Hillary

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 1 Exponents to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 1 Answer Key Exponents

Math in Focus Grade 8 Chapter 1 Quick Check Answer Key

Identify the integers in the list below.

Question 1.
4.31, -6, 8.12, 58, -211,43
Answer:
-211,-6,43 and 58,

Explanation:
Integers are positive,negative and zero but not decimal numbers therefore
from the given numbers 4.31 and 8.12 are not integers
the integers are -211,-6,43 and 58.

Write each rational number as a terminating decimal.

Question 2.
\(\frac{3}{5}\)
Answer:
0.6,

Explanation:

when we divide 3 by 5 we get terminating decimal as 0.6.

Question 3.
4\(\frac{1}{4}\)
Answer:
4.25,

Explanation:
Given 4\(\frac{1}{4}\) we get \(\frac{4 x 4 + 1}{4}\) =
\(\frac{17}{4}\)
the terminating decimal is 4.25.

Write each rational number as a repeating decimal.

Question 4.
\(\frac{6}{11}\)
Answer:
The repeating decimal of 6/11 is 0.5363636 and so on,

Explanation:
Given to find \(\frac{6}{11}\) we get

the repeating decimal of 6/11 is 0.5363636 and so on.

Question 5.
\(\frac{5}{12}\)
Answer:
The repeating decimal of 5/12 is 0.4166 so on,

Explanation:

The repeating decimal of 5/12 is 0.4166 so on.

Locate each irrational number on a number line using rational approximation.

Question 6.

Answer:

Explanation:
Located irrational number cube root of -47 on the
number line – 3.6 using rational approximation.

Question 7.
\(\sqrt{19}\)
Answer:

Explanation:
Located irrational number square root of 19 on the
number line 4.35 using rational approximation.

Evaluate each expression.

Question 8.
-3 + (-4)
Answer:
-7,

Explanation:
Adding  -3 to -4 we get -7.

Question 4.
-4 – (-2)
Answer:
-6,

Explanation:
Adding -4 to -2 we get -6.

Evaluate each expression.

Question 10.
(-7) • (-3)
Answer:
21,

Explanation:
Given -7 multiplied by -3 we get – 7 X -3 = 21.

Question 11.
(-12) ÷ 3
Answer:
-4,

Explanation:
Given to divide (-12) with 3 we get -4.

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 1 Lesson 1.4 The Power of a Product and the Power of a Quotient to finish your assignments.

Math in Focus Grade 7 Course 3 A Chapter 1 Lesson 1.4 Answer Key The Power of a Product and the Power of a Quotient

Math in Focus Grade 8 Chapter 1 Lesson 1.4 Guided Practice Answer Key

Simplify each expression. Write your answer in exponential notation.

Question 1.
6³ • 7³
6³ • 7³ = Use the power of a property.
= Simplify.
Answer:
(6 X 7)³ = (42)³,

Explanation:
Given 6³ • 7³  when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property,
(6 X 7)³ = (42)³.

Question 2.

Answer:
(5/24)⁴,

Explanation:
Given (-5/6)⁴. (- 1/4)⁴ when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product and quotients of a property,
(-5/6 X -1/4)⁴ = (5/24)⁴.

Question 3.
(1.8)² • (0.75)²
(1.8)² • (0.75)² = Use the power of a property.
= Simplify.
Answer:
((1.8) • (0.75))²,

Explanation:
Given (1.8)² • (0.75)² when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product of a property,
((1.8) • (0.75))².

Simplify each expression. Write your answer in exponential notation.

Question 4.
p⁶ • q⁶
p⁶ • q⁶ = Use the power of a property.
= Simplify.
Answer:
(p X q)⁶ ,

Explanation:
Given p⁶ • q⁶  when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property,
(p X q)⁶ .

Question 5.
(3a)⁴ • (4b)⁴
(3a)⁴ • (4b)⁴ = Use the power of a property.
= Simplify.
Answer:
(3a X 4b)⁴ ,

Explanation:
Given (3a)⁴ • (4b)⁴  when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property we get (3aX 4b)⁴.

Question 6.
(-3y²)³ • \(\left(\frac{1}{12 y}\right)^{3}\)
Answer:
\(\left(-\frac{y}{4}\right)^{3}\),

Explanation:
Given (-3y²)³ • \(\left(\frac{1}{12 y}\right)^{3}\) on simplification
(-y)^{2 X 3} • \(\left(\frac{1}{4 y}\right)^{3}\) = (-y)^6-3 • \(\left(\frac{1}{4}\right)^{3}\) =
(-y)³ • \(\left(\frac{1}{4}\right)^{3}\) = \(\left(-\frac{y}{4}\right)^{3}\).

Simplify each expression. Write your answer in exponential notation.

Question 7.
2⁵ ÷ 4⁵

Answer:
\(\left(\frac{1}{2}\right)^{5}\),

Explanation:
Given 2⁵ ÷ 4⁵  as 4 can be written as (2 X 2) so 2⁵ ÷ (2 X 2)⁵ upon simplification
we get 1/2⁵ or \(\left(\frac{1}{2}\right)^{5}\).

Question 8.
(-9)³ ÷ (-3)³

Answer:
(3)³,

Explanation:
Given(-9)³ ÷ (-3)³  as (-9) can be written as (-3 X 3) so (-3 X 3)³ ÷ (-3)³
on simplification we get (3)³.

Simplify each expression. Write your answer in exponential notation.

Question 9.
x⁴ ÷ y⁴

Answer:
\(\left(\frac{x}{y}\right)^{4}\),

Explanation:
Given x⁴ ÷ y⁴ when the variable bases are different and the
powers are the same, the bases are simplified first as (x ÷ y)⁴ and solved.

Question 10.
(8p)⁵ ÷ (3q)⁵
Answer:
(8p ÷ 3q)⁵,

Explanation:
Given (8p)⁵ ÷ (3q)⁵ when the variable bases are different and the
powers are the same, the bases are simplified first as (8p ÷ 3q)⁵ and solved.

Simplify each expression. Write your answer in exponential notation.

Question 11.

Answer:
(2)⁷,

Explanation:

Question 12.

Answer:
(4)³ X (3)³,

Explanation:

Question 13.

Answer:
(6)⁸,

Explanation:

Question 14.
\(\frac{2^{5} \cdot 3^{12} \cdot 2^{7}}{6^{7}}\)
Answer:
(6)⁵,

Explanation:
Given \(\frac{2^{5} \cdot 3^{12} \cdot 2^{7}}{6^{7}}\),
6 can be written as 2 X 3 so \(\frac{2^{5} \cdot 3^{12} \cdot 2^{7}}{({2 X 3})^{7}}\),

Question 15.
\(\frac{\left(25^{3}\right)^{2} \cdot 7^{6}}{5^{6}}\)
Answer:
(35)⁶,

Explanation:
Given \(\frac{\left(25^{3}\right)^{2} \cdot 7^{6}}{5^{6}}\)

Math in Focus Course 3A Practice 1.4 Answer Key

Simplify each expression. Write your answer in exponential notation.

Question 1.
5⁴ • 6⁴
Answer:
(5 X 6)⁴ = (30)⁴,

Explanation:
Given 5⁴ • 6⁴  when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property,
(5 X 6)⁴ = (30)⁴.

Question 2.
5.4³ • 4.5³
Answer:
(5.4 X 4.5)³ = (24.3)³,

Explanation:
Given 5.4³ • 4.5³  when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property,
(5.4 X 4.5)³ = (24.3)³.

Question 3.
2⁵ • 10⁵
Answer:
(2 X 10)⁵ = (20)⁵,

Explanation:
Given 2⁵ • 10⁵  when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property,
(2 X 10)⁵ = (20)⁵.

Question 4.
a³ • b³
Answer:
(a X b)³ = (ab)³,

Explanation:
Given a³ • b³  when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property,
(a X b)³ = (ab)³.

Question 5.
(2x)⁵ • (3y)⁵
Answer:
(2x X 3y)⁵ = (6xy)⁵,

Explanation:
Given (2x)⁵ • (3y)⁵  when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property,
(2x X3y)⁵ = (6xy)⁵.

Question 6.
(2.5a)⁶ • (1.6b)⁶
Answer:
(2.5a X 1.6b)⁶ = (4ab)⁶,

Explanation:
Given (2.5a)⁶ • (1.6b)⁶  when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property,
(2.5a X 1.6b)⁶ = (4ab)⁶.

Question 7.
\(\left(-\frac{1}{3}\right)^{4} \cdot\left(-\frac{2}{5}\right)^{4}\)
Answer:
\(\left(\frac{2}{15}\right)^{4}\),

Explanation:
Given \(\left(-\frac{1}{3}\right)^{4} \cdot\left(-\frac{2}{5}\right)^{4}\)
when the variable bases are different and the powers are the same,
the bases are multiplied first,Used the powers of product property,
\(\left(\frac{-1 X -2}{3X 5}\right)^{4}\) = \(\left(\frac{2}{15}\right)^{4}\).

Question 8.
9² ÷ 3²
Answer:
(3)²,

Explanation:
Given (9)² ÷ (3)²  as 9 = 3 X 3 can be written as (3²)² so (3)⁴ ÷ (3)²
on simplification we get (3)^4-2 = (3)².

Question 9.
10⁶ ÷ 5⁶
Answer:
(2)⁶,

Explanation:
Given (10)⁶ ÷ (5)⁶  as 10 can be written as (5 x 2)⁶ so (5)⁶ .(2)⁶ ÷ (5)⁶
on simplification we get (2)⁶.

Question 10.
2.8⁷ ÷ 0.7⁷
Answer:
(4)⁷,

Explanation:
Given (2.8)⁷ ÷ (0.7)⁷  as 2.8 ÷ 0.7 = 4 so can be written as (4)⁷ .

Question 11.
15² ÷ 25²
Answer:
\(\left(\frac{3}{5}\right)^{2}\),

Explanation:
Given 15² ÷ 25², 15² can be written as (5 X 3)² and 25²  as (5 X 5)²  ,
upon simplification we get \(\left(\frac{3}{5}\right)^{2}\).

Question 12.
7.2⁹ ÷ 2.4⁹
Answer:
(3)⁹,

Explanation:
Given (7.2)⁹ ÷ (2.4)⁹  as 7.2 ÷ 2.4 = 3, so can be written as (3)⁹.

Question 13.
(3.2x)⁹ ÷ (1.1y)⁹
Answer:
(3.2x ÷ 1.1y)⁹,

Explanation:
Given (3.2x)⁹ ÷ (1.1y)⁹ when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property (3.2x ÷ 1.1y)⁹.

Question 14.
(-6)⁸ ÷ (-2)⁸
Answer:
(3)⁸,

Explanation:
Given (-6)⁸ ÷ (-2)⁸  as -6 can be written as (-2 x 3)⁸ so (-2)⁸ .(3)⁸ ÷ (-2)⁸
on simplification we get (3)⁸.

Question 15.
s⁵ ÷ r⁵
Answer:
\(\left(\frac{s}{r}\right)^{5}\),

Explanation:
Given s⁵ ÷ r⁵ the variable bases are different and the
powers are the same, the bases are divided first,upon simplification we get
\(\left(\frac{s}{r}\right)^{5}\).

Question 16.
(3a)⁶ ÷ (2b)⁶
Answer:
\(\left(\frac{3a}{2b}\right)^{6}\),

Explanation:
Given (3a)⁶ ÷ (2b)⁶ the variable bases are different and the
powers are the same, the bases are divided first, upon simplification we get
\(\left(\frac{3a}{2b}\right)^{6}\).

Question 17.
(h²k⁵)⁴
Answer:
h⁸k²⁰,

Explanation:
Given (h²k⁵)⁴ when the variable bases are different and the
powers are the same,Used the powers of product and
quotients of a property we get h^{2X 4} X k^5X4 = h⁸k²⁰.

Question 18.
\(\left(\frac{32 m^{6}}{4 n^{4}}\right)^{2}\)
Answer:
\(\left(\frac{8 m^{6}}{4n^{4}}\right)^{2}\),

Explanation:
Given \(\left(\frac{32 m^{6}}{4 n^{4}}\right)^{2}\) upon simplification
we get \(\left(\frac{8 m^{6}}{4n^{4}}\right)^{2}\).

