) and radius 1 unit. Find the center of dilation.
Answer:
To identify the center of dilation, let x₁ be the x-coordinate of the center of dilation and y₂ for the y-coordinate. Then, solving the x-coordinate of the center of dilation is given by the scale factor, f multiplied by the x-coordinate of the original point denoted by x₂ minus x-coordinate of the dilated image denoted by x₃ divided by scale factor minus 1.
x₁ = \(\frac{f x_{2}-x_{3}}{f-1}\)    Formula.
= \(\frac{3(2)-6}{3-1}\)    Substitution.
= \(\frac{6-6}{3-1}\)    Multiply.
= \(\frac{0}{2}\)   Subtract.
= 0    Simplify.
Thus, the center of dilation is on 0 of x Next, solving the y-coordinate of the center of dilation is given by the scale factor, f multiplied by the y-coordinate of the original point denoted by y₂ minus y-coordinate of the dilated image denoted by y₃ divided by scale factor minus 1.
y₁   = \(\frac{f y_{2}-y_{3}}{f-1}\)    Formula.
= \(\frac{3(2)-2}{3-1}\)    Substitution.
= \(\frac{6-2}{3-1}\)    Multiply.
= \(\frac{4}{2}\)   Subtract.
= 2    Simplify.
Thus, the center of dilation is on 2 of y. Therefore, the center of dilation lies on point (0, 2).

Question 7.
Each figure is mapped into its image by a dilation. Draw each figure and its image on the coordinate plane. Then mark and label C as the center of dilation. Find the scale factor for each figure. Use 1 grid square on both axes to represent 1 unit for the interval from —8 to 6.

a) Quadrilateral WXYZ

Answer: C(1, 2) Scale factor = -0.5

b) Triangle PQR

Answer:

C(1, 2), Scale factor is 2.5

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Lesson 8.2 Reflections to score better marks in the exam.

Math in Focus Grade 7 Course 3 B Chapter 8 Lesson 8.2 Answer Key Reflections

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

Solve.

Question 1.
Andrew wants to hang a square poster on his bedroom wall. He knows that to get the poster to balance properly, he needs to place a second picture hanger at W’, with the dotted vertical line as the line of symmetry. If the distance between W and the vertical line is 3 inches, find WW’.

Answer:
W’ is the reflection of W so if W is 3 inches from the vertical then W’ is also 3 inches away from the vertical line. Thus WW’is the sum of the length of W and the length of W’. Therefore, WW’ is 6 inches.

Technology Activity

Explore The Properties Of Reflections With Geometry Software

Step 1.
Draw a vertical line segment using a geometry software program.

Step 2.
Select the y-axis as the line of reflection. Reflect the line segment. How are the line and its image related to the line of symmetry?

Step 3.
How are the lengths of the segment and its image relate?

Step 4.
Repeat Step 1 to Step 3 first using a horizontal segment and then a segment that is neither vertical nor horizontal.

Step 5.
Repeat Step 1 to Step 4 using the x-axis as the line of reflection.

Step 6.
Draw a rectangle and reflect it over the y-axis. Do the parallel lines of the rectangle stay parallel? Do the perpendicular sides of the rectangle stay perpendicular?

Step 7.
Repeat Step 6 using the x-axis as the line of reflection.

Math Journal Which of these properties does a reflection seem to preserve: length, shape, parallel lines, or perpendicular lines? Explain.

Draw Images After Reflections.

In the activity, you may have observed the following:

Reflections preserve shape and size. They also preserve parallelism and perpendicularity.
All polygons and combinations of polygons are made up of line segments joined at their endpoints. When you reflect these figures, you reflect all the line segments joined at their endpoints. The images of all these line segments and their endpoints combine to form the whole image.

Copy and complete on graph paper.

Question 2.
Each line segment is reflected in \(\overleftrightarrow{M N}\). On a copy of the diagram, draw each image.

a)

Answer:

b)

Answer:

c)

Answer:

d)

Answer:

Copy and complete on graph paper.

Question 3.
Layla is designing a star-shaped figure for a stencil. She wants the bottom half to be a reflection of the top half. She will reflect it across the x-axis to draw the other half. Complete the design for her.

Answer:
P(-4, 0), Q(0, 4) and R(4, 0) are plotted on the plot. Put additional points M and N on the coordinates (-1, 1) and (1, 1) to make the plotting easier. Since the reflection is in the x-axis then all the coordinates will not change. For P(-4, 0) and R(4, 0), since 0 is 0 units away from x then for image P’ and R’ will be along the x-axis with coordinates (-4, 0) and (-4, 0) respectively. Point Q(0, 4) on the other hand is 4 units above the x-axis so its R’ units below the axis. For M(-1, 1) and N(1, 1)since it is 1 unit above the x-axis then their image will be 1 unit beneath that axis. Thus M’ will be on (-1, 1) and N’ will be on (1, -1).

Complete.

Question 4.
A figure has vertices P (0, 2), Q (-1, 0), R (-2, 1), S (-1, -2), and T (0, -2), is reflected in the y-axis. Draw the figure and its image on the coordinate plane. Use 1 grid square on both axes to represent 1 unit for the interval from -2 to 2.
Answer:

Complete.

Question 5.
Mr. Patterson is building a double bird house, one next to the other. The vertices of the front of one houses have coordinates P (3, 0), Q (5, 3), R (3, 6), and S (1, 3).
The front of the other bird house, P’Q’R’S’, is a reflection of the first one in the y-axis.
The x-coordinates of vertices of PQRS and P’Q’R’S’ are , and their y-coordinates are .
P(3, 0) is mapped onto P'(, ).
Q (5, 3) is mapped onto Q’ (, ).
R (3, 6) is mapped onto R’ (, ).
S (1, 3) is mapped onto S’ (, ).
Any point (x, y) is mapped onto (, ) when reflected in the y-axis.

Answer:

The x-coordinates of vertices of PQRS and P’Q’R’S’ are negative of each other and their y-coordinates are did not change.
P(3, 0) is mapped onto P'(-3, 0).
Q (5, 3) is mapped onto Q’ (-5, 3).
R (3, 6) is mapped onto R’ (-3, 6).
S (1, 3) is mapped onto S’ (-1, 3).
Any point (x, y) is mapped onto (-x, y) when reflected in the y-axis

Math in Focus Course 3B Practice 8.2 Answer Key

Copy each diagram on graph paper and draw the image using the given reflection.

Question 1.
In the x-axis

Answer:

Question 2.
In the y-axis

Answer:

Question 3.
Ethan placed six sticks on a table. Three of the sticks, \(\overline{P Q}\), \(\overline{R S}\), and \(\overline{T U}\) are shown on the coordinate plane. The other sticks are images of the three sticks, with x = 0 as the line of reflection. On a copy of the graph, draw the sticks not shown on the coordinate plane.

Answer:

Solve.

Question 4.
A pattern is drawn on the coordinate plane and then repeated by first reflecting it in the x-axis and reflecting the original pattern in the y-axis.

a) Copy and complete the table by finding the position of each of the other tiles. On a copy of the coordinate plane, indicate the positions of the images.

Answer:
Note that if A, B, C, D, E, and Fis reflected in x-axis, the coordinates wiLl be in the form (x, -y) while the image after applying the reflection in y-axis has coordinates in the form (-x, y) then and in the coordinate plane.

b) Reflect A’, B’, and C’ in the y-axis. What are the coordinates of the image?
Answer:
If A’, B’, and C’ is reflected in y-axis, the coordinates wiLL be in the form (-x, y) then the ¡mage of A’(4, -5) will be on (-4, -5), B’(4, -4) will have an image on (-4, -4) and the image of C’(5, -4) will be plotted on (-5, -4).

c) Reflect A”, B”, and C” in the x-axis. What are the coordinates of the image? How do these coordinates compare to those in b)?
Answer:
If A”,B”, and C” ¡s refLected in x-axis, the coordinates will be in the form (x, -y) then the images of A”(-4, 5), B”(-4, 4), and C”(-5, 4) will be on (-4, -5), (-4, -4) and (-5, -4). Observe that A’, B’, and C’ if reflected in y-axis has the same image with A”, B”, and C” if it is reflected in x-axis.

Copy and complete on graph paper.