Question 19.
\(\frac{9^{2} \cdot 9^{7}}{3^{5} \cdot 3^{4}}\)
Answer:
(3)⁹,

Explanation:
Given \(\frac{9^{2} \cdot 9^{7}}{3^{5} \cdot 3^{4}}\)
9 can be written as \(\frac{3 X 3 ^{2} \cdot 3 X 3^{7}}{3^{5} \cdot 3^{4}}\)
when bases are same powers are added \(\frac{3^{2+2+7+7}}{3^{5+4}}\) =
\(\frac{3^{18}}{3^{9}}\) on further simplification we get (3)^18-9 = (3)^9.

Question 20.
\(\frac{6^{5} \cdot 2^{3} \cdot 6^{4}}{12^{3}}\)
Answer:
(6)⁶,

Explanation:
Given\(\frac{6^{5} \cdot 2^{3} \cdot 6^{4}}{12^{3}}\) upon simplification
\(\frac{6^{5+4} \cdot 2^{3} }{6 X 2^{3}}\) = \(\frac{6^{9} \cdot 2^{3} }{(6 X 2)^{3}}\)=
(6)^9-3 = (6)⁶.

Question 21.
\(\frac{\left(5^{4}\right)^{2} \cdot 6^{8}}{10^{8}}\)
Answer:
(3)⁸,

Explanation:
Given \(\frac{\left(5^{4}\right)^{2} \cdot 6^{8}}{10^{8}}\) as denominator
10 can be written as 5 X 2 so \(\frac{\left(5\right)^{8} \cdot 6^{8}}{(5 X 2)^{8}}\) we get
\(\frac{6^{8}}{2^{8}}\) = \(\frac{(2 X 3)^{8}}{2^{8}}\) = (3)⁸.

Question 22.
\(\frac{\left(6^{3}\right)^{3} \cdot 4^{9}}{8^{9}}\)
Answer:
(3)⁹,

Explanation:
Given \(\frac{\left(6^{3}\right)^{3} \cdot 4^{9}}{8^{9}}\) as denominator
8 can be written as 4 X 2 so  we get
\(\frac{\left((2 X 3)^{3}\right)^{3} \cdot 4^{9}}{(4 X 2)^{9}}\) = (3)⁹.

Question 23.
\(\frac{24^{9}}{4^{3} \cdot 6^{2} \cdot 4^{6}}\)
Answer:
(6)⁷,

Explanation:
Given \(\frac{24^{9}}{4^{3} \cdot 6^{2} \cdot 4^{6}}\) as
24 in numerator can be written as 4 X 6 so

Question 24.
\(\frac{9^{12}}{\left(3^{3}\right)^{3} \cdot 3^{3}}\)
Answer:
(3)¹²,

Explanation:

Question 25.
Math Joural Charles thinks that a³ • b³ = ab⁶. Is he correct? Why?
Answer:
No, Charles is incorrect,

Explanation:
a³ • b³ ≠ ab⁶ it should be (ab)³ as when the variable bases are different and the
powers are the same, the bases are multiplied first,
Used the powers of product property we get (aX b)³ ≠ ab⁶.

Solve. Show your work.

Question 26.
At the beginning of January, Mr. Howard gives his niece $1 to start a savings account.
For each month that she can triple the amount in the account,
Mr. Howard will double the amount in the account at the end of each month.
How much does Mr. Howard’s niece have in her account at the beginning of May?
Answer:
$7,776 Mr. Howard’s niece have in her account at the beginning of May,

Explanation:
Given at the beginning of January, Mr. Howard gives his niece $1 to start a savings account.
For each month that she can triple the amount in the account,
Mr. Howard will double the amount in the account at the end of each month.
So at the beginning of January it is

therefore,$7,776 Mr. Howard’s niece have in her account at the beginning of May.

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 1 Lesson 1.6 Real-World Problems: Squares and Cubes to finish your assignments.

Math in Focus Grade 7 Course 3 A Chapter 1 Lesson 1.6 Answer Key Real-World Problems: Squares and Cubes

Math in Focus Grade 8 Chapter 1 Lesson 1.6 Guided Practice Answer Key

Solve. Show your work.

Question 1.
Find the two square roots of 169.

Answer:
Explanation:
Given to find two square roots of 169 and -169,So using prime factorization
we get for 169 it is 13 and for – 169 it is -13

Solve. Show your work.

Question 2.
Find the cube root of \(\frac{1}{729}\).

Answer:
1/9,

Explanation:

Solve. Show your work.

Question 3.
x² = 2.25
x² = 2.25

Answer:
1.5,

Explanation:

Solve. Show your work.

Question 4.

Answer:
x = (1/2),

Explanation:

Question 5.
A square field has an area of 98.01 square meters. Find the length of each side of the field.
Let the length of each side be x meters.
x² = Translate into an equation.
= Solve for x by taking the positive root of both sides.
x = m Use a calculator to find the square root.
The length of each side is meters.
Answer:
x² = 98.01 We translate into an equation:
\(\sqrt{x^{2}}\) = \(\sqrt{98.01}\) We solve for x by taking the positive square root of both sides:
x = 9.9m We use a calculator to find the square root:
The length of each side is 9.9 meters.
9.9 meters

Solve. Show your work.

Question 6.
Robin bought a crystal globe that has a volume of 1,774\(\frac{2}{3} \pi\)ir cubic centimeters. Find the radius of the crystal globe.
Let the radius of the crystal globe be r centimeters.

Answer:
The radius of the crystal globe is 29.7 cms,

Explanation:

Question 7.
A spherical waterme’on has a volume of 562.5π cubic centimeters. What is the diameter of the watermelon?

Let the radius of the watermelon be r centimeters.

Answer:
The diameter of the watermelon is 15 centimeters,

Explanation:

Math in Focus Course 3A Practice 1.6 Answer Key

Find the two square roots of each number. Round your answer to the nearest tenth when you can.

Question 1.
25
Answer:
5, or -5,

Explanation:
Given to find sqaure roots of 25
1. 5 X 5 = 25,
2. -5 X -5 = 25,
so 5, or -5.

Question 2.
64
Answer:
8 or -8,

Explanation:
Given to find sqaure roots of 64
1. 8 X 8 = 64,
2. -8 X -8 = 64,
so 8, or -8.

Question 3.
80
Answer:
8.944 or -8.944,

Explanation:
Given to find sqaure roots of 64
1. 8.944 X 8.944 = 80,
2. -8.944 X -8.944 = 80,
so 8.944, or -8.944.

Question 4.
120
Answer:
10.9544 or -10.9544,

Explanation:
Given to find sqaure roots of 64
1. 10.9554 X 10.9554 = 120,
2. -10.9554X -10.9554 = 120,
so 10.9554, or -10.5994.

Find the cube root of each number. Round your answer to the nearest tenth when you can.

Question 5.
512
Answer:
8

Explanation:

Question 6.
1,000
Answer:
10

Explanation:

Question 7.
999
Answer:
9.997,

Explanation:

Question 8.
\(\frac{64}{343}\)
Answer:
\(\frac{4}{7}\),

Explanation:
As

So Cube root of \(\frac{64}{343}\) is \(\frac{4}{7}\).

Solve each equation involving a variable that ¡s squared.
Round your answer to the nearest tenth when you can.

Question 9.
a² = 46.24
Answer:
a= 6.8,

Explanation:

Question 10.
b² = \(\frac{25}{49}\)
Answer:
b= \(\frac{5}{7}\),

Explanation:
Given b² = \(\frac{25}{49}\) as b² = \(\frac{5X5}{7X7}\),
therefore b = square root of \(\frac{5X 5}{7 X 7}\) = \(\frac{5}{7}\),

Question 11.
m² = 196
Answer:
m = 14,

Explanation:

Therefore m = 14.

Question 12.
n² = 35o
Answer:
n= 18.708,

Explanation:

therefore n = 18.708.

Solve each equation involving a variable that is cubed.
Write fractions in simplest form, and round decimal answers
to the nearest tenth.

Question 13.
x³ = 74.088
Answer:
x =4.2,

Explanation:

the cube root of 74.088 is 4.2.

Question 14.
x³ = \(\frac{216}{729}\)
Answer:
x = \(\frac{6}{9}\),

Explanation:

the cube root of x³ = \(\frac{216}{729}\) so x = \(\frac{6}{9}\).

Question 15.
x³ = 1,728
Answer:
x= 12,

Explanation:

Question 16.
x³ = 2,500
Answer:
x= 13.572,

Explanation:

the cube root x = 13.572088083.

Solve. Show your work. Round to the nearest tenth.

Question 17.
The volume of a spherical tank is 790.272π cubic feet.
What is the diameter of the container?

Answer:
The diameter of the container is 16.8 ft,

Explanation:
Given the volume of a spherical tank is 790.272π cubic feet.
so radius is 4/3 pi r³ = 790.272π,
r³ = (790.272 X 3)/4 = 592.704,
r= 8.4,

diameter of the container is 2r = 2 X 8.4 = 16.8 ft.

Question 18.
An orchard planted on a square plot of land has 3,136 apple trees.
If each tree requires an area of 4 square meters to grow,
find the length of each side of the plot of land.

Answer:
The length of each side of plot is 112 meters,

Explanation:
Given an orchard planted on a square plot of land has 3,136 apple trees.
If each tree requires an area of 4 square meters to grow,
the length of each side of the plot of land of square root of 3136 X 4
sqaure root of 12544 = 112 meters.

Question 19.
Mr. Berman deposited $2,500 in a savings account. Three years later there
was $2,812.16 in the savings account. Use the formula A = P(1 + r)ⁿ to find
the rate of interest, r percent, that he was paid. A represents the final amount of the
investment, P is the original principal, and n is the number of years it was invested.
Answer:
Rate of intreset is 5.78%,

Explanation:
Given Mr. Berman deposited $2,500 in a savings account. Three years later there
was $2,812.16 in the savings account. Using the formula A = P(1 + r)ⁿ to find
the rate of interest, r percent, that he was paid. A represents the final amount of the
investment, P is the original principal, and n is the number of years it was invested,
$2812.16 = $2,500(1 + r)³
$2812.16 – $2,500 = (1 + r)³
(1 + r)³ = $312.16
1 + r = cube root of $312.16,

1 + r = 6.7835,
therefore r = 6.7835 – 1 = 5.7835.

Brain @ Work

Question 1.
Evaluate \(\frac{4^{3} \cdot 10^{4}}{5^{2}}\) without using a calculator.
Answer:
2¹⁰ X 5²,

Explanation:

Question 2.
Find the values of x and y that make the equation \(\frac{81 x^{4} \cdot 16 y^{4}}{\left[(2 y)^{2}\right]^{2}}\) = 1,296 true,
Answer:
X = 1.414 and y = 1,

Explanation:

Question 3.
Use each of the numbers 1 to 9 exactly once to fill in the blanks.

Answer:
3¹¹,

Explanation:
Used each of the numbers 1 to 9 exactly once to fill the blanks

Question 4.
Jeremy wants to measure the radius of a marble. He uses a tank and
filled the tank with 360 identical marbles shown below, If the volume of the
tank is 9,720 cubic inches, find the radius of each marble.

Answer:
Radius is 3 inches,

Explanation:
Given Jeremy wants to measure the radius of a marble. He uses a tank and
filled the tank with 360 identical marbles shown below, If the volume of the
tank is 9,720 cubic inches, volume of each cube is 9,720 ÷ 360 = 27,
radius is cube root of 27 = 3 inches.

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 1 Lesson 1.5 Zero and Negative Exponents to finish your assignments.

Math in Focus Grade 7 Course 3 A Chapter 1 Lesson 1.5 Answer Key Zero and Negative Exponents

Math in Focus Grade 8 Chapter 1 Lesson 1.5 Guided Practice Answer Key

Hands-On Activity

Understand the zero exponent

Work individually.

Step 1.
Use the power of a quotient property to simplify each expression. Write the quotient as an exponent.

Step 2.
In factored form, the quotient \(\frac{3^{5}}{3^{5}}\) is \(\frac{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3}{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3}\). If you divide out all the common factors in the numerator and denominator, what is the value of \(\frac{3^{5}}{3^{5}}\)?

Step 3.
Based on your findings, what can you conclude about the value of 3°?

Step 4.
Make a prediction about the value of any nonzero number raised to the zero power. Then, use a calculator to check your prediction for several numbers. For example, to raise the number —2 to the zero power, you can enter these keystrokes:

Does your prediction hold true?
Answer:
Step 1:

Wrote the quotient as an exponent of 3° used exponent \(\frac{3^{5}}{3^{5}}\),
Step 2:
Value is 1,
Step 3:
The value of 3° is 1.
Step 4:
Yes,

Explanation:
Step 1:
Used the power of a quotient property to simplify each expression.
Wrote the quotient as an exponent of 3° used exponent \(\frac{3^{5}}{3^{5}}\),
Step2:
In factored form, the quotient \(\frac{3^{5}}{3^{5}}\) is \(\frac{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3}{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3}\). If you divide out all the common factors in the numerator and denominator the value of \(\frac{3^{5}}{3^{5}}\) is 1 ,
Step 3 :
Any number to power 0 the value is 1, so the value of 3° is 1,
Step 4:
Makeing a prediction about the value of any nonzero number raised to the zero power.
Then, using a calculator to check your prediction for several numbers.
For example, to raise the number —2 to the zero power, you can enter these keystrokes:

Yes my predictions hold true as
Any number to power 0 the value is 1.

Simplify each expression and evaluate where applicable.

Question 1.

Answer:
(0.4)^-2,

Explanation:

Question 2.

Answer:
1,

Explanation:

Question 3.
\(\frac{t^{0} \cdot t^{7}}{t^{5}}\)
Alternatively, you can also apply the product of powers property to the expression t° ∙ t⁷ to get t⁰⁺⁷ = t⁷.
Answer:
t²,

Explanation:
Given\(\frac{t^{0} \cdot t^{7}}{t^{5}}\) as t⁰ is equal to 1  so
we get 1 X t^7-5= t².

Hands-On Activity

Work individually.

Step 1.
Use the quotient of powers property to simplify each expression. Write the quotient in exponential notation.

Answer:

Explanation:
Wrote the quotient in exponential notation above,
As (4)⁵÷ (4)⁶ = (4)^5-6= (4)^-1 used negative exponent.

Step 2.
In factored form, the quotient \(\frac{4^{5}}{4^{6}}\) is \(\frac{4 \cdot 4 \cdot 4 \cdot 4 \cdot 4}{4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4}\). If you divide out all the common factors in the numerator and denominator, what is the value of
\(\frac{4^{5}}{4^{6}}\)?

Answer:
The value is (4)^-1,

Explanantion:
(4)⁵÷ (4)⁶ = (4)^5-6= (4)^-1 used negative exponent.

Step 3.
Repeat Step 2 for \(\frac{4^{5}}{4^{7}}\). What is the value of \(\frac{4^{5}}{4^{7}}\)?

Answer:
The value is (4)^-2,

Explanantion:
(4)⁵÷ (4)⁷ = (4)^5-7= (4)^-2.

Math Journal
Suppose a represents any nonzero number. How would you write a^-3 using a positive exponent?

Simplify each expression and evaluate where applicable.
Answer:
1 ÷ a³ ,

Explanation:
Given  a^-3 using a positive exponent for any nonzero real
number a and any integer 3, a^-3 = 1 ÷ a³ ,a ≠ 0.

Question 4.

Answer:
(2.5)^-3,

Explanantion:

Question 5.

Answer:
(-6)^-1,

Explanation:

Simplify each expression. Write your answer using a positive exponent.

Question 6.
14a^-5 ÷ (7a • 2a^-4)
14a^-5 ÷ (7a • 2a^-4) = Write the expression as a .
=
Answer:
1,

Explanation:

Usually you should write your answer using a positive exponent
unless asked to use a negative exponent.

Math in Focus Course 3A Practice 1.5 Answer Key

Simplify each expression and evaluate.

Question 1.
8³ • 8°
Answer:
8³,

Explanation:
Given 8³ • 8° any number to power 0 the value is 1, so the value of 8° is 1,
so 8³ • 8° = 8³ • 1 = 8³.

Question 2.
5⁴ • (-5)°
Answer:
5⁴,

Explanation:
Given 5⁴ • (-5)° any number to power 0 the value is 1, so the value of (-5)° is 1,
so 5⁴ • (-5)° = 5⁴ • 1 = 5⁴.

Question 3.
\(\left(\frac{1}{3}\right)^{4} \cdot\left(\frac{1}{3}\right)^{0}\)
Answer:
\(\left(\frac{1}{3}\right)^{4}\),

Explanation:
Given \(\left(\frac{1}{3}\right)^{4} \cdot\left(\frac{1}{3}\right)^{0}\) as
\(\left(\frac{1}{3}\right)^{0}\) = 1 therefore \(\left(\frac{1}{3}\right)^{4}\) X 1 =
\(\left(\frac{1}{3}\right)^{4}\).

Question 4.
7 • 10³ + 4² • 10² + 5 • 10°
Answer:
8,605,

Explanation:
Given 7 • 10³ + 4² • 10² + 5 • 10° solving as 7 X 10 X 10 X 10 + 4 X 4 X 10 X 10 + 5 X 1 =
7 X 1,000 + 16 X 100 + 5 = 7,000 + 1,600 + 5 = 8,605.

Question 5.
(2.3) • 10² + 5 • 10¹ + 1 • 10°
Answer:
281,

Explanation:
Given (2.3) • 10² + 5 • 10¹ + 1 • 10° solving as (2.3) X 10 X 10 + 5 X 10 + 1 X 1 =
230 + 50 + 1 =281.

Question 6.
\(\frac{7^{4} \cdot 7^{5}}{7^{9}}\)
Answer:
1,

Explanation:
Given \(\frac{7^{4} \cdot 7^{5}}{7^{9}}\) solving
\(\frac{7^{4+5}}{7^{9}}\) = \(\frac{7^{9}}{7^{9}}\)=
(7)^9-9 =(7)⁰ = 1.

Question 7.
(9^-3)° • 5²
Answer:
5²,

Explanation:
Given (9^-3)° • 5² any number to power 0 the value is 1,
so the value of (9^-3)°  is 1, therefore 1 X 5² = 5².

Question 8.
\(\frac{\left(6^{-3}\right)^{-2} \cdot 8^{6}}{48^{6}}\)
Answer:
1

Explanation:

Simplify each expression. Write your answer using a negative exponent.

Question 9.
7³ • 7^-4
Answer:
7^-1,

Explanation:
Given 7³ • 7^-4  using products of powers property we get
7^3-4 = 7^-1.

Question 10.
\(\frac{(-5)^{-2}}{(-5)^{3}}\)
Answer:
\(\frac{(-5)^{-2}}{(-5)^{3}}\) simplify
= (-5)^-1

Question 11.

Answer:

Question 12.
\(\left(\frac{2}{5}\right)^{-4} \cdot\left(\frac{2}{5}\right)^{-1} \div\left(\frac{2}{5}\right)^{-3}\)
Answer:
\(\left(\frac{2}{5}\right)^{-2}\),

Explanation:

Question 13.
\(\frac{x^{0}}{x^{2} \cdot x^{3}}\)
Answer:
x ^-5,

Explanation:

Question 14.
\(\frac{4 h^{-5} \cdot 6 h^{-2}}{3 h^{-3}}\)
Answer:
8(h)^-4,

Explanation:

Simplify each expression. Write your answer ustng a positive exponent.

Question 15.
1.2° ÷ 1.8²
Answer:
0.308,

Explanation:
Given 1.2° ÷ 1.8² any number to power 0 the value is 1,
so the value of (1.2)° is 1, therefore 1÷1.8² = 1 ÷ 3.24 = 0.308.

Question 16.
5.2^-3 ÷ 2.6^-3
Answer:
0.125,

Explanation:
Given 5.2^-3 ÷ 2.6^-3  we get  2.6³ ÷ 5.2³ upon solving
(2.6 X 2.6 X 2.6) ÷ (5.2 X 5.2 X 5.2) = 0.125.

Question 17.
\(\frac{(-3)^{-4}}{(-3)^{2}}\)
Answer:
\(\frac{(1)}{(-3)^{6}}\),

Explanation:
Given

Question 18.
\(\left(\frac{5}{6}\right)^{-4} \cdot\left(\frac{5}{6}\right)^{-2} \div\left(\frac{5}{6}\right)^{-3}\)
Answer:
\(\frac{(1)}{\frac{5}{6}^{3}}\),

Explanation:

Question 19.
\(\frac{9 k^{-1} \cdot 2 k^{-3}}{27 k^{-6}}\)
Answer:
2k² ÷ 3,

Explanation:

Question 20.
\(\frac{c^{-4} \cdot c^{12}}{c^{-7}}\)
Answer:
(c)¹⁵,

Explanation:

Evaluate each numeric expression.

Question 21.
\(\frac{7^{-2} \cdot 7^{0}}{8^{3} \cdot 8^{-5}}\)
Answer:
(8/7)²,

Explanation:

Question 22.
\(\frac{\left(7^{-2}\right)^{2} \cdot 9^{-4}}{21^{-4}}\)
Answer:
(3)^-4,

Explanation:

Question 23.
\(\frac{10^{0}}{2^{-2} \cdot\left(5^{-1}\right)^{2}}\)
Answer:
100,

Explanation:

Question 24.
\(\frac{\left(3^{6}\right)^{-2}}{6^{-9}\left(-2^{10}\right)}\)
Answer:
(-1/54),

Explanation:

Simplify each algebraic expression.

Question 25.
\(\left(\frac{7 m^{3}}{-49 n^{0}}\right)^{-1}\)
Answer:
(-7)m^-3,

Explanation:

Question 26.
\(\frac{8 r^{2} s}{4 s^{-3} r^{4}}\)
Answer:
2r^-2s⁴,

Explanation:

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 1 Review Test to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 1 Review Test Answer Key

Concepts and Skills

Identify the base and exponent in each expression.

Question 1.
\(\left(-\frac{1}{5}\right)^{-3}\)
Answer:
base is \(\left(-\frac{1}{5}\right)\) and exponent is -3,

Explanation:
Given \(\left(-\frac{1}{5}\right)^{-3}\) so base is
base value is \(\left(-\frac{1}{5}\right)\) and exponent value is -3,

Question 2.
-0.92⁴
Answer:
base is (-0.92) and exponent is 4,

Explanation:
Given -0.92⁴ here the base is -0.92 and exponent is 4.

Tell whether each statement is correct. If it is incorrect, state the reason.

Question 3.
-0.7³ = -0.7 • 0.7 • 0.7
Answer:
Statement is incorrect,

Explanation:
Given -0.7³ = -0.7 • 0.7 • 0.7 which is incorrect the correct statement is
-0.7³ = -0.7 • -0.7 • -0.7.

Question 4.
5^-4 = (-5) • (-5) • (-5) • (-5)
Answer:
Statement is correct,

Explanation:
Given statement is
5^-4 = (-5) • (-5) • (-5) • (-5) is correct.

Write in exponential notation.

Question 5.
2 • 2 • 2 • 2
Answer:
2⁴,

Explanation:
Given 2 • 2 • 2 • 2 here base is 2 and exponent is 4 therefore
the exponential notation is 2⁴.

Question 6.
4.8 • 4.8
Answer:
4.8²,

Explanation:
Given 4.8 • 4.8 here base is 4.8 and exponent is 2 therefore
the exponential notation is 4.8².

Question 7.
\(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\)
Answer:
\(\frac{1}{2}\),

Explanation:
Given \(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\) here base is \(\frac{1}{2}\) and exponent is 3 therefore the exponential notation is \(\left(\frac{1}{2}\right)^{3}\).