Question 5.
Isabella painted a water color design on graph paper. Some of the points were at A (-4, 8), 8 (-2, 8), C(-1, 6), D (-2, 4), E (-4, 4), and F(-5, 6). She folded the paper along y = 3 to reflect the design.
The image points are A’, 8′, C’, D’, E’, and F’.

a) Draw the line y = 3.
Answer:

The line y equals to 3 is a line segment whose coordinates are (0,3) then

b) Find the coordinates of A’, B’, C’, D’, E’, and F’.
Answer:
A coordinates are (-4, 8) first know the position of the point on the coordinate plane. Notice that A is 5 units above y = 3, so the y coordinate for the image A is 5 units below y = 3 which is -2. Therefore, the coordinates of A’ is (-4 ,-2).
B coordinates are (-2, 8) first locate the point on the coordinate plane. Notice that B is 5 units above y = 3, so the y coordinate for the image B is 5 units below y = 3 which is -2. Therefore, the coordinates of A’ is (-2, -2).

Locating C(-1, 6) observe that C is 3 units above y = 3 then, y coordinate for the image C is 3 units under y = 3 which is 0. Therefore, the coordinates of C’ is (-1, 0).

For D whose coordinates are (-2, 4), the first step is to locate the point notice that D is 1 unit above y = 3 so the y coordinate for the image of D is 1 unit below y = 3 which is 2. Therefore, the coordinates of D’ is (-2, 2).

Determine the position of E (-4, 4) note that E is 1 unit above y = 3 thus, the y coordinate for the image of E is 1 unit beneath y = 3 which is 2. Therefore, the coordinates of E’ is (-4, 2).

Lastly, Locate F(-5, 6). Observe that F is 3 units above y = 3 then, the y coordinate for the image of F is 3 units under y = 3 which is 0 . Therefore, the coordinates of F’ is (-5, 0).

c) Draw the image and label A’, B’, C’, D’, E‘, and F’.
Answer:

Solve.

Question 6.
The image of a butterfly with its wings symmetrically spread out is outlined on the coordinate plane. The uppermost tips of the wings are at (4, 5) and (-2, 5). The lower most tip of one wing is at (2, 0).

a) Find an equation of the line of reflection.
Answer: Plot the points on the coordinate plane. Let these points (4, 5) and (-2, 5) be an original point and its image. The y coordinates did not change, the reflection must be in x. Also, they are 6 points away from each other, the points are 3 units away from x = 1. Since the line of reflection is exactly in between the original point and its image, therefore, the line of reflection for this case is x =1.

b) Find the position of the lower tip of the other wing.
Answer: The line of reflection x = 1, let (2, 0) be an original point and the image be the point where the last tip of the wing is. Observe that the line of reflection is a vertical line then the image of (2, 0) will be on the same horizontal line where (2, 0). Next, the original point is 1 unit away from the line of reflection thus its image is 1 unit from the opposite side of x =1. Therefore, on the same horizontal line, move 1 unit to the left the line of reflection, the other tip of the wings is on (0, 0).

Question 7.
Math Journal Point A’ is the image of point A under a reflection. How do you find the line of reflection, without the use of a coordinate grid?
Answer: The reflection of A has the same distance from the line of reflection. Therefore, the line of reflection can be determined by adding the lengths of the original point and the image then dividing the sum by 2. Their midpoint must be the line of reflection.

Question 8.
A tablecloth has two red dots on it. They are at positions (-3, -1) and (-1, -3). The cloth is folded in half, so that the dots touch each other. What is an equation for the line along which the tablecloth was folded?
Answer:
x = y
The point (-2,-2) is the reflection of the points (-3,-1) and (-1,-3) since it is exactly in between the points. Both are 1 diagonal unit away from the point. Thus, the diagonal line whose points are (-3,-3),(-2,-2),(-1,-1)is the line of reflection of the points. When y is 1, x is equal to 1, if y is equal to 2 x is 2.

Question 9.
A leaf is symmetric about its midvein, the central vein that runs the length of the leaf. The leaf is outlined in the coordinate plane with its midvein on the line y = -x.
a) A side vein has a length of 6 units on the grid. What is the length of its symmetric counterpart?
Answer:
6 units
The line of reflection is exactly in the middle of the original point or line and its image. If the original points or lines are 10 units away from the line of reflection, then so is the image, if it is 1 unit away, then so is their reflection. Let the side vein be the original line, the grid be the reflection of the line and the image be the symmetric counterpart of the vein. From the above explanation, since the original line is 6 units away from the grid then the image is also 6 units away.

b) The endpoint of another side vein is at (4, 3). What is the endpoint of its symmetric counterpart?
Answer: (-3,-4)
Explanation:
The line of reflection according to its equation is a diagonal line. Plotting (4,3) notice that the point is 3 and a half diagonal units from the line of reflection thus its image also has the same distance on the opposite side. So from the line of reflection, move 3 and half diagonal units. Therefore, the endpoint of the corresponding symmetric side of the vein is (-3,-4).

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Geometric Transformations to score better marks in the exam.

Math in Focus Grade 8 Course 3 B Chapter 8 Answer Key Geometric Transformations

Math in Focus Grade 8 Chapter 8 Quick Check Answer Key

Solve. Show your work.

Question 1.
Points W and Z are reflections of each other on the x-axis. If W is the point (4, -3), what are the coordinates of Z?
Answer:
Given that the points W and Z are reflections of each other on the x-axis.
The points of W are (4,-3)
The points (4,-3) are in the 4th quadrant.
The coordinates of Z are (4,3) because the reflection to the fourth quadrant in the x axis is the first quadrant. The points (4,3) lie in the first quadrant.

Question 2.
Points P (-2, 5) and Q (-2, -5) are reflections of each other in the line k. What is an equation of line k?
Answer:
Given that the points are P(-2,5) and Q(-2,-5)
(x1, y1) = (-2,5)
(x2,y2) = (-2,-5)
We know that, Equation of the line formula is y2 – y1/x2 – x1
Equation of the line k = y2 – y1/x2 – x1
k = -5-(5)/-2-(-2)
k = -5-5/-2+2
k = -10/0
k = -10.

Question 3.
Points M (2, -1) and N (-2, -1) are reflections of each other in the line m. What is an equation of line m?
Answer:
Given points are M(2,-1) and N(-2,-1)
Here (x1,y1) = (2,-1)
(x2,y2) = (-2,-1)
We know that, Equation of the line formula is y2 – y1/x2 – x1
equation of the line m = y2 – y1/x2 – x1
m = -1-(-1)/-2-(2)
m = -1+1/-4
m = 0/-4
Therefore m = 0

Question 4.
If a point S is the reflection of R (1,4) in the x-axis, what is the length of \(\overline{R S}\)?
Answer:
Given point is R(1,4)
The point R is reflection to S(1,-4)
The distance between the RS is 8 grids.

State whether x and y are directly proportional.

Question 5.
y = 1.5x
Answer:
Given equation is y = 1.5x
Suppose x,y = 2
Substitute in the above equation
2 = 1.5 × 2
2 = 3
Here x increases then y also increases so, the given equation is directly proportional.

Question 6.
y = –\(\frac{x}{2}\)
Answer:
Given equation is y = -x/2
Suppose x,y = 2
Substitute in the above equation
2 = -2/2
2 = -1
Here x increases then y decreases so, it is not directly proportional.

Question 7.

Answer:
x = y²
We take x = 2 and y = 2
Then 2 = 2²
2 = 4
Here x increases then y decreases so, it is directly proportional.

State whether \(\overleftrightarrow{Y Z}\) is a perpendicular bisector of the given segment. Justify your answer. If \(\overleftrightarrow{Y Z}\) is a perpendicular bisector, name two pairs of distances that are equal.

Question 8.

Answer:
In the given figure CD and ZY are not perpendicular bisectors.
Perpendicular bisector means it is a line or a segment that passes through the midpoint of the segment. And any point on the perpendicular bisector is equidistant from the endpoints of the line segment.
Here the two pairs CD and ZY distances are not equal.

Question 9.

Answer:
In the given figure EF and YZ are perpendicular bisectors.
Perpendicular bisector means it is a line or a segment that passes through the midpoint of the segment. And any point on the perpendicular bisector is equidistant from the endpoints of the line segment.
Here the two pairs EF and YZ distances are equal.