Question 8.
c • c • c • c • c • c
Answer:
c⁶,

Explanation:
Given c • c • c • c • c • c here base is c and exponent is 6 therefore
the exponential notation is c⁶.

Question 9.
\(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\)
Answer:
\(\frac{1}{2}\),

Explanation:
Given \(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\) here base is \(\frac{1}{2}\) and exponent is 3 therefore the exponential notation is \(\left(\frac{1}{2}\right)^{3}\).

Question 10.
(-1.2)(-1.2)(-1.2)(-1.2)
Answer:
(-1.2)⁴,

Explanation:
Given (-1.2)(-1.2)(-1.2)(-1.2)  here base is (-1.2) and exponent is 4 therefore
the exponential notation is (-1.2)⁴.

Write the prime factorization of each number in exponential notation.

Question 11.
3,780
Answer:
Prime factorization of 3,780 = 2² X 3³ X 5 X 7,

Explanation:

Question 12.
27,720
Answer:
Prime factorization of 27,720 = 2³ X 3² X 5 X 7 X 11,

Explanation:

Expand and evaluate each expressions.

Question 13.
(-6)²
Answer:
36,

Explanation:
Given (-6)² so upon expanding we get (-6) X (-6) = 36.

Question 14.
1.1²
Answer:
1.21,

Explanation:
Given (1.1)² so upon expanding we get (1.1) X (1.1) = 1.21.

Question 15.
10⁵
Answer:
100,000

Explanation:
Given 10⁵ so upon expanding we get 10 X 10 X 10 X 10 X 10 = 100,000.

Question 16.
\(\left(\frac{2}{3}\right)^{3}\)
Answer:
\((\frac{8}{27}\),

Explanation:
Given\(\left (\frac{2}{3}\right)^{3}\) so upon expanding we get
\((\frac{2}{3})\) X \((\frac{2}{3})\) X \((\frac{2}{3})\) =
\((\frac{2X 2 X 2}{3 X 3 X 3})\) = \((\frac{8}{27})\).

Simplify each expression. Write your answer using a positive exponent.

Question 17.
(-3)^-1 . (-3)°
Answer:
\(\frac{1}{-3}\),

Explanation:
Given (-3)^-1 . (-3)° as (-3)°  is 1 and (-3)^-1 value in positive exponent is
\(\frac{1}{-3}\).

Question 18.
\(\left(\frac{5}{6}\right)^{4} \cdot\left(\frac{5}{6}\right)^{3}\)
Answer:

Question 19.
5m³n⁴ . 4m⁵n²
Answer:
20m⁸n⁶,

Explanation:
Given 5m³n⁴ . 4m⁵n² on solving 5 X 4 X m³⁺⁵ X n⁴⁺² = 20m⁸n⁶.

Question 20.
\(\left(\frac{7}{8}\right) \div\left(\frac{7}{8}\right)^{3}\)
Answer:

Question 21.
(-3)^-1 • (-3)°
Answer:
\(\frac{1}{-3}\),

Explanation:
Given (-3)^-1 . (-3)° as (-3)°  is 1 and (-3)^-1 value in positive exponent is
\(\frac{1}{-3}\).

Question 22.
x⁸z⁵ ÷ x³z⁹
Answer:
x⁵z^-4,

Explanation:
Given x⁸z⁵ ÷ x³z⁹ using division of product property
x^8-3z^5-9 we get x⁵z^-4.

Question 23.
25p⁶q⁹ ÷ 45p⁸q⁴
Answer:
5p^-2 q⁵ ÷ 9,

Explanation:
Given 25p⁶q⁹ ÷ 45p⁸q⁴
= 5p^6-8 q^9-4 ÷ 9,
= 5p^-2 q⁵ ÷ 9,

Question 24.
\(\left[\left(\frac{2}{3}\right)^{2} \cdot\left(\frac{2}{3}\right)^{-1}\right]^{3}\)
Answer:

Question 25.
40c⁵d³ ÷ 10c⁹d²
Answer:
4c^-4d,

Explanation:
Given 40c⁵d³ ÷ 10c⁹d² =4 X 10 X c^5-9 X d^3-2 ÷ 10 = 4c^-4d.

Question 26.

\(\left(\frac{72 b^{-1}}{32 c^{-1}}\right)^{-2}\)
Answer:

Question 27.
\(\frac{\left(9^{-2}\right)^{-2} \cdot 2^{2}}{9^{2}}\)
Answer:
9² X 2²,

Explanation:
Given

Question 28.
\(\frac{6^{8} \cdot 56^{-3}}{6^{5} \cdot 7^{-3}}\)
Answer:
6³ X 8^-3,

Explanation:

Question 29.
\(\frac{42^{-1}}{\left(2^{0}\right)^{12} \cdot 21^{-1}}\)
Answer:
2^-1,

Explanation:

Question 30.
\(\frac{\left(3^{5} \cdot 3^{4}\right)^{2}}{\left(3^{3}\right)^{6}}\)
Answer:
1,

Explanation:

Solve each equation involving a variable that is squared.

Question 31.
r² = 256
Answer:
r = 16,

Explanation:

Question 32.
c² = \(\frac{121}{169}\)
Answer:
c = \(\frac{11}{13}\),

Explanation:
Given c² = \(\frac{121}{169}\)= \(\frac{11 X 11}{13 X 13}\),
c is sqaure root of \(\frac{11 X 11}{13 X 13}\) = \(\frac{11}{13}\).

Solve each equation involving a variable that is cubed.

Question 33.
x^{3 = 32.768
Answer:}
x = 3.2,

Explanation:

Question 34.
c³ = –\(\frac{27}{343}\)
Answer:
c= – \(\frac{3}{7}\),

Explanation:
Given c³ = –\(\frac{27}{343}\) can be written as
c³ = –\(\frac{3 X 3 X 3}{7 X 7 X 7}\) therefore c is cube root of
–\(\frac{3 X 3 X 3}{7 X 7 X 7}\) = – \(\frac{3}{7}\)

Question 35.
The expanded form of a number is 5 • 10¹ + 8 • 10° + 1 • 10^-1 + 9 • 10^-2. What is this number in standard form?
Answer:
The number in standard form is 58.19,

Explanation:
Given 5 • 10¹ + 8 • 10° + 1 • 10^-1 + 9 • 10^-2= 5 X 10 + 8 X 1 + 1 X 0.1 + 9 X 0.01 =
50 + 8 +0.1 + 0.09 = 58.19.

Question 36.
The pattern of triangles shown is called the Sierpinski’s gasket.
a) Find a pattern in the number of shaded triangles.
Answer:
(3/4)ⁿ,

Explanation:
The concept of the Sierpinski triangle is very simple:
Take a triangle. Create four triangles out of that one by connecting the centres of each side.
So basically each triangle cuts into four and each again cuts four and so on.
So, there are an infinite number of triangles pattern is (3/4)ⁿ.

b) How many shaded triangles will be in the fifth diagram of this pattern?
Write an exponential expression for this number.

Answer:
Shaded triangles will be in the fifth diagram of this pattern are 243.
Explanation:
We have pattern(3/4)ⁿ here in first shape we have
3 shaded triangles out of 4, 5th pattern will have (3/4)⁵ so
shaded triangles will be in the fifth diagram of this pattern
3 X 3 X 3 X 3 X 3 = 243.

Problem Solving

Solve. Show your work. Round your answer to the nearest tenth.

Question 37.
The Reina Sofia Museum in Spain has a glass elevator.
The floor of the elevator shaft is a square with an area of about 42.25 square feet.
Find the length of a side of the floor.
Answer:
The length of a side of the floor is 6.5 feet,

Explanation:
Given the Reina Sofia Museum in Spain has a glass elevator.
The floor of the elevator shaft is a square with an area of about 42.25 square feet.
The length of a side of the floor is 6.5 feet.

Question 38.
Earth’s volume is approximately 1,083,210,000,000 cubic kilometers.
What is the diameter of the Earth in kilometers?

Answer:
The diameter of the earth is 12,562 km ,

Explanation:
Given Earth’s volume is approximately 1,083,210,000,000 cubic kilometers.,
4/3pir³ = 1,08321 X 10^{7 ,}
we get r = 6281 kms,
therefore the diameter of the earth is 2 X 6281 km =  12,562 km.

Question 39.
Koch’s snowflake is a pattern that starts with an equilateral triangle.
Then, in successive images, each line segment is replaced by 4 line segments, as shown below.
How many line segments make up the seventh image of this pattern?
Write this number in exponential form and evaluate.

Answer:
3 X 4⁶,

Explanation:
Given Koch’s snowflake is a pattern that starts with an equilateral triangle.
Then, in successive images, each line segment is replaced by 4 line segments, as shown above.
we can express it in sequence of numbers
a1=3,
a2 = 3 X 4 = 12,
a3 = 3 X 4 X 4 = 48,
an = a^(n-1) X 4 = 4 a^(n-1),(geometric progression)
an = 3 x 4^(n-1),
n= 7
a7 = 3 X 4⁶.

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 9 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 9 Review Test Answer Key

Concepts and Skills

Solve.

Question 1.
Name the triangle congruent to ∆ABC.

Answer:
∆XYZ
Explanation:
Given,
∆ABC is congruent to ∆XYZ

Question 2.
ABCDEF is a regular hexagon. Name a quadrilateral congruent to ADCB.

Answer:
ADEF
Explanation:
Given,
ABCDEF is a regular hexagon,
ADCB is congruent to ADEF

Question 3.
\(\overline{S Q}\) and \(\overline{T P}\) are straight lines. Name the triangle similar to ∆PQR.

Answer:
∆TRS
Explanation:
Given,
∆PQR is congruent to ∆TRS

Solve. Show your work.

Question 4.
The two figures are congruent. Find the value of each variable,
a)

b)

Answer:
a) x = 8, y = 6, z = 104 degrees, b) x = 6, y = 4, z = 2
Explanation:
a) Since the triangles are congruent to each other, the longest side of the triangle measures 8 cm which is corresponding to the longest side of the second triangle which is x, therefore x = 8. While y is corresponding to the segment measuring 6 cm, therefore y = 6.
Solving z, the angle of 47 degrees corresponds to the other angle of the second triangle which is already having 29 degress, therefore the other triangle has 47 and 29 degress.
Sum of all the angles = 180 degrees
180 = 47 + 29 + z
180 = 76 + z
180 – 76 = z
104 = z
Therefore, x = 8, y = 6 and z = 104 degrees.
b)Since both the triangles are congruent to each other, the corresponding side of the second triangle is measuring 6 cm which is congruent to the first triangle of x, therefore x is measured as 6 cm. While y on the second triangle which is congruent to the first triangle which is measuring 4 cm, therefore y is measured as 4 cm. While z on the first triangle which is congruent to the first triangle’s measure of 2 cm, therefore, z is measured as 2 cm.
Therefore, x = 6, y = 4 and z = 2.