Question 10.

Answer:
In the given figure GH and YZ are perpendicular bisectors.
Perpendicular bisector means it is a line or a segment that passes through the midpoint of the segment. And any point on the perpendicular bisector is equidistant from the endpoints of the line segment.
Here the two pairs GH and YZ distances are equal.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Lesson 8.1 Translations to score better marks in the exam.

Math in Focus Grade 7 Course 3 B Chapter 8 Lesson 8.1 Answer Key Translations

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

Copy and complete on graph paper.

Question 1.
Abigail jogs from point H (3, 2) in a park to point H’, as described by a translation of 5 units to the left and 3 units up. Mark the position of H’ on the coordinate plane.

Answer:

Therefore, H point is on (-2,5)

Copy and complete on graph paper.

Question 2.
Mr. McBride wanted to set up a barbeque pit in his backyard. He had to move a swing set represented by \(\overline{H L} \text {. }\). He decided to move the swing set by a translation of 4 units to the right and 2 units down to \(\overline{H^{\prime} L^{\prime}}\). Draw and mark \(\overline{H^{\prime} L^{\prime}}\) on the coordinate plane.

Answer:

Technology Activity

Explore The Properties Of Translations With Geometry Software

Step 1.
Draw a line segment using a geometry software program.
Step 2.
Select the Translate function, within the Transform menu. Enter the number of units by which you wish to translate the line segment. Translate the line segment to its new position.
Step 3.
How is the length and position of the new line segment related to the original line segment?

Think Math
A horizontal segment is translated horizontally. Is the image parallel to the original segment? Does a translation always map a segment onto a parallel segment? Explain.

Step 4.
Repeat Step 2 using a triangle and then a rectangle. Observe how each figure is related to its image.

Step 5.
How do the size and shape of the figures change under a translation? Do the parallel sides of the rectangle remain parallel? Do the perpendicular sides of the rectangle remain perpendicular?

Math Journal
Which of these properties does a translation seem to preserve: lengths, shapes, parallel lines, or perpendicular lines? Explain.

Draw Images After Translations.

From the activity, you have observed that translations preserve shape, size, parallelism, and perpendicularity.

When a figure is drawn on a coordinate plane, you can easily draw its image. You need only to find the images of the vertices and connect them appropriately.

Complete on graph paper.

Question 3.
Tim is creating a wrapping paper design on his computer. He wants to move figure EFGH by translating it 5 units to the left and 4 units down. E, F, G, and FI have the coordinates (1, 2), (3, 2), (3, 3), and (1, 3). Draw EFGH and E’F’G’H’ on the same coordinate plane. Use 1 grid square on both axes to represent 1 unit for the interval from —4 to 3. Then complete the following statements.
E (1, 2) is mapped onto E’ (, ).
F(3, 2) is mapped onto F’ (, ).
G (3, 3) is mapped onto G’ (, ).
H (1, 3) is mapped onto H’ (, ).
Answer:
E (1, 2) is mapped onto E’ (-4,-2).
F(3, 2) is mapped onto F’ (-2,-2).
G (3, 3) is mapped onto G’ (-2,-1).
H (1, 3) is mapped onto H’ (-4,-1).

Complete.

Question 4.
A triangle has coordinates A (2, 1), B (3, 2), and C (1, 4). It is moved under the translation 2 units to the left and 3 units up. Find the coordinates of the image triangle A’B’C’. Then state the new coordinates for any point (x, y) under this translation.

To find the coordinates of A’, B’, and C’, subtract 2 units from the x-coordinate and add 3 units to the y-coordinate of A, B, and C.

Answer:

Math in Focus Course 3B Practice 8.1 Answer Key

Find the coordinates of the image under each translation.

Question 1.
P (0, 2) is translated by 8 units to the left.
Answer: P is mapped onto P’ (-8,2).

Question 2.
Q(-3, 5) is translated by 3 units to the right and 10 units up.
Answer: Q will be (0, 15) of the coordinate plane.

Question 3.
R(-4, -2) is translated by 1 unit to the left and 6 units up.
Answer: R(-4, -2) is mapped onto R'(-5, 4).

Copy each diagram on graph paper and draw the ¡mage under each translation.

Question 4.
\(\overline{A B}\) is translated 5 units to the right and 1 unit down.

Answer:

Question 5.
Triangle DEF is translated 3 units to the left and 2 units up.

Answer:

Find the coordinates of each point using the given translation. Label the images on a coordinate plane.

Question 6.
Jon’s apartment is located at A (2, 2). He uses the translations described in a) to d) to visit each of his neighbors.

a) From A (2, 2), translate by 3 units to the right, 2 units up to B.
Answer:

b) From B, translate by 2 units to the left, 1 unit up to C.
Answer:

c) From C, translate by 1 unit to the right, 2 units down to D.
Answer:

d) From D, translate by 2 units to the left, 3 units down to E.
Answer:

Solve. Show your work.

Question 7.
The base of a box is at ABCD. It is moved by a translation to a new position A’B’C’D’. The table shows the position to which A was mapped. Find the new position of the other three vertices of the base. Copy and complete the table.

Answer:

Question 8.
A crane moved a cargo pallet from ABCD to other positions on the ship’s deck.
a) Find the coordinates of A’B’C’D’ under a translation that moves each point (p, q) to (p + 4, q + 1). Copy and complete the table. Draw A’B’C’D’ on a graph paper.

Answer:

b) The position of A”B”C”D” is shown on the coordinate plane. State the new coordinates of any point (x, y) under the translation from ABCD to A”B”C”D”.

Answer:
A,B,C and D are on (-2, 1), (0, 3), (2, 1) and (0, -1), A” will have coordinates (-6, 4), and B” has (-4, 6). On the other hand, C” will be lying on (-2, 4) and D” is on (-4, 2).

Question 9.
A computer program T instructs a robotic arm to move an object on the coordinate plane 2 units to the right and 3 units down. The object at point P is translated by T to point P’. Find the coordinates of P if point P’ is (3, 3).
Answer: The original point for P'(3, 3) is P(1, 6).

Question 10.
A line has the equation y = x. It is translated up by 3 units. What is the equation of the new line? How do the slopes of the line and its image compare?
Answer:

Both the original and new line segment are linear but y + 3 = x is 3 units above y = x.

Question 11.
In a wallpaper pattern, a vertical stripe at x = -1 is copied by moving it to x = 1. Describe the translation of this stripe both verbally and algebraically.
Answer:
First locate at x at -1 on the coordinate plane, then know the position of the point at 1. The distance of the two would determine the number of units it move, so the translation is 2 units. Also the original points is x at -1 and then it moved it to right. Therefore, the translation is 2 units to the right.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 7 Review Test to score better marks in the exam.

Math in Focus Grade 8 Course 3 B Chapter 7 Review Test Answer Key

Concepts and Skills

For this review, you may use a calculator. Use 3.14 as an approximation for π.
Round your answer to the nearest tenth where necessary.

Question 1.
Find the value of x in each of the following diagrams.
a)

Answer:
The value of x is 17.69 approximately 18 in,

Explanation:
First we find the altitude a for smaller right traiangle,
a² + 6² = 10²,
a² = 100 -36 = 64 so a is square root of 64 is 8 inches as bigger right triangle
altitude is twice the smaller one so it is 2 X 8 =16 inches now
x² = 12² + 16²,
x² = 144 + 169 = 313, therefore x is square root of 17.69 approximately 18 in.

b)

Answer:
The value of x is 12 cm,

Explanation:
Given two right triangles first we calculate hypotenuse be y of
right triangle of width 12.8 cm and height  9.6 cm so
y ² = 9.6² + 12.8² = 92.16 + 163.84 = 256,
so the value of y is square root of 256 is 16 cm,
Now the value of x is  x² + 16² = 20²,
x² = 400 – 256 = 144 , the value of x is square root of 144 is 12 cm.

c)

Answer:
The value of x is approximately equal to 19 cm,

Explanation:
Applying pythagorean we have
x² + 7² = 20²,
x² = 400 – 49 = 351, the value of x is square root of 351 is 18.734,
so x is approximately equal to 19 cm.

Question 2.
Find the distance between each pair of points. Which pair of points are the greatest distance apart?
a) A(5, 2), B(8, 5)
Answer:
To find the distance of two points, use the distance formula given by the square root of the sum of the x-coordinate of the second point minus the x-coordinate of the first point and the y-coordinate of the second point minus the y-coordinate of the first point Also. note that there is no negative magnitude(distance) Then, let A(5, 2) be (x₁, y₁) whiLe B(8, 2) be (x₂, y₂) and e be the distance of A to B.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(8-5)^{2}+(2-2)^{2}}\)    Substitution.
= \(\sqrt{3^{2}+0^{2}}\)    Subtract.
= \(\sqrt{9+0}\)    Evaluate the squares.
= \(\sqrt{9}\)    Add.
= ±3 Evaluate.
Therefore, A is 3 units away from B.

b) C(-3, 2), D(2, 3)
Answer:
Apply the distance formula then Let C(-3, 2) be (x₁, y₁) while D(2, 3) be (x₂, y₂) and f be the distance of C to D.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(2-(-3))^{2}+(3-2)^{2}}\)    Substitution.
= \(\sqrt{5^{2}+1^{2}}\)    Subtract.
= \(\sqrt{25+1}\)    Evaluate the squares.
= \(\sqrt{26}\)    Add.
≈ ±5.1    Evaluate.
Therefore, the distance of C to D is 5.1 units

c) L(1, -3), N(-2, -1)
Answer:
Use the distance formula then let L(1, -3) be (x₁, y₁) white N(-2, -1) be (x₂, y₂) and g be the distance of L to N.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{((-2)-1)^{2}+((-1)-(-3))^{2}}\)    Substitution.
= \(\sqrt{(-3)^{2}+2^{2}}\)    Subtract.
= \(\sqrt{9+4}\)    Evaluate the squares.
= \(\sqrt{13}\)    Add.
≈ ± 3.6     Evaluate.
Therefore, L is 3.6 units away from N.

d) Y(-3, -3), Z(0, 4)
Answer:
Let Y(-3, -3) be (x₁, y₁) while Z(0, 4) be (x₂, y₂) and h be the distance of Y to Z then apply the distance formula.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(0-(-3))^{2}+(4-(-3))^{2}}\)    Substitution.
= \(\sqrt{3^{2}+7^{2}}\)    Subtract.
= \(\sqrt{9+49}\)    Evaluate the squares.
= \(\sqrt{58}\)    Add.
≈ ± 7.6     Evaluate.
Therefore, Y is 7.6 units away from Z.

Since A is 3 units away from B. the distance of C to D is 5.1 units, L is 3.6 units away from N. and Y is 7.6 units away from Z. Therefore, the points with largest distance is Y and Z.

Question 3.
Find the volume of each of the following composite solids.
a)

Answer:
The volume of given solid is 120 in³,

Explanation:
First we calculate x,
x² = 6² + 8²,
x² = 36 + 64 = 100, the value of x is square root of 100 is 10 in,
Now the volume of given soild is \(\frac{1}{3}\) • (base edge)² X h =
\(\frac{1}{3}\) • (6)² X 10 in = 120 in³.

b)

Answer:
The volume is 3704.094 in³,

Explanation:
First we calculate value of  x which is height of cylinder,
Applying pythagorean theorem 16² = 9.6² + x²,
x² = 256 – 92.16 = 163.84, the value of x is square root of
163.84 is 12.8 in now volume of cylinder is πr²h =
3.14 X 9.6 X 9.6 X 12.8 = 3704.094 in³.

c)

Answer:
The volume is 113.04 in³,

Explanation:
We have height of sphere is 2r, so r = 6/2 = 3 in,
Now volume of sphere is \(\frac{4}{3}\) • πr³ = \(\frac{4}{3}\) X 3.14 X 3 X 3 X 3 = 113.04 in³.

Question 4.
Find the value of x in each of the following diagrams.
a)

Answer:
To solve for the voLume of the composite soLid, add the volume of the cyLinder to the voLume of the cone. For the volume of the cylinder,
Volume of Cylinder = Area of the base · height of Cylinder
= 20 · 0.5
= 10
Thus, the cylinder has a volume of 10 cubic centimeters. Next, solve for the volume of the cone. First find the radius of its base. Notice that the radius r of the base, the slanted side s and the height j of the cone forms a right triangle with the slanted side as the hypotenuse. Use the Pythagorean Theorem which describes the sum of the square of the legs of a right triangle to be equal to the square of the hypotenuse.
j² + r² = s²      Use the Pythagorean Theorem.
(2.8)² + r² = (3.2)²      Substitution
7.84 + r² = 10.24    Multiply.
7.84 + r² – 7.84 = 10.24 – 7.84     Subtract 7.84 to both sides.
r² = 2.4    Simplify.
r = \(\sqrt{2.4}\)    Find the positive square root.
r ≈ 1.5     Simplify.
Thus, radius r measures around 1.5 cm. Now solving for the volume of the cone, note that the area of a circle (which is the base of the cone) is pi multiplied to the square of the radius and use 3.14 as the value of π.
Volume of Cone = \(\frac{1}{3}\) · Area of the base · Height of Cone
= \(\frac{1}{3}\) πr² · j
= \(\frac{1}{3}\) (3.14) · (1.5)² · 2.8
≈ 6.6
So, the volume of the cone is about 6.6 cubic centimeters. Now for the total volume of the composite solid,
Volume of the Composite Solid = Volume of the Cone + Volume of the Cylinder
= 6.6 + 10
= 16.6
Therefore, the composite solid has a 16.6 cm³ volume.

b)

Answer:
To solve for the volume of the composite solid, add the volume of the cube to the volume of the triangular prism. For the volume of the cube,
let s be the measure of the sides.
Volume of Cube = s³  Formula.
= 6³     Substitution.
= 216    Evaluate.
Thus, the volume of the cube is 216 cubic centimeters. Next for the triangular prism, first solve for the measure of its base. Notice that the edges of the triangular prism has a 1.5 centimeters distance from the ends of the side of the cube. Then the length of the side of the triangular base of the prism can be computed.
Measure of the Base = Measure of the Side of the Cube + 1.5 + 1.5
= 6 + 1.5 + 1.5
= 9
So, the side of the base of the triangular prism measures 9cm. Then, find the altitude of the triangular base of the prism. Notice that half of the length of the base side of the prism denoted as e, the slanted side s, and the altitude a creates a right triangle with the slanted side as the hypotenuse. Then, applying Pythagorean Theorem states that the sum of the square of the legs of a right triangle is equal to the square of the hypotenuse,
(\(\frac{e}{2}\))² + a² = s²    Use the Pythagorean Theorem.
(\(\frac{9}{2}\)) + a² = (7.5)²     Substitution.
(4.5)² + a² = (7.5)²    Divide.
20.25 + a² = 56.25    Multiply.
20.25 + a² – 20.25 = 56.25 – 20.25   Subtract 20.25 to both sides.
a² = 36    Evaluate.
a = \(\sqrt{36}\)    Find the positive square root.
a = 6 Simplify.
Thus, the altitude of the triangLe is 6 centimeters Then, note that the area of a triangle (which is the base of this prism) is half of its length multiplied by the aLtitude. Also, notice that the side of the cube is equal to the height of the triangular prism so the height h of the prism is equal to 6 centimeters. Solving for the volume of the triangular prism,
Volume of Triangular Prism = Area of the Base · Height of the Prism
= \(\frac{1}{2}\) · e a · h
= (\(\frac{1}{2}\) · 9 · 6) · 6
= 27 · 6
= 162
Thus, the triangular prism has a volume of 162 cubic centimeters. Now, for the total volume of the composite solid,
Volume of the Composite Solid = Volume of the Cube + Volume of the Triangular Prism
= 216 + 162
= 378
Therefore, tne composite solid has a 378 cm³ volume.

Problem Solving

Solve. Show your work.

An interactive game teaches users how to find the distance between two points on a coordinate plane. Each unit on the grid equals 1 kilometer. Use the grid for questions 5 and 6.