Question 5.
The two figures are similar. Find the value of each variable.
a)

b)

Answer:
a) x = 2.5 inches, y = 10 inches and z = 30 degrees, b) w = 1.6, x = 13.3 inches and y = 1.8 inches.
Explanation:
a) Since the two triangles are similar the ratio of the corresponding lengths are equal, the side of the bigger triangle corresponds to x of smaller triangle.The side measuring 8 inches in the bigger triangle corresponds to the side of 4 inches in the smaller triangle.
Ratio of corresponding lengths are equal
5/x = 8/4
cross multiplying,
5 x 4 = 8 x X
20 = 8x
20/8 = x
2.5 = x
The side which measures y in the bigger triangle corresponds to the measure of 5 inches in the smaller triangle and the side of 8 inches in the bigger triangle corresponds to the side of 4 inches in the smaller triangle.
Ratio of corresponding lengths are equal
y/5 = 8/4
cross multiplying,
y x 4 = 8 x 5
4y = 40
y = 40/4
y = 10
Since the sum of all the angle in the triangle = 180 degrees
z + 52 degrees + p = 180 degrees
The other angle p whose measure is unknown corresponds to the angle in the larger triangle whose measure is 98 degrees, two triangles are similar, thus their angles are equal. So p is 98 degrees then,
Sum of the angles in the triangle = 180 degrees
180 = z + 52 + p
Substituting p = 98
180 = z + 52 + 98
180 = z + 150
z = 180 – 150
z = 30
Therefore, x = 2.5, y = 10, z = 30
b) Since both the triangles are similar, the ratio of their corresponding lengths are equal. The side which measures 10 inches in the bigger triangle corresponds to the side which measures 6 inches in the smaller triangle.The side which measures 8 1/3 in the smaller triangle corresponds to the w in the bigger triangle.
Ratio of corresponding lengths are equal
6/10 = w/ 8 1/3
cross multiplying,
6 x 8 1/3 = w x 10
16 = 10 w
w = 16/10
w = 1.6
The side which measures 10 inches in the bigger triangle corresponds to the side which measures 6 inches in the smaller triangle. The side which measures x inches in the bigger triangle corresponds to the side which measures 8 inches in the smaller triangle.
Ratio of corresponding lengths are equal
6/10 = 8/x
cross multiplying,
6 x X = 8 x 10
6x = 80
x = 80/6
x = 13.33
The side which measures 10 inches in the bigger triangle corresponds to the side which measures 6 inches in the smaller triangle. The side which measures y inches in the bigger triangle corresponds to the side which measures 3 inches in the smaller triangle.
Ratio of corresponding lengths are equal
6/10 = y/3
cross multiplying,
6 x 3 = y x 10
18 = 10y
18/10 = y
y = 1.8
Therefore, x = 13.33, y = 1.8 and w = 1.6

Question 6.
State whether the triangles are congruent. If they are congruent, write the statement of congruence and state the test used.
a)

b)

Answer:

a) The triangles are not congruent
Explanation:
The side BC and DA are congruent, however, the sides are not corresponding.
The side BA corresponds to DA but lines are not congruent,
BC corresponds to DC but lines are not congruent, therefore triangles are not congruent.

b) ∠C from ∆TRS corresponds to ∠A from ∆BCA and the two angles are congruent.
∠A from ∆DAC corresponds to ∠C from ∆BCA and the two angles are congruent.
Side AC from ∆DAC is congruent to side CA from ∆BCA.
By the Angle – Side – Angle Congruence Theorem, triangles are congruent.
This theorem states that two triangles are congruent and the sides in between these angles are congruent and corresponds to each other, then the triangle is congruent.

Question 7.
State whether the triangles are similar. Explain with a test for similar triangles.
a)

b)

Answer:

a) Triangles are similar
Explanation:
Both the triangles are similar triangles by Angle-Angle postulate, that states that triangles are similar if two pairs of their corresponding angles are congruent. For example, 60 degree angle of first triangle is corresponding the 60 degree angle of second triangle and also 50 degree angle of first triangle is corresponding the 50 degree angle of second triangle, then the two triangles are similar.
Therefore, triangles are similar.

b) Triangles are not similar
Explanation:
If the triangles are similar triangles by Side-Angle-Side postulate, then their two corresponding sides are proportional and their included angle is congruent. The included angles of two corresponding sides are congruent. However, the corresponding are not proportional, then the side measuring 5 cm corresponds to side measuring 6 cm, while side measuring 8 cm corresponds to side measuring 9 cm.
5/6 not equal to 8/9
0.83 not equal to 0.88
Therefore, Triangles are not similar.

Question 8.
∆ABC undergoes two reflections to be mapped onto ∆A”B”C”.

a) Describe one such pair of reflections.
Answer:
Since there are two reflections showing that A(-4,4) has been reflected into the x-axis and was mapped on to A'(-4,-4). The image of the first reflection is reflected in to the y-axis and shows A”(4, -4)

b) Describe a single transformation that would also map ∆ABC onto ∆A”B”C”.
Answer:
180 degree rotation changes the coordinates from (x,y) to (-x,-y).
Then A(-4, 4) mapped to (4, -4) B(-4, 1) mapped to (4, -1) and C(-2, 1) mapped to C(2, -1). The coordinates of the image of ∆ABC under the rotation of 180 degrees is actually ∆A”B”C”. Therefore, 180 degrees rotation maps the original triangle to ∆A”B”C” in a single transformation.

Describe the single transformations that maps ∆ABC onto ∆A’B’C’, ∆ A’B’C’ onto ∆A”B”C”, and ∆ABC onto ∆A”B”C”.

Question 9.

Answer:
i) For ∆ABC mapped to ∆A’B’C’, that the original points A(-5,4), B(-4,6) and C(-3,4). All the points move 3 units to the right. Therefore, transformation took place is translation.
ii) In ∆ABC mapped to ∆A”B”C”, the original points A(-5,4), B(-4,6) and C(-3,4) mapped to A”(-1, 2), B”(-0.5, 3) and C”(0, 2). ∆A”B”C” is smaller version of ∆ABC. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet on (2,0). Therefore, this transformation is dilation.
iii) For ∆A’B’C’ mapped to ∆A”B”C” original points A'(-2,4), B'(-1,6) and C(0,4) are mapped to A”(-1,2),B”(-0.5,3) and C”(0,2). ∆A”B”C” is a smaller version of ∆ABC. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet on (0,0). Therefore, this transformation is dilation.

Question 10.

Answer:
i) For ∆ABC mapped to ∆A’B’C’, that the original points A(1,2), B(2,1) and C(3,1) are mapped to A'(3,6), B'(6,3) and C'(9,3).∆A’B’C’ is bigger version of ∆ABC If points A is connected on A”, B to B” and C on C” ” by the line, then the line should be extended until it makes an intersection, then this will meet on (0,0). Therefore, this transformation is dilation.
ii) In ∆A’B’C’ mapped to ∆A”B”C”, the original points A'(3,6), B'(6,3) and C'(9,3) mapped to A”(2,4), B”(4,2) and C”(6, 2). ∆A”B”C” is larger version of ∆A’B’C’. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet at the origin. Therefore, this transformation is dilation.
iii) For ∆ABC mapped to ∆A”B”C” original points A(1,2),B(2,1) and C(3,1) are mapped to A”(2,4),B”(4,2) and C”(6,2). ∆A”B”C” is a larger version of ∆ABC. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet on (0,0). Therefore, this transformation is dilation.

Question 11.

Answer:
i) For ∆ABC mapped to ∆A’B’C’, that the original points A(-6,2), B(-6,1) and C(-4,1) are mapped to A'(2,2), B'(2,1) and C'(4,1). All the points move 8 units to the right. Therefore, transformation took place is translation.
ii) In ∆A’B’C’ mapped to ∆A”B”C”, the original points A'(2,2), B'(2,1) and C'(4,1) mapped to A”(-2,2), B”(-2,1) and C”(-4,1). ∆A”B”C” is larger version of ∆A’B’C’. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet at the origin. Therefore, this transformation is dilation.
iii) For ∆ABC mapped to ∆A”B”C” original points A(-6,2),B(-6,1) and C(-4,1) are mapped to A”(-2,2),B”(-2,1) and C”(-4,1). The transformation is a reflection at x = -4. Both C and C” lies at -4 because 0 unit away, then so it is its image. A and B is 2 units to the left of -4 so A” and B” is 2 units to the right of it.

Problem Solving

Solve. Show your work.

Question 12.
Two geometrically similar buckets have bases with radii in the ratio 2 : 5. The top of the small bucket has radius 7 inches.

a) Find the radius of the top of the big bucket.
Answer:
The radius of the top of the big bucket
Explanation:
Given,
2 : 5 = 2/5
Let the radius be p,
7/p = 2/5
cross multiplying, 7 x 5 = 2 x p
35 = 2p
p = 35/2
p = 17.5

b) Write the ratio of the height of the small bucket to the height of the big bucket.
Answer:
Height of the small bucket/ Height of the big bucket
2 : 5

Question 13.
A 5-foot tall vertical rod and a 25-foot tall vertical flagpole stand on horizontal ground. At a certain time of the day, the flagpole casts a shadow 30 feet long.

a) Write a ratio comparing the height of the rod to the height of the flagpole.
Answer:
Height of rod/ Height of the flagpole
5/25 = 1/5

b) Calculate the length of the shadow cast by the rod.
Answer:
Let the length of the shadow of the rod be a,
Height of the rod/Height of flapole = Shadow length of the rod/Shadow length of the flagpole
5/25 = a/30
Cross multiplying, 5 x 30 = a x 25
125 = 25a
a = 125/25
a = 6

Question 14.
Two barrels are geometrically similar. Their heights are 16 inches and 20 inches. The diameter of the base of the larger barrel is 8 inches. Calculate the diameter of the base of the smaller barrel.
Answer:
Let t be the diameter of the base of smaller barrel,
Height of small barrel/ Height of large barrel = Diameter of small barrel/Diameter of large barrel
16 / 20 = t / 8
Cross multiplying, 16 x 8 = t x 20
128 = 20 t
128/20 = t
6.4 = t
Therefore, Diameter of the small barrel = 6.4 inches.

Question 15.
A photograph is 4 inches wide and 6 inches long. It is enlarged to a width 6 inches.

a) What is the length of the enlarged photo?
Answer:
The length of the enlarged photo is 9 inches
Explanation:
Let the length of the enlarged photo is r,
Width of the original photo/ Length of original photo = Width of enlarged photo/ Length of enlarged photo
4/6 = 6/r
cross multiplying, 4r = 6 x 6
4r = 36
r = 36/4
r = 9

b) If the original photo is enlarged to a length 12 inches, find the new width.
Answer:
New width = 8 inches
Explanation:
Let new width of the photo is i,
Width of the original photo/ length of the original photo = new width of the photo/ length of the photo
4/6 = i/12
Cross multiplying, 4 x 12 = i x 6
48 = 6i
i = 48/6
i = 8
Therefore, new width of the photo = 8 inches.

Question 16.
Anthony made an exact copy of the shape ABCDE using a photocopier.

a) Name the shape congruent to ABCDE on the photocopy.
Answer:
GHIJF
Explanation:
Point A is congruent to point to G, Point B is congruent to point H, Point C is congruent to point I, Point D is congruent to point J, Point E is congruent to point F. Therefore, ABCDE is congruent to GHIJF.

b) Find the values for w, x, y, and z.
Answer:
w = 8, x = 5, y = 3, z = 6
Explanation:-
Solving x, IJ from GHIJF measuring x inches is congruent to CD of ABCDE which measures 5 inches.
IJ = CD
x = 5
Solving y, GF from GHIJF measuring 3y inches corresponds to AE from ABCDE which is measuring 9 inches.
GF = AE
3y = 9
divide 3 on both the sides
3y/3 = 9/3
y = 3
Solving z, HG from GHIJF measuring 4 inches corresponds to BA of ABCDE which is measuring (z – 2).
HG = BA
4 = (z – 2)
Adding 2 on both the sides
4 + 2 = z – 2 + 2
6 = z
Solving w, HI from GHIJH measuring 11 inches corresponds to BC of ABCDE which is measuring (w + y) inches.
HI = BC
11 = w + y
substituting y = 3,
11 = w + 3
Subtract 3 on both the sides,
11 – 3 = w + 3 – 3
8 = w.

Question 17.
A framed poster has a length of 28 inches and a perimeter of 88 inches. A similar poster has a perimeter of 66 inches. Find the area of the similar poster.
Answer:
Area of the similar poster = 252 sq inches
Explanation:
Let the length of the similar poster = p
length of similar poster/ length of framed poster = perimeter of similar poster/ perimeter of framed poster
substituting the values, p/28 = 66/88
cross multiplying, 88 x p = 66 x 28
88p = 1848
dividing 88 on both the sides, 88p/88 = 1848/88
p = 21
Therefore, length of the similar poster = 21 inches.
Perimeter of rectangle = 2 (l + w)
66 = 2 (21 + w)
66 = 42 + 2w
Subtracting 42 from both the sides,
66 – 42 = 42 + 2w – 42
24 = 2w
Divide 2 on both the sides,
24/2 = 2w/2
12 = w
Therefore, Width = 12 inches.
Area of rectangle = length x width
= 21 x 12
= 252 sq inches.