Question 5.
Find the distance between the farmer and each location.
a) River
b) Pond
c) Town
d) Orchard

Answer:
a) To find the distance of two points, use the distance formula given by the square root of the sum of the x-coordinate of the second point minus the x-coordinate of the first point and the y-coordinate of the second point minus the y-coordinate of the first point Also, note that there is no negative magnitude(distance). Then, let Mr. Farmer at (1, 2) be (x₁, y₁) white the river at (5,4) be (x₂, y₂) and e be the distance of Mr Farmer and the river.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(5-1)^{2}+(4-2)^{2}}\)     Substitution.
= \(\sqrt{4^{2}+2^{2}}\)      Subtract.
= \(\sqrt{16+4}\)     Evaluate the squares.
= \(\sqrt{20}\)     Add.
≈ ± 4.5 Evaluate.
Therefore, Mr. Farmer is approximate[y 4.5 kilometers away from the river.

b. Apply the distance formula again letting Mr. Farmer at (1, 2) be (x₁, y₁) and pond (-4, -2) be (x₂, y₂) and f be the distance of the pond and Mr. Farmer
f = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{((-4)-1)^{2}+((-2)-2)^{2}}\)     Substitution.
= \(\sqrt{(-5)^{2}+(-4)^{2}}\)      Subtract.
= \(\sqrt{25+16}\)     Evaluate the squares.
= \(\sqrt{41}\)     Add.
≈ ± 6.4   Evaluate.
Therefore, the distance of Mr. Farmer to the pond is about 6.4 kilometers.

c) Use the distance formula then let Mr. Farmer at (1, 2) be (x₁, y₁) while the town at (-4, 5) be (x₂, y₂) and g be the distance of Mr. Framer to the town.
g = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{((-4)-1)^{2}+(5-2)^{2}}\)     Substitution.
= \(\sqrt{(-5)^{2}+3^{2}}\)      Subtract.
= \(\sqrt{25+9}\)     Evaluate the squares.
= \(\sqrt{34}\)     Add.
≈ ± 5.8     Evaluate.
Therefore, Mr Farmer is around 5.8 kilometers away from the town

d) Let Mr Farmer (1, 2) be (x₁, y₁) while Orchard (6, -4) be (x₂, y₂) and h be the distance of Mr. Farmer and the orchard then apply the distance formula.
h = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(6-1)^{2}+((-4)-2)^{2}}\)     Substitution.
= \(\sqrt{5^{2}+(-6)^{2}}\)      Subtract.
= \(\sqrt{25+36}\)     Evaluate the squares.
= \(\sqrt{61}\)     Add.
≈ ± 7.8     Evaluate.
Therefore, Mr Farmer is approximately 7.8 kilometers away from the orchard.

Question 6.
The farmer wants to go fishing in either the river or the pond. Which one is closer to his current position? How many kilometers fewer will he walk if he chooses the source that is closer?
Answer:
The objective is to determine the which body of water is closer to Mr. Farmer at (1, 2) and the difference of the distance of the bodies of water to one another.

First, determine the distance of the farmer to the first body of water which is the river. Apply the distance formula given by the square root of the sum of the x-coordinate of the second point minus the x-coordinate of the first point and the y-coordinate of the second point minus the y-coordinate of the first point. Also, note that there is no negative magnitude(distance). Then, Let Mr. Farmer at (1, 2) be (x₁, y₁) while the river at (5, 4) be (x₂, y₂) and p be the distance of Mr. Farmer and the river
p = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\) Formula.
= \(\sqrt{(5-1)^{2}+(4-2)^{2}}\)    Substitution.
= \(\sqrt{4^{2}+2^{2}}\)      Subtract.
= \(\sqrt{16+4}\)      Evaluate the squares.
= \(\sqrt{20}\)    Add.
≈ ± 4.5 Evaluate.
Thus, Mr. Farmer is about 4.5 units away from the river

Next, determine the distance of Mr. Farmer to the pond by using the distance formula again Letting Mr. Farmer at (1, 2) be (x₁, y₁) pond (-4, -2) be (x₂, y₂) and y be the distance of the pond and Mr. Farmer
y = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\) Formula.
= \(\sqrt{((-4)-1)^{2}+((-2)-2)^{2}}\)    Substitution.
= \(\sqrt{(-5)^{2}+(-4)^{2}}\)      Subtract.
= \(\sqrt{25+16}\)      Evaluate the squares.
= \(\sqrt{41}\)    Add.
≈ ± 6.4    Evaluate.
Thus, the distance of Mr. Farmer to the pond is around 6.4 units. Since the distance of the river to Mr. Farmer is smaller than the distance of the pond to Mr Farmer, therefore, the river is cLoser to his position.

Lastly, to determine the difference of the distance, subtract the distance of the farmer to the river to the distance of the farmer to pond.
Distance to the Pond – Distance to the River
6.4 – 4.5
1.9
Therefore, the difference of the distance of the pond and river to Mr. Farmer is 1.9 units.

Question 7.
Candy builds a cage for some chickens out of wood and wire mesh, as shown below. The two slanted sides of the cage are the same size and shape. Find the height of the cage.

Answer:
The height of the cage is 24 in,

Explanation:
Given Candy builds a cage for some chickens out of wood and
wire mesh, as shown the two slanted sides of the cage are
the same size and shape, we take width as 64/2= 32 in and consider
right triangle and apply pythagorean theorem to find height h of the cage as
h² + 32² = 40²,
h² = 1600 – 1024 = 576, so h is square root of 576 is 24 in.

Question 8.
A straw that is ‘12 centimeters long fits inside a glass. The height of the glass is 7.5 centimeters. Find the radius of the glass.

Answer:
The radius of the glass is 4.68 centimeters,

Explanation:
Given a straw that is ‘12 centimeters long fits inside a glass.
The height of the glass is 7.5 centimeters, let b be the base of the glass,
applying pythagorean theorem
b² + 7.5² = 12²,
b² = 144 – 56.25 = 87.75,therefore b is square root of 87.75 is 9.36 centimeters,
Now the radius of glass is half base is 9.36/2  = 4.68 centimeters.

Question 9.
Naomi makes a lampshade by making a cone and then cutting off
part of the cone, as shown. Find the diameter of the larger opening in the lampshade.

Answer:
The diameter of the larger opening in the lampshade is 5.71 in,

Explanation:
Applying pythagorean theorem let d be diameter of the
larger opening in the lampshade so d² + 20² = 20.8²,
d² = 432.64 – 400 = 32.64 ,so d is square root of 32.64 is 5.71 in.

Question 10.
Four identical pieces are cut off the sides of a cylinder, as shown. The remaining shape is a square prism. The diagonal of the prism’s square base is as long as the diameter of the cylinder. The diameter of the cylinder is 12 centimeters, and the height of the cylinder is 9 centimeters. Find the length of a side of the square base of the prism. Then find an approximate volume of the prism.

Answer:
The length of a side of the square base of the prism 3.82 cm,
Approximate volume of the prism is 67 cm³,

Explanation:
Given Four identical pieces are cut off the sides of a cylinder, as shown.
The remaining shape is a square prism. The diagonal of the prism’s
square base is as long as the diameter of the cylinder.
The diameter of the cylinder is 12 centimeters, and
the height of the cylinder is 9 centimeters.
we have Diameter of cylinder = diagonal of base of square prism
=√2 a, so a = 12/3.14 = 3.82 cm,,
now volume of prism is v = a²h = 3.82 X 3.82 X 4.6 = 67.125 cm³.

Question 11.
Ellen’s computer screen is a 20 inch screen, which means that the length of the diagonal of the rectangular screen is 20 inches. The screen can be laid flat in a box that is 12 inches wide. How long is the box?
Answer:
The box is 16 inch long,

Explanation:
Given Ellen’s computer screen is a 20 inch screen,
which means that the length of the diagonal of the rectangular screen is 20 inches.
The screen can be laid flat in a box that is 12 inches wide. Applying
pythagorean theorem let l is the length of the box so
l² + 12² = 20²,
l² = 400 -144 = 256, so length is square root of 256 is 16 inch.