Question 18.
A robotic arm with two circular ends moves through a sequence of translations and rotations. The diagram shows two positions of the arm. All dimensions are in centimeters.

a) Find the values of x and y.
Answer:
x = 6 cm and y = 12 cm
Explanation:
The original image is exactly the same as transformed image, then the measure of the side of the original robotic arm is congruent to corresponding side of its image.
Solving for y:-
Side of original arm = Side of the image
substituting the expressions, y – 1 = 11
Adding 1 on both the sides, y – 1 + 1 = 11 + 1
y = 12
Therefore, y = 12 cm.
Solving for x:-
Other side of the original arm = other side of the image
Substituting the expressions, 2x = y
Substitute y = 12, 2x = 12
Divide 2 on both the sides, 2x/2 = 12/2
x = 6
Therefore, x = 6 cm

b) Calculate the area of the robotic arm. Use \(\frac{22}{7}\) as an approximation for π.
Answer:
Area of the robotic arm = 678.9 sq cm
Explanation:
Area of the arm = pi x diameter (diameter + length)
= 22/7 x 12 (12 + 6)
= 22/7 x 12 (18)
= 22/7 x 216
= 678.9 sq cm.

Question 19.
A table, which is in front of a mirror, has the side view ABCDEFGH shown below (all dimensions are in inches). ABCH and DEFG are rectangles. The side view is reflected in the mirror to form the congruent image STUVWXYZ.

a) State the corresponding angle of ∠CDE and ∠AHG.
Answer: ∠V corresponds to ∠D and ∠Z corresponds to ∠H

Explanation:
Since ABCDEFGH is congruent to STUVWXYZ, the points A,B,C,D,E,F,G and H corresponds to S,T,U,V,W,X,Y and Z respectively.
Therefore, ∠V corresponds to ∠D and ∠Z corresponds to ∠H.

b) Write the length of \(\overline{T U}\), \(\overline{W X}\), and \(\overline{Y Z}\).
Answer:
TU = 3 inches, WX = 16 inches and YZ = 4.5 inches
Explanation:
i) As TU is congruent to BC of table ABCDEFGH, since BC measures 3 inches then TU also has 3 inches.
ii) WX is congruent to EF of table ABCDEFGH, to know the value of EF subtract the two measures of CD, GH to AB,Let the measure of AB = 25 inches and CD and GH = 4.5 inches then,
EF = AB – (CD + GH)
= 25 – ( 4.5 + 4.5 )
= 25 – 9
= 16
Since, EF measures 16 inches therefore, WX also measures 16 inches.
iii) YZ is congruent to GH of table ABCDEFGH, Since GH measures 4.5 inches therefore, YZ also measures 4.5 inches.

c) Find the perimeter of the image.
Answer:
Perimeter of STUVWXYZ = 106 inches.
Explanation:
Perimeter of ABCDEFGH = sum of all the lengths
Substituting all the sides of the table
Perimeter = AB + BC + CD + DF + EF + FG + GH + HA
Substituting all the values
Perimeter = 25 + 3 + 4.5 + 25 + 16 + 25 + 4.5 + 3
= 106
Therefore, the perimeter of ABCDEFGH is 106 inches and ABCDEFGH is congruent to STUVWXYZ then perimeter of STUVWXYZ is 106 inches

Question 20.
Gisele reduced the size of ∆PQR with area 14.4 square centimeters using a photocopier. ∆PQR ~ ∆STU.

a) What is the scale factor of dilation? Express your answer as a percent.
Answer:
The scale factor of dilation is defined as the ratio of the size of the new image to the size of the old image.
Scale factor is 62.5%
Explanation:
PQ measuring 8 cm corresponds to side ST measuring 5 cm, to know the scale factor need to find the ratio of length of ST to length of PQ
Scale factor = length of ST/ length of PQ
= 5/8
= 0.625
Therefore, scale factor is 0.625
To convert it into percentage, multiply it with 100
Scale factor = 0.625 x 100
= 62.5%

b) Find the area of ∆STU.
Answer:
Area of ∆STU = 9 sq cm.
Explanation:
Two triangle are similar to each other,
Scale factor is 0.625,
Area of ∆PQR = 14.4 sq cm,
The area of ∆STU is given by the area of ∆PQR multiplied with scale factor.
Area of ∆STU = ∆PQR x scale factor
= 14.4 x 0.625
= 9

This handy Math in Focus Grade 8 Workbook Answer Key Cumulative Review Chapters 7-9 detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 7-9 Answer Key

Concepts and Skills

Find the value of each variable. (Lesson 7.1)

Question 1.

Answer:
Pythagorean Theorem states that the sum of the squares of the legs is the square of hypotenuse of a right angle. Let x be the unknown length in the right angle so,
x² + 15² = 17²
x² + 225 = 289
x² = 289 – 225
x² = 64
x = 8 (Square root of 64)
Therefore, the leg is 8 inches.

Question 2.

Answer:
(3.6)² + (2.7)² = x²
12.96 + 7.29 = x²
20.25 = x²
4.5 = x (Square root of 20.25)
Therefore, x is 4.5 inches.

The side lengths of a triangle are given. Decide whether each triangle is a right triangle. (Lesson 7.1)

Question 3.
6 cm, 8 cm, 12 cm
Answer:
6² + 8² = 12²
36 + 64 = 144
100 = 144 (100 is not equal to 144)
Since the sum of the square of the legs is not equal to the square of hypotenuse, h=the triangle is not a right triangle.

Question 4.
9 in., 7.2 in., 5.4 in.
Answer:
(7.2)² + (5.4)² = (9)²
51.84 + 29.16 = 81
81 = 81
Therefore, the triangle is a right triangle.

Find the distance between each pair of points. If necessary, round your answer to the nearest tenth of a unit. (Lesson 7.2)

Question 5.
P(2, 5), Q(4, 13)
Answer:
d = √(x2 – x1)² + (y2 – y1)²
d = √(4 – 2)² + (13 – 5)²
d = √(2)² + (8)²
d = √4+64
d = √68
d = ±8.2

Question 6.
X(3, -1), Y(4, 2)
Answer:
d = √(x2 – x1)² + (y2 – y1)²
d = √(4 – 2)² + (2 – (-1))²
d = √(2)² + (3)²
d = √4+9
d = √13

For each solid, find the unknown dimension. If necessary, round your answer to the nearest tenth of a unit. (Lesson 7.3)

Question 7.

Answer:
Using pythagorean theorem
18² + w² = 30²
324 + w² = 900
w² = 900 – 324
w² = 536
w ~ 23.2 (Square root of 536)
Therefore, the height w of the cone measures 23.2 cm.

Question 8.

Answer:
Using pythagorean theorem,
z² + 7² = 21²
z² + 49 = 441
z² = 441 – 49
z² = 392
z = 19.8 (Square root of 392)
Therefore, the height is 19.8 inches.

Find the volume of each composite solid. Use 3.14 as an approximation for π. If necessary, round your answer to the nearest tenth. (Lesson 7.4)

Question 9.

Answer:
Finding the radius of the cone:
4² + n² = 5²
16 + n² = 25
n² = 25 – 16
n² = 9
n = 3 ( square root of 9 )
Therefore, the radius of the cone is 3 inches. Deducting 2 inches to radius of the cone determines the radius of the cylinder according to the diagram. Then the radius of the cylinder is 1 inch.
Finding volume of the cone:
For the volume of the cone, using 3.14 as the value of π and let h be the height,
Volume of the cone = 1/3x Area of the base x Height of the cone
= 1/3 πr² x h
= 1/3 x 3.14 x (3)² x h
= 37.68
Therefore, the cone has a volume of 37.68 cubic meters.
Finding the volume of cylinder:
For the volume of the cylinder, using 3.14 as the value of π and the height j measures 10 inches.
Volume of the cylinder = Area of the base x height of cylinder
= πr² x j
= 3.14 x (1)² x 10
= 31.4
Therefore, the cylinder has a volume of 31.4 cubic inches.
By adding two volumes will get the total volume of the figure,
Total volume = Volume of the cone + Volume of the cylinder
= 37.68 + 31.4
= 69.08
Therefore, the total volume is 69.08 cubic meters.

Question 10.

Answer:
Finding the sides of the cube:
g² + h² = 3²
g² + h² = 9
g² + g² = 9 (since g = h)
2 g² = 9
g² = 9/2
g² = 4.5
g ~ 2.1(Square root of 4.5)
Therefore, the sides of the cube measures 2.1 cm.
Finding the volume of the cube:
Volume of the cube = s³
= (2.1)³
~ 9.3
Therefore, the volume of the cube is 9.3 cubic cm.
Finding the volume of the sphere:
Volume of the sphere = 4/3 πr³
= 4/3 x 3.14 x (1.05)³
= 4/3 x 3.14 x 1.2
= 4.8
Volume of the hemisphere = Volume of the sphere/2
= 4.8/2
= 2.4
Therefore, the hemisphere has a volume of 2.4 cubic cm.

Find the coordinates of the image under each translation. (Lesson 8.1)

Question 11.
A (3, -2) is translated 1 unit to the right and 8 units up.
Answer: Locate the point (3, -2) in the coordinate plane then move it 1 unit to the right and 8 times upwards as described by the translation. Thus the image for A will be on points (4,6) on the coordinate plane. Therefore, A(3, -2) is mapped onto A'(4,6).

Question 12.
B(-1, -6) is translated 5 units to the left and 3 units down.
Answer: Locate the point (-1,-6) in the coordinate plane then move it 5 units to the left and 3 units downwards as stated by the translation. Thus the image for B will be on points (-6,-9) of the coordinate plane.

Copy each diagram on graph paper and draw the image under each translation. (Lesson 8.1)

Question 13.
\(\overline{X Y}\) is translated 3 units to the left and 1 unit up.

Answer:

Question 14.
Square ABCD is translated 2 units to the right and 2 units down.

Answer:

Solve. Show your work.

Question 15.
On a coordinate plane, the coordinates of two points are A (3, 4) and B (3, 0). Point A’ is the image of point A and point 8′ is the image of point 8 under a reflection in the y-axis. (Lesson 8.2)

a) Find the coordinates of A’and B’.
Answer:

b) Draw the image of the line segment OA, where O is the origin, under a reflection in the x-axis. Use 1 grid square on both axes to represent 1 unit for the interval from -4 to 4.
Answer:

Question 16.
Triangle ABC and trapezoid STUV are shown on the coordinate plane. On a copy of the graph, draw the images of triangle ABC under a reflection in the y-axis. Then draw the image of trapezoid STUV under a reflection in the x-axis. (Lesson 8.2)

Answer:

Question 17.
A rotation of point A about the origin maps the point onto A’. State the angle and direction of rotation. (Lesson 8.3)

a)

Answer: The point A is on (1,1). The rotation of 90 degree clockwise, the coordinate should have changed from (x,y) to (y, -x). Since A’ is on (1, -1) from the original point (1,1) then it follows the rule for the 90 degree rotation in the clockwise direction.

b)

Answer: Point A is on (-2,-2). The rotation of 90 degree counterclockwise, the coordinate shifts from (x,y) to (-y,x). Since A’ is on (2,-2) from the original point (-2,-2) then it follows the rule for the 90 degree rotation in the counterclockwise direction.

Solve. Show your work.

Question 18.
Triangle ABC has vertices A (2, 3), 8 (3, 1) and C (4, 2). Draw the triangle ABC and its image under a rotation of 180° about the origin. Use 1 grid square on both axes to represent 1 unit for the interval from -4 to 4. (Lesson 8.3)
Answer:

Tell whether each transformation is a dilation. Explain. (Lesson 8.4)

Solve. Show your work.

Question 19.

Answer: The center of dilation is point C(or C’) since it is the point that did not change evn after the transformation. Also note that a transformation in dilation has a scale factor. The triangle A’B’C’ is 1/3 size compared to the original triangle, so the scale factor is 1/3. Therefore, the transformation is dilation.

Question 20.

Answer: Creating lines to connect points of the original rectangle to their corresponding points in their image to pinpoint the intersection of all points, the line won’t ever intersect. Thus, the center of dilation does not exist. Therefore, the transformation is not a dilation.

Solve. Show your work.

Question 21.
The figure described is mapped onto its image by a dilation with its center at the origin, O. Draw each figure and its image on a coordinate plane. (Lesson 8.4)

a) The vertices of triangle DEF are D (-3, -1), E (-3, -4), and F (-1, -3). Triangle DEF is mapped onto triangle D’E’F’ with scale factor -2.
Answer:
Coordinates of D x scale factor
(-3,-1) x -2
(6,2)
Thus the coordinate of D’ is (6,2). Next E’ corresponds to E, then use the coordinates of E to determine the coordinates E’ after the dilation.
Coordinates of E x scale factor
(-3,-4) x -2
(6,8)
Thus, the coordinate of E’ is (6,8). Lastly, F’ corresponds to F so use the coordinates of F to determine the coordinates of F’ after the dilation.
Coordinates of F x scale factor
(-1,-3) x -2
(2,6)
Thus the F’ coordinate (6,6)

b) The vertices of trapezoid ABCD are A (-2, -2), B (-2, -4), C (4, -4) and D (2, -2). Trapezoid ABCD is mapped onto trapezoid A’B’C’D’ with scale factor \(\frac{1}{2}\).
Answer:
Coordinate of A x scale factor
(-2,-2) x 1/2
= (-1,-1)
The coordinate of A’ is (-1,-1)
Coordinates of B x scale factor
(-2,-4) x 1/2
= (-1,-2)
The coordinate of B’ is (-1,-2)
Coordinates of C x scale factor
(4,-4) x 1/2
= (2,-2)
The coordinate of C’ is (2,-2)
Coordinates of D x scale factor
(2,-2) x 1/2
= (1,-1)
The coordinate of D is (1,-1)

Question 22.
The side lengths of a triangle are 3 inches, 6 inches, and 8 inches. The triangle undergoes a dilation. Find the side lengths of the ¡mage of the triangle for each of the scale factors in a) to d). Tell whether each dilation is an enlargement or a reduction of the original triangle. (Lesson 8.4)
a) 2
b) \(\frac{1}{4}\)
c) 120%
d) 1.6
Answer:

a) Side length of 8 inches x scale factor
8 x 2
16
Side length of 6 inches x scale factor
6 x 2
12
Side length of 3 inches x scale factor
3 x 2
6
Therefore, the 3 inches side of the triangle grows into a 6 inches side.

b) At a scale of 1/4
Side length of 8 inches x scale factor
8 x 1/4
2
Side length of 6 inches x scale factor
6 x 1/4
1.5
Side length of 3 inches x scale factor
3 x 1/4
0.75
Therefore, the 3 inches side of the triangle shrinks into a 0.75 inch side.

c) The scale factor is in a percentage form, divide the give percent into 100 to convert it into decimal.
120/100 = 1.2
Thus, the scale factor is 1.3. So, at a scale of 1.2
Side length of 8 inches x scale factor
8 x 1.2
9.6
Therefore, the 8 inches side of the triangle grows into a 9.6 inches side.
Side length of 6 inches x scale factor
6 x 1/2
7.2
Therefore, the 6 inches side of the triangle grows into a 7.2 inches side.
3 x 1.2
3.6
Therefore, the 3 inches side of the triangle grows into a 3.6 inches side.

d) Scale factor is 1.6
Side length of 8 inches x scale factor
8 x 1.6
12.8
Therefore, the 8 inches side of the triangle grows into 12.8 side.
Side length of 6 inches x scale factor
6 x 1.6
9.6
Therefore, the 6 inches side of the triangle grows into 9.6 inches side
Side length of 3 inches x scale factor
3 x 1.6
4.8
Therefore, the 3 inches side of the triangle grows into a 4.8 inches side.

Question 23.
Triangle A with vertices (-1, 3), (-1, 4) and (-3, 3) is mapped onto triangle B. Then triangle B is mapped onto triangle C as shown on the coordinate plane. (Lesson 8.5)

a) Describe the transformation that maps triangle A onto triangle B.
Answer: The point (-3,3) of triangle A is mapped onto (4,4) by moving 1 unit up and 7 units to the right. For the points of triangle A, it also ascends once and moves 7 times to the right. This transformation that moves the original figure up or down or left or right is translation.

b) Describe the transformation that maps triangle B onto triangle C.
Answer: Points (4,4), (6,4) and (6,5) mapped onto (-4,4),(-4,6) and (-5,6) changes the coordinates from its original for (x,y) into (-y,x). The transformation that changes the form of the coordinates from (x,y) into (-y,x) is a 90-degree rotation at the origin in a counterclockwise direction.

Solve. Show your work.

Question 24.
State whether the triangles are congruent. If they are congruent, write the statement of congruence and state the test used. (Lesson 9.1)

Answer: AB corresponds to XY and both the measures 10 cm, Triangle A corresponds to Triangle Y while Triangle C corresponds to Triangle Z and this two corresponding angles are congruent by Angle Angle Side Test which states that if the two corresponding angles of two triangles are congruent and any of their side are also congruent, then the triangles are congruent

Question 25.
∆AMS is congruent to ∆ERN. Find the values of x and y. (Lesson 9.1)

Answer:
To angle A of Triangle AMS and angle E of Triangle ERN are equal.
Angle A = Angle E
The measure of angle A is equals to 112 degree while measure of angle E is 16x degree
112 = 16x
112/16 = x
7 = x
Therefore, x is equal to 7
SM of triangle ASM and NR of triangle ERN are equal.
SM = NR
Measure of SM equals to 41 inches while measure of NR is (3x+5y)inches.
41 = (3x+5y)
using x = 7,
41 = 3(7) + 5y
41 = 21 + 5y
5y = 41 – 21
5y = 20
y = 20/5
y = 4

Triangle ABC is similar to triangle POR. Find the value of a. (Lesson 9.2)

Question 26.

Answer:
AC measuring 3 cm corresponds to PQ whose measure is 4.5 cm and CB measuring 5.2 corresponds to QR which measure a. Since the triangles are similar then their corresponding sides are proportional thus,
AC/PQ = CB/QR
3/4.5 = 5.2/a
3a = 23.4
a = 23.4/3
a = 7.8
Therefore, a is equal to 7.8

Name the test you can use to determine whether the two triangles are similar. Then find the value of x. (Lesson 9.2)

Question 27.

Answer:
Triangle ROS and POQ are vertical angles, and pair of opposite angles are formed due to an intersection of line is equal. Observe that angle R is congruent to angle P. Third corresponding angle is also equal. As there are two pairs of corresponding angles in Triangle ROS and POQ, third corresponding angle is also equal. By the Angle-Angle-Angle similarity theorem, which states ate if all the corresponding angles of two triangles are similar, that two triangles are similar.
Similarity of the triangles are proven, to solve x, RS which measures 12 km corresponds to PQ measuring x km and RO whose length is 13 km corresponds to PO measuring 27.3 km
RS/PQ = RO/PO
12/x = 13/27.3
12 x 27.3 = 13x
327.6 = 13x
x = 327.6/13
x = 25.2
Therefore, x = 25.2 km

Solve on a coordinate grid. (Lesson 9.3)

Question 28.
∆ABC undergoes two transformations to form the image ∆A”B”C”.
a) What are the coordinates of ∆A”B”C” if ∆ABC is first translated 2 units to the right and 3 units down, and then reflected in the line x = 2?
Answer:

b) What are the coordinates of ∆A”B”C” if ∆ABC is first reflected in the line x = 2, and then translated 2 units to the right and 3 units down?
Answer:

c) Do the two triangles ∆A”B”C” described in a) and b) have the same coordinates? Are they congruent? Explain.
Answer:
The two traingles are congruent since the only transformation is a reflection and transformation which is known for flipping the original images while translation is known for moving image up and down and left to right. So the size of triangle is still the same and are just plotted at different locations. For their coordinate, the locations are transformed twice, both depends on the first trnasformations that the triangles undergo. Triangle A”B”C” are on Quadrant III while the second one is on Quadrant IV. Therefore, their coordinates are not congruent.

Problem Solving

Solve. Show your work.

Question 29.
A pool table has a width of 5.5 inches. A ball travels a distance of 12.3 inches diagonally from one corner to the other corner of the table. Find the length of the pool table, rounded to the nearest tenth. (Chapter 7)
Answer:
Using pythagorean theorem,
(5.5)² + j² = (12.3)²
30.25 + j² = 151.29
j² = 151.29 – 30.25
j² = 121.04
j ~ 11 (Square root of 121.04)
Therefore, the length of the pool table is 11 inches.

Question 30.
When a drawbridge is open, the bridge forms a right angle with the doorway it leads to. A straight chain connects the far end of the bridge to the top of the doorway. If the doorway is 2 meters high and the chain is 2.5 meters long, find the length of the bridge, rounded to the nearest tenth. (Chapter 7)
Answer:
Using the pythagorean theorem,
y² + z² = x²
(2.5)² + 2² = x²
6.25 + 4 = x²
10.25 = x²
x ~ 3.2 (square root of 10.25)
Therefore, the drawbridge is around 3.2 m in length

Question 31.
Gabriel photocopied an exact copy of the shape ABCDE. The photocopied shape is shown on the right. (Chapter 9)

a) Name the shape congruent to ABCDE.
Answer: Point A corresponds to G , B point to H and point C corresponds to I, D corresponds to J and E corresponds to F. Therefore, the congruent shape to ABCDE is GHIJF.

b) Find the values of w, x, and y.
Answer:
Since ABCDE is congruent to GHIJF, their corresponding angles are equal. Thus, side AB of ABCDE is equal to GH of GHIJF, since these two sides corresponds to each other. So,
AB = GH
The measure of the lengths of AB is 2 feet while GH is 4y feet.
2 = 4y
y = 2/4
y = 0.5
Therefore, y is equal to 0.5
Side EA of ABCDE is equal to FG of GHIJF
EA = FG
The measure of the lengths of EA is (y+2x) feet while FG is 3.5 feet
y + 2x = 3.5
0.5 + 2x = 3.5
2x = 3.5 – 0.5
2x = 3
x = 3/2
x = 1.5
Therefore, x is equal to 1.5
Side BC of ABCDE is equal to HI of GHIJF
BC = HI
The measure of lengths of BC is (x + w) feet while HI is 5.6 feet
x + w = 5.6
1.5 + w = 5.6 (As x = 1.5)
w = 5.6 – 1.5
w = 4.1
Therefore, w is equal to 4.1

Solve. Show your work. (Chapters 7, 9)

Question 32.
Mary wants to pour 130 cm³ of water into the conical cup shown.

a) Find the volume of the conical cup. Use 3.14 as an approximation for π. Round your answer to the nearest tenth.
Answer:

Using pythagorean theorem,
(4.5)² + m² = (7.5)²
20.25 + m² = 56.25
m² = 56.25 – 20.25
m² = 36
m = 6
Therefore, the height of the cone is 6 cm
For the volume of the cone, 3.14 is the value of π and m is the height,
Volume of the cone = 1/3 Area of the base x Height of the cone
= 1/3πr² x m
= 1/3 x (3.14) x (4.5)² x (6)
= 127.17
Therefore, cone has a volume of 127.17 cubic cm.

b) Will the water overflow? Explain.
Answer: Since Mary wants to pour 127.17 cubic centimeters, the water will spill. The conical cup can only accommodate up to 127.17 cubic centimeters of liquid.