Question 12.
An 18-centimeter tall suitcase is shaped like a rectangular prism. The length of the diagonal of one of its largest faces is 30 centimeters. Find the width of the suitcase.
Answer:
The width of the suitcase is 24 centimeter,

Explanation:
Given an 18-centimeter tall suitcase is shaped like a rectangular prism.
The length of the diagonal of one of its largest faces is 30 centimeters.
Let w be the width of the suitcase applying pythagores theorem
w² + 18² = 30²,
w² = 900 – 324 = 576, so width is square root of 576 is 24 centimeter.

Question 13.
Vera kicks a ball 123 feet diagonally across a rectangular playground of width 85 feet. Find the length of the playground.
Answer:
The length of the playground is 88.9 feet,

Explanation:
Given Vera kicks a ball 123 feet diagonally across a
rectangular playground of width 85 feet.
Let l be the length of the playground applying pythagorean theorem
l² + 85² = 123²,
l² = 15129 – 7225 = 7904, the value of l is square root of 7904 is 88.9 feet.

Question 14.
A river is 10.5 meters wide and its banks are parallel to each other. John tries to swim straight across, but the current pushes him downstream so that he lands 14 meters from the spot he wanted to reach. How far did he swim?
Answer:
Far did Jhon swim is 17.5 meters,

Explanation:
Given a river is 10.5 meters wide and its banks are parallel to each other.
John tries to swim straight across, but the current pushes him downstream so that he lands 14 meters from the spot he wanted to reach. Applying pythagorean theorem to find how far did he swim s is
s² = 14² + 10.5² = 196 + 110.25=306.25, the value is s square root of 306.25 is 17.5 meters.

Question 15.
An extendable ladder 9.8 feet long leans against a wall with the base 3 feet from the wall. Assuming the base of the ladder does not move, about how much higher will the top of the ladder be above the ground when the ladder extended to twice its original length?
Answer:
higher will the top of the ladder above the ground is 10.04 feet,

Explanation:
Given an extendable ladder 9.8 feet long leans against a wall with the base 3 feet from the wall and assuming the base of the ladder does not move, about how much higher will the top of the ladder be above the ground when the ladder extended to twice its original length, Let h1 is 9.8 feet and h2 is extended length is 2 X 9.8 = 19.6, Applying pythagorean theorem
h₁² =(9.8) ² -(3)²  ( For right triangle)  = 96.04 – 9 = 87.04, so h1 is square root
of 87.04 is 9.32 feet
and h₂² = (19.6)² – (3)² = 384.16 – 9 = 375.16, So h2 is the square root of
375.16 is 19.36 feet, therefore higher will the top of the ladder above
the ground is 19.36 feet – 9.32 feet = 10.04 feet.

Question 16.
A doorway is 80 inches tall and 32 inches wide. Can a round tabletop with a diameter of 90 inches fit through the doorway? If not, what is the greatest possible diameter that will fit through the doorway? Explain.
Answer:
The round tabletop does not fit through the doorway even diagonally,

Explanation:
Given a doorway is 80 inches tall and 32 inches wide,
the diagonal measure d² = 80² + 32² = 6400 + 1024 = 7424,
So d is squareroot of 7424 is 86.16 inches,
So, the round tabletop does not fit through the doorway even diagonally,
The greatest possible diameter of the round doorway is about 86 inches.

Go through the Math in Focus Grade 4 Workbook Answer Key Chapter 5 Practice 5 Probability as a Fraction to finish your assignments.

Math in Focus Grade 4 Chapter 5 Practice 5 Answer Key Probability as a Fraction

Find the probability as a fraction in simplest form.

Jake spins the spinner once. He wants to land on these numbers. What is the probability of a favorable outcome? Find the probability as a fraction in simplest form for each outcome.

Example
He wants to land on a number less than 3.
There are 2 favorable outcomes: 1 and 2
There are 8’ possible outcomes: 1, 2, 3, 4, 5, 6, 7, and 8

Question 1.
He wants to land on the number 7.
Answer:
The probability of a favorable outcome is \(\frac{7}{8}\).

Explanation:
Here, the possible outcomes are \(\frac{1}{8}\),\(\frac{2}{8}\),\(\frac{3}{8}\),\(\frac{4}{8}\),\(\frac{5}{8}\),\(\frac{6}{8}\),\(\frac{7}{8}\),\(\frac{8}{8}\). As he wants to land on the number 7 it will be \(\frac{7}{8}\).

Question 2.
He wants to land on an odd number.
Answer:
The possible outcomes are \(\frac{1}{8}\),\(\frac{3}{8}\),\(\frac{5}{8}\),\(\frac{7}{8}\).

Explanation:
Here, the possible outcomes are \(\frac{1}{8}\),\(\frac{2}{8}\),\(\frac{3}{8}\),\(\frac{4}{8}\),\(\frac{5}{8}\),\(\frac{6}{8}\),\(\frac{7}{8}\),\(\frac{8}{8}\). As he wants to land on an odd numbers so the outcomes will be \(\frac{1}{8}\),\(\frac{3}{8}\),\(\frac{5}{8}\),\(\frac{7}{8}\).

Find the probability as a fraction in simplest form for each outcome.

A coin is tossed once. The probability of getting

Question 3.
heads is

Answer:
\(\frac{1}{2}\).

Explanation:
The probability of getting heads is \(\frac{1}{2}\).

Question 4.
tails is
Answer:
\(\frac{1}{2}\).

Explanation:
The probability of getting tails is \(\frac{1}{2}\).

A number cube numbered 1 to 6 is tossed once. The probability of getting

Question 5.
the number 2 is
Answer:
The probability is \(\frac{2}{6}\).

Explanation:
As the number was tossed once, so the probability of getting the number 2 is \(\frac{2}{6}\).

Question 6.
the number 0 is
Answer:
No possible outcomes.

Explanation:
There will be no possibility of getting the number 0 as the number cube numbered 1 to 6.

Question 7.
an even number is
Answer:
The possibilities of getting even number is \(\frac{2}{6}\),\(\frac{4}{6}\),\(\frac{6}{6}\).

Explanation:
The possible outcomes are \(\frac{1}{6}\),\(\frac{2}{6}\),\(\frac{3}{6}\),\(\frac{4}{6}\),\(\frac{5}{6}\),\(\frac{6}{6}\). So the possibilities of getting even number is \(\frac{2}{6}\),\(\frac{4}{6}\),\(\frac{6}{6}\).

Question 8.
a number greater than 4 is
Answer:
The possibilities of getting a number greater than 4 is \(\frac{5}{6}\),\(\frac{6}{6}\).

Explanation:
The possible outcomes are \(\frac{1}{6}\),\(\frac{2}{6}\),\(\frac{3}{6}\),\(\frac{4}{6}\),\(\frac{5}{6}\),\(\frac{6}{6}\). So the possibilities of getting a number greater than 4 is \(\frac{5}{6}\),\(\frac{6}{6}\).

A circular spinner has 4 equal parts. The parts are colored red, blue, green, and yellow. The spinner is spun once. The probability of landing on

Question 9.
red is
Answer:
The probability of landing on red is \(\frac{r}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on red is \(\frac{r}{4}\).

Question 10.
blue is
Answer:
The probability of landing on blue is \(\frac{b}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on blue is \(\frac{b}{4}\).

Question 11.
purple is
Answer:
The probability of landing on purple is \(\frac{p}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on blue is \(\frac{p}{4}\).

Question 12.
green, red, or yellow is
Answer:
The probability of landing on purple is \(\frac{p}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on blue is \(\frac{p}{4}\).

Question 13.
red, blue, green, or yellow is
Answer:
The probability of landing on red, blue, green, or yellow is \(\frac{r}{4}\), \(\frac{b}{4}\), \(\frac{g}{4}\), \(\frac{y}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on red, blue, green, or yellow is \(\frac{r}{4}\), \(\frac{b}{4}\), \(\frac{g}{4}\), \(\frac{y}{4}\).

Find the probability as a fraction in simplest form for each outcome.

A bag contains 10 discs numbered 1 to 10. A disc is drawn from the bag. The probability of drawing

Question 14.
the number 10 is
Answer:
The probability of drawing the number 10 is \(\frac{10}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10. So the probability of drawing the number 10 is \(\frac{10}{10}\).

Question 15.
a number less than 5 is
Answer:
The probability of drawing the number 5 is \(\frac{5}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10. So the probability of drawing the number 5 is \(\frac{5}{10}\).

Question 16.
on odd number is .
Answer:
The probability of drawing the odd number is \(\frac{1}{10}\), \(\frac{3}{10}\), \(\frac{5}{10}\), \(\frac{7}{10}\), \(\frac{9}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10 the probability outcomes are \(\frac{1}{10}\), \(\frac{2}{10}\), \(\frac{3}{10}\), \(\frac{4}{10}\), \(\frac{5}{10}\), \(\frac{6}{10}\), \(\frac{7}{10}\), \(\frac{8}{10}\), \(\frac{9}{10}\), \(\frac{10}{10}\). So the probability of drawing the odd number is \(\frac{1}{10}\),\(\frac{3}{10}\), \(\frac{5}{10}\), \(\frac{7}{10}\), \(\frac{9}{10}\).

Question 17.
a number divisible by 3 is
Answer:
The probability of drawing the number divisible by 3 is \(\frac{3}{10}\), \(\frac{6}{10}\), \(\frac{9}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10 the probability outcomes are \(\frac{1}{10}\), \(\frac{2}{10}\), \(\frac{3}{10}\), \(\frac{4}{10}\), \(\frac{5}{10}\), \(\frac{6}{10}\), \(\frac{7}{10}\), \(\frac{8}{10}\), \(\frac{9}{10}\), \(\frac{10}{10}\). So the probability of drawing the number divisible by 3 is \(\frac{3}{10}\), \(\frac{6}{10}\), \(\frac{9}{10}\).

Question 18.
a number greater than 8 is
Answer:
The probability of drawing the number greater than 8 is \(\frac{9}{10}\), \(\frac{10}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10 the probability outcomes are \(\frac{1}{10}\), \(\frac{2}{10}\), \(\frac{3}{10}\), \(\frac{4}{10}\), \(\frac{5}{10}\), \(\frac{6}{10}\), \(\frac{7}{10}\), \(\frac{8}{10}\), \(\frac{9}{10}\), \(\frac{10}{10}\). So the probability of drawing the number greater than 8 is \(\frac{9}{10}\), \(\frac{10}{10}\).

Question 19.
the number 12 is .
Answer:
There is no possibility of drawing number 12

Explanation:
Here, a bag contains 10 discs numbered 1 to 10 the probability outcomes are \(\frac{1}{10}\), \(\frac{2}{10}\), \(\frac{3}{10}\), \(\frac{4}{10}\), \(\frac{5}{10}\), \(\frac{6}{10}\), \(\frac{7}{10}\), \(\frac{8}{10}\), \(\frac{9}{10}\), \(\frac{10}{10}\). So there is no possibility of drawing number 12.

A bag contains 3 white marbles, 3 blue marbles, and 6 red marbles. A marble is drawn from the bag. The probability of getting

Question 20.
a white marble is
Answer:
The probability of getting a white marble is \(\frac{3}{12}\).

Explanation:
Given that a bag contains 3 white marbles, 3 blue marbles, and 6 red marbles. So the total number of marbles are 3+3+6 which is 12 marbles. So the probability of getting a white marble is \(\frac{3}{12}\).

Question 21.
a blue marble is
Answer:
The probability of getting a blue marble is \(\frac{3}{12}\).

Explanation:
Given that a bag contains 3 white marbles, 3 blue marbles, and 6 red marbles. So the total number of marbles are 3+3+6 which is 12 marbles. So the probability of getting a white marble is \(\frac{3}{12}\).

Question 22.
Which is more likely: drawing a red marble or drawing a blue marble? Explain.
Answer:
Red marbles.

Explanation:
As there are 3 white marbles, 3 blue marbles, and 6 red marbles. So the more likely drawing marble will be red marbles.

Find the probability of each outcome on the number line. Then describe the outcome as more likely, less likely, certain, impossible, or equally likely.

Example
A box contains 4 red pencils, 1 blue pencil, and 1 black pencil. Find the probability of picking a red pencil.

The closer the probability of an outcome is to 1, the more likely the outcome is to occur.

The probability of picking a red pencil is \(\frac{4}{6}\) or \(\frac{2}{3}\)
\(\frac{2}{3}\) is closer to 1 than to 0 on the number line, So, the likelihood of picking a red pencil is more likely.

Each card in a set of 8 cards has a picture of a fruit. There are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. The cards are shuffled, placed in a stack, and one card is picked.

Question 23.
An orange card: ____
Answer:
The probability of picking the orange card is \(\frac{3}{8}\).

Explanation:
Given that each card in a set of 8 cards has a picture of a fruit and there are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. So the probability of picking the orange card is \(\frac{3}{8}\).

Question 24.
An apple card: ____
Answer:
The probability of picking the apple card is \(\frac{2}{8}\).

Explanation:
Given that each card in a set of 8 cards has a picture of a fruit and there are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. So the probability of picking the apple card is \(\frac{2}{8}\).

Question 25.
An apple, peach, or pear card: ____
Answer:
The probability of picking an apple, peach, or pear card \(\frac{2}{8}\), \(\frac{1}{8}\), \(\frac{2}{8}\).

Explanation:
Given that each card in a set of 8 cards has a picture of a fruit and there are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. So the probability of picking an apple, peach, or pear card \(\frac{2}{8}\), \(\frac{1}{8}\), \(\frac{2}{8}\).

Question 26.
An apple, orange, peach, or pear card: ____
Answer:
The probability of picking an apple, peach, or pear card\(\frac{2}{8}\), \(\frac{3}{8}\), \(\frac{1}{8}\) and \(\frac{2}{8}\).

Explanation:
Given that each card in a set of 8 cards has a picture of a fruit and there are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. So the probability of picking an apple, orange, peach, or pear card \(\frac{2}{8}\), \(\frac{3}{8}\), \(\frac{1}{8}\) and \(\frac{2}{8}\).

Go through the Math in Focus Grade 4 Workbook Answer Key Chapter 5 Data and Probability to finish your assignments.

Math in Focus Grade 4 Chapter 5 Answer Key Data and Probability

Math Journal

Write the steps to solve the problem.

Neil bought 5 books. The average price of 2 of the books is $5. The average price of the rest of the books is $4. Find the total amount of money Neil paid for the 5 books.

Then, following your steps above, solve the problem.

Answer:
The total amount of money Neil paid for the 5 books is $16.

Explanation:
Given that Neil bought 5 books and the average price of 2 of the books is $5 and the average price of the rest of the books is $4. So the total amount of money Neil paid for the 5 books is
as the price of 2 books is $5 and the price of 1 book is $4. So 5-2 is 3, and the price of the 4 books will be 4×3 which is $12. So the cost for 5 books it will be $12+$4 which is $16.

Put On Your Thinking Cap!

Challenging Practice

Question 1.
Michelle got an average score of 80 on two tests. What score must she get on the third test so that her average score for the three tests is the same as the average score for the first two tests?
Answer:
The score she got on the third test is 80.

Explanation:
Given that Michelle got an average score of 80 on two tests and the sum of score in 2 tests will be 80×2 which is 160. Let the score for third test be x,
so the new average will be \(\frac{160+x}{3}\),
and the average score is 80, so \(\frac{160+x}{3}\) = 80
160+x = 80×3
160+x = 240
x = 240 – 160
= 80.

Question 2.

The line plot shows the shoe sizes of students in Ms. George’s class.

a. How many students are in the class?
Answer:
25 students.

Explanation:
The total number of students are in the class is 25 students.

b. What is the mode of the set of data?
Answer:
3\(\frac{1}{2}\).

Explanation:
The mode of the set of data is 3\(\frac{1}{2}\) as the number that appears most often.

c. How many students in the class wear a size 3\(\frac{1}{2}\) shoe?
Answer:
10 students.

Explanation:
The number of students in the class wear a size 3\(\frac{1}{2}\) shoe is 10 students.

d. Suppose you looked at 100 pairs of shoes for the grade, which includes 3 other classes. How many pairs of size 3\(\frac{1}{2}\) would there be? Explain your answer.
Answer:

Put On Your Thinking Cap!

Problem Solving

Question 1.
The average height of Andy, Chen, and Chelsea is 145 centimeters. Andy and Chen are of the same height and Chelsea is 15 centimeters taller than Andy. Find Andy’s height and Chelsea’s height.
Answer:
The Andy’s height and Chelsea’s height is 140 cm.

Explanation:
Given that the average height of Andy, Chen, and Chelsea is 145 centimeters and Andy and Chen are of the same height and Chelsea is 15 centimeters taller than Andy, so let the height of Andy and Chen be x and the height of Chelsea is 15 centimeters taller than Andy which is x+15. So Andy’s height and Chelsea’s height will be
\(\frac{x+x+x+15}{3}\) = 145
\(\frac{3x+15}{3}\) = 145
3x+15 = 145×3
3x+15 = 435
3x = 435-15
3x = 420
x = 420÷3
= 140.
So the Andy’s height and Chelsea’s height is 140 cm.

Question 2.
Eduardo has 3 times as many stamps as Sally. The average number of stamps they have is 450. How many more stamps does Eduardo have than Sally?
Answer:
\(\frac{1}{2}\) of total number of stamps extra.

Explanation:
Given that Eduardo has 3 times as many stamps as Sally and the average number of stamps they have is 450. Here, if Sally has 1 stamp then Eduardo has 3 stamps. So total stamps will be 4, Eduardo has 2 extra and Eduardo has \(\frac{2}{4}\) which is \(\frac{1}{2}\) of total number of stamps extra.

Question 3.
Bag A and Bag B each contain 2 marbles — 1 white and 1 red. Troy picks 1 marble from Bag A and 1 from Bag B. What is the probability that the following are picked?

a. 2 white marbles
Answer:
\(\frac{2}{4}\)

Explanation:
The probability of picking up 2 white marbles is \(\frac{2}{4}\).

b. 1 red and 1 white marble
Answer:

Explanation:
The probability of picking up 1 red and 1 white marble is

Practice the problems of Math in Focus Grade 4 Workbook Answer Key Chapter 6 Practice 3 Mixed Numbers to score better marks in the exam.

Math in Focus Grade 4 Chapter 6 Practice 3 Answer Key Mixed Numbers

Write a mixed number lor each model.

Example
Write a mixed number for each model

Question 1.

Answer:

Explanation:
1 + 1 + 1 + \(\frac{3}{4}\) = 3 + \(\frac{3}{4}\)
= 3\(\frac{3}{4}\)

Question 2.

Answer:

Explanation:
1 + 1 + 1 + \(\frac{2}{5}\) = 3 + \(\frac{2}{5}\)
= 3\(\frac{2}{5}\)

Write a mixed number for each model.
Question 3.

Answer:

Explanation:
1 + 1 + 1 + \(\frac{1}{2}\) = 3 + \(\frac{1}{2}\)
= 3\(\frac{1}{2}\)

Question 4.

Answer:

Explanation:
1 + \(\frac{3}{5}\) = 1 + \(\frac{3}{5}\)
= 1\(\frac{3}{5}\)

Question 5.

Answer:

Explanation:
1 + 1 + 1 + \(\frac{5}{9}\) = 3 +\(\frac{5}{9}\)
= 3\(\frac{5}{9}\)

Check (✓) the correct model.
Question 6.
Which model shows 1\(\frac{3}{4}\) shaded?

Answer:

Explanation:
Model 1:
1 + 1 + 1 + \(\frac{1}{4}\) = 3 + \(\frac{1}{4}\)
= 3\(\frac{1}{4}\)

Model 2:
1 +  \(\frac{1}{4}\) = 1 +\(\frac{1}{4}\)
= 1\(\frac{1}{4}\)

Question 7.
Which model shows 2\(\frac{3}{5}\) shaded?

Answer:

Explanation:
Model 1:
1 + 1 + \(\frac{3}{5}\) = 2 + \(\frac{3}{5}\)
= 2\(\frac{3}{5}\)
Model 2:
1 + 1 + \(\frac{2}{5}\)  = 2 + \(\frac{2}{5}\)
= 2\(\frac{2}{5}\)

Write each answer as a mixed number.
Question 8.
4 + \(\frac{1}{4}\) = ____
Answer:
4 + \(\frac{1}{4}\) = 4\(\frac{1}{4}\)

4 + \(\frac{1}{4}\)
=4\(\frac{1}{4}\)

Question 9.
3 + \(\frac{5}{9}\) = ____
Answer:
3 + \(\frac{5}{9}\) = 3\(\frac{5}{9}\)

Explanation:
3 + \(\frac{5}{9}\)
= 3\(\frac{5}{9}\)

Question 10.
\(\frac{5}{8}\) + 2 = ____
Answer:
\(\frac{5}{8}\) + 2 = 2\(\frac{5}{8}\)

Explanation:
\(\frac{5}{8}\) + 2
= 2\(\frac{5}{8}\)

Question 11.
\(\frac{3}{5}\) + 4 = ____
Answer:
\(\frac{3}{5}\) + 4 = 4\(\frac{3}{5}\)

Explanation:
\(\frac{3}{5}\) + 4
= 4\(\frac{3}{5}\)

Write the correct mixed number in each box.
Question 12.

Answer:

Explanation:
1 + \(\frac{4}{7}\) = 1 \(\frac{4}{7}\)
2 + \(\frac{6}{7}\) = 2\(\frac{6}{7}\)

Write a mixed number for each item.
Question 13.
The pears have a weight of

Answer:

Explanation:
Weight of the pears = 1 + \(\frac{2}{4}\)
= 1 \(\frac{1}{2}\) pounds.

Question 14.
The worm started crawling from 0 centimeters.
It has crawled _____ centimeters.

Answer:
It has crawled 7\(\frac{7}{1}\) or 7.7 centimeters.

Explanation:
Number of centimeters the worm travelled = 0 + 7\(\frac{14}{2}\)
= 7\(\frac{7}{1}\) or 7.7 centimeters.

Write each mixed number in simplest form.
Example

Question 15.

Answer:

Explanation:
2\(\frac{4}{6}\) = 2\(\frac{2}{3}\)

Question 16.
3\(\frac{4}{8}\) = ____
Answer:
3\(\frac{4}{8}\) = 3\(\frac{1}{2}\)

Explanation:
3\(\frac{4}{8}\) = 3\(\frac{1}{2}\)

Question 17.
5\(\frac{6}{9}\) = ____
Answer:
5\(\frac{6}{9}\) = 5\(\frac{2}{3}\)

Explanation:
5\(\frac{6}{9}\) = 5\(\frac{2}{3}\)

Question 18.
6\(\frac{4}{12}\) = ____
Answer:
6\(\frac{4}{12}\) = 6\(\frac{1}{3}\)

Explanation:
6\(\frac{4}{12}\) = 6\(\frac{1}{3}\)

Question 19.
4\(\frac{3}{6}\) = ____
Answer:
4\(\frac{3}{6}\) = 4\(\frac{1}{2}\)

Explanation:
4\(\frac{3}{6}\) = 4\(\frac{1}{2}\)

Write each fraction and mixed number in a box to show its correct location on the number line.
Question 20.
1\(\frac{1}{2}\)
Answer:

Explanation:
1\(\frac{1}{2}\) = (2 + 1) ÷ 2
= 3 ÷ 2
= 1.5.

Question 21.
\(\frac{1}{2}\)
Answer:

Explanation:
\(\frac{1}{2}\) = 0.5.

Question 22.
1\(\frac{3}{4}\)
Answer:

Explanation:
1\(\frac{3}{4}\) = (4 + 3) ÷ 4
= 7 ÷ 4
= 1.75.

Fill in the boxes with fractions or mixed numbers. Express each answer in simplest form.
Example

Question 23.

Answer:

Explanation:
2 + \(\frac{1}{2}\)
= 2\(\frac{2}{4}\)
= 2\(\frac{1}{2}\)

3 + \(\frac{3}{4}\)
= 3 \(\frac{3}{4}\)

Question 24.

Answer:

Explanation:
3 +  \(\frac{1}{4}\)
= 3 \(\frac{1}{4}\)
4 + \(\frac{1}{2}\)
= 4 \(\frac{1}{2}\)

Question 25.

Answer:

Explanation:
5 +  \(\frac{1}{2}\)
= 5 \(\frac{1}{2}\)
5 + \(\frac{3}{5}\)
= 5 \(\frac{3}{5}\)