Question 33.
Two cones are similar in shape. The ratio of the diameters of their bases is 2 : 7. The radius of the smaller cone is 4.5 inches.

a) Find the radius of the larger cone.
Answer:
Convert the ratio 2:7 into fractions, then the result will be 2/7. Let s be the unknown measure of the radius, so
2/7 = 4.5/s
2 x s = 4.5 x 7
2s = 31.5
s = 31.5/2
s = 15.75
Therefore, the larger cone has a radius measuring 15.75 inches.

b) Write the ratio of the height of the smaller cone to the height of the larger cone.
Answer: Height of the smaller cone / Height of the larger cone

Solve. Show your work. (Chapters 8, 9)

Question 34.
A car is 25 feet long and 6 feet wide. A model of the same car is 9 inches long. What is the width of the model of the car if the model is a dilation of the car?
Answer:
Scale factor = Length of the model car/ Length of the actual car
= 0.75/25
= 0.03
Thus, the scale factor is 0.03
To solve width, multiplying the actual width to the scale factor,
Width of the model car = Width of the car x scale factor
= 6 x 0.03
= 0.18
Therefore, the width of the model car is 0.18 feet,
Converting this into inches, multiplying 0.18 with 12
Thus, the model car has a width of 2.16 inches.

Question 35.
Figure PQRST is dilated with center P and a scale factor of 1.45. Under this dilation, it is mapped onto figure UVWXY. Figure UVWXYis then mapped onto ABCDE by a reflection about segment UV.

a) Tell whether PQRST and ABCDE are congruent or similar figures. Explain.
Answer: The reflection that maps UVWXYZ onto ABCDE does not change the size and shape of the figure. Thus, they are congruent. The dilation that maps PQRST onto UVWXYZ however enlarges the size of PQRST. Therefore, UVWXYZ and PQRST are similar.

b) Find m∠UYX.
Answer: UVWXYZ and PQRST are congruent, then their corresponding angles are equal. The angle P corresponds to angle U, Q to angle V and angle R corresponds to W. The corresponding angle of S is angle X and angle T corresponds to Y . Since angle T corresponds to angle Y, then T is equal to Y. Angle T is equal to 120 degree. Therefore, angle Y is equal to 120 degree.

c) The area of PQRST is 20 square inches. Find the area of ABCDE.
Answer: The area of ABCDE is also a dilated image of PQRST. So, by multiplying the area of the actual image which is 20 square inches to the scale factor, the area of ABCDE
Area of ABCED = Area of PQRST x Scale factor
= 20 x 1.45
= 29
Therefore, ABCDE has an area of 29 square inches.

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 9 Lesson 9.3 Relating Congruent and Similar Figures to Geometric Transformations detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 9 Lesson 9.3 Answer Key Relating Congruent and Similar Figures to Geometric Transformations

Math in Focus Grade 8 Chapter 9 Lesson 9.3 Guided Practice Answer Key

Solve.

Question 1.
△ABC and △A’B’C’ are congruent isosceles triangles. Describe a rotation that maps △ABC onto △A’B’C’.

Answer:

Complete each with a value or word and each with +, —, x, or ÷.

Question 2.
△JKL is dilated by a scale factor of 1.2 to form △J’K’L’.

a) Find m∠J’.
△JKL and △J’K’L’ are triangles.
So, m∠J’ = m∠
– 50° – 30° = ∠ sum of triangle
m∠J’ is
Answer:
△JKL and △J’K’L’ are similar triangles.
So, m∠J’ = m∠J
180° – 50° – 30° = 100° ∠ sum of triangle
m∠J’ is 100°

b) Find the length of \(\overline{K^{\prime} L^{\prime}}\).
K’L’ = 2.5 1.2
= cm
Answer:
K’L’ = 2.5 × 1.2
We have to multiply the length with the scale factor to find the scaled length of \(\overline{K^{\prime} L^{\prime}}\).
= 3 cm

Question 3.
△UVW undergoes a geometric transformation to form △XYZ.

a) Identify whether △UVW ~ △XYZ.
m∠V 180° – – ∠ sum of triangle
=
Since m∠W = m∠Z and m∠V = m∠, two pairs of corresponding angles have equal measures. So, △UVW is to △XYZ.
Answer:
m∠V 180° – 110° – 40° ∠ sum of triangle
= 30°
Since m∠W = m∠Z and m∠V = m∠Y, two pairs of corresponding angles have equal measures. So, △UVW is ∼ to △XYZ.

b) If △UVW ~ △XYZ, what is the geometric transformation and scale factor?
Scale factor = \(\frac{Y Z}{?}\)
= \(\frac{6}{?}\)
=
△UVW undergoes a by a scale factor to form △XYZ.
Answer:
Scale factor = \(\frac{Y Z}{WV}\)
= \(\frac{6}{12}\)
= \(\frac{1}{2}\)
△UVW undergoes a dilation by a scale factor \(\frac{1}{2}\) to form △XYZ.

Copy and complete.

Question 4.
An engineer designs a manufacturing machine to transfer a food package GHIJ onto G’H’I’J’. Describe a sequence of two transformations that maps GHIJ onto G’H’l’J’.
GHIJ is mapped onto G’H’l’J’ by a translation of 2 units up and a reflection in the line .
Think Math
Does the order in which you do the transformations affect the image in Example 11 ? in 4?

Answer:
GHIJ is mapped onto G’H’l’J’ by a translation of 2 units up and a reflection in the line 2.

Technology Activity

Materials
geometry software

EXPLORE SEQUENCES OF TRANSFORMATIONS

STEP 1: Draw a triangle on a coordinate system using a geometry software. Record the position of each vertex.

STEP 2: Translate the triangle 1 unit down. Then reflect it in the x-axis. Select the Translate function and Reflect function, within the Transform menus. Record the position of each vertex of the image triangle.

STEP 3: Change the order of transformations in STEP 2. Record the position of each vertex of this image triangle.

STEP 4: Draw another triangle on a coordinate system using a geometry software. Record the position of each vertex.

STEP 5: Translate the triangle 1 unit to the right. Then dilate it with the center at the origin and scale factor 2. Select the Translate function and Dilate function, within the Transform menus. Record the position of each vertex of the image triangle.

STEP 6: Change the order of transformations in STEP 5. Record the position of each vertex of this image triangle.

Math journal Compare the positions of the original and transformed triangles. Does the order in which you perform two transformations make a difference in the position of the image triangle? Explain.
Answer:

Copy and complete on a graph paper.

Question 5.
DEFG is mapped onto D’E’F’G’ under a transformation. D”E”F”G” is the image of D’E’F’G’ under another transformation.

a) Describe the transformations that map DEFG onto D’E’F’G’ and D’E’F’G’ onto D”E”F”G”. Then describe a single transformation that maps DEFG onto D”E”F”G”.
DEFG is mapped onto D’E’F’G’ by using a reflection in the . D’E’F’G’ is mapped onto D”E”F”G” by using a rotation of about the origin. DEFG is mapped onto D”E”F”G” by using a single transformation, which is .
Answer:
DEFG is mapped onto D’E’F’G’ by using a reflection in the y-axis. D’E’F’G’ is mapped onto D”E”F”G” by using a rotation of 180 degrees about the origin. DEFG is mapped onto D”E”F”G” by using a single transformation, which is reflection in the x-axis.

b) Suppose the order of the transformations is reversed. Draw DEFG, D’E’F’G’, and D”E”F”G” on a coordinate plane. Does the order of the transformations affect the position of D”E”F”G”?
Answer: No, the order of the transformations affects the position of D”E”F”G” if you change the position in the y-axis.

Question 6.
△PQR is mapped onto △P’Q’R’ under a transformation. △P”Q”R” is the image of △P’Q’R’ under another transformation.
a) Describe the transformation that maps △PQR onto △P’Q’R’ and △P’Q’R’ onto △P”Q”R”.

Answer: △PQR is mapped on to △P’Q’R’ by the reflection in the line y = -1 and △P’Q’R’ is mapped onto △P”Q”R” on a coordinate plane by using the anticlockwise rotation of 90 degree about (-3, -3)

b) Suppose the order of the transformations is reversed. Draw △PQR, △P’Q’R’, and △P”Q”R” on a coordinate plane. Does the order of the transformations affect the position of △P”Q”R”?
Answer:

Question 7.
△PQR is mapped onto △P’Q’R’ under a transformation. △P”Q”R” is the image of △P’Q’R’ under another transformation.

a) Describe the transformations that map △PQR onto △P’Q’R’ and △P’Q’R’ onto △P”Q”R”. Then describe a single transformation that maps △PQR onto △P”Q”R”.
△PQR is mapped onto △P’Q’R’ by using a dilation with center (, ) and scale factor . △P’Q’R’ is mapped onto △P”Q” R” by using a rotation of about the point (, ) . △PQR can be mapped onto △P”Q”R” by a single dilation with center (imgg 2, ) and scale factor .
Answer:
△PQR is mapped onto △P’Q’R’ by using a dilation with center (0, 1) and scale factor 1. △P’Q’R’ is mapped onto △P”Q” R” by using a rotation of 180 degrees about the point (0, 1) . △PQR can be mapped onto △P”Q”R” by a single dilation with center (imgg 2, 1) and scale factor 1.

b) If the order of transformations is reversed, draw △PQR, △P’Q’R’, and △P”Q”R” on a coordinate plane.
Answer:

Math in Focus Course 3B Practice 9.3 Answer Key

State whether the figure and image are congruent or similar.

Question 1.
A triangle is rotated 180° about the origin.
Answer: If a triangle is rotated 180° about the origin then the figure is congruent.

Question 2.
A pentagon is translated 1 unit to the left and 5 units up.
Answer: Similar figure

Question 3.
A projector dilates a picture by a scale factor of 10, and projects the image on a screen.
Answer: Similar figure

Question 4.
A parallelogram is dilated with center (-2, 4) and scale factor 3.5, and rotated 90° clockwise.
Answer: Similar figure

Question 5.
A cartoon character is reflected in the y-axis and translated to the right.
Answer: Congruent figure

Solve on a coordinate grid.

Question 6.
△ABC undergoes two transformations.

a) What would be the coordinates of △A”B”C” if △ABC is first translated 2 units up and 3 units left, and then reflected in the line x = 1?
Answer:

b) What would be the coordinates of △A”B”C” if △ABC is first reflected in the line x = 1, and then translated 2 units up and 3 units left?
Answer:

c) Do the two triangles △A”B”C” have the same coordinates? Are they congruent? Explain.
Answer:

Solve.

Question 7.
△PQR is mapped onto triangle △P’Q’R’ by using a transformation. △P”Q”R” is the image of △P’Q’R’ by using another transformation. Describe the sequence of transformations from △PQR to △P”Q”R”.
a)

Answer: The above figure has dilation with center (0, 1) and scale factor of 0.5 and also we can see the reflection in the y-axis.

b)

Answer: From the above figure we can see the translation of 1 unit up and then a rotation of 90 degrees counterclockwise about the origin.

c)

Answer: P’Q’R” is the reflection in the x-axis and P”Q”R” is the reflection in the y-axis.

Question 8.
△ABC is mapped onto △A’B’C’, which is then mapped onto △A”B”C”. △ABC and △A”B”C” are shown in each diagram. Describe the sequence of transformations from △ABC to △A”B”C”. Then describe a single transformation from △ABC to △A”B”C”, if any.
a) △ABC is mapped onto △A’B’C’ by a rotation of 180° about (0, 0).

Answer:

b) △ABC is mapped onto △A’B’C’ by a reflection in the line y = 1.

Answer:

Solve. Show your work.

Question 9.
Tom is at position A of a hall whose floor is a square with a bay of window across the front of the room. The length of the square is 30 meters. Describe how he gets to position C by a translation followed by a rotation about the center of the square.

Answer:
Given,
Tom is at position A of a hall whose floor is a square with a bay of window across the front of the room.
The length of the square is 30 meters.
A translation of 30 meters forward followed by a rotation of 90 degrees clockwise direction about the center of the square.

Question 10.
The figure PQRST is dilated with center P and a scale factor 1.2. PQRST is mapped onto UVWXY. The area of PQRST is 24 square inches.

a) Find m∠WVX.
Answer: 75 degrees

b) Find the area of UVWXY.
Answer:
The area of PQRST is 24 square inches.
scale factor = 1.2
24 × 1.2 = 28.8 square inches.

Question 11.
Jack walks into a dark room. The area of the pupils of his eyes dilates to 3 times their normal area to allow more light into his eyes. By what scale factor is the diameter of his pupils enlarged?
Answer:
Given,
Jack walks into a dark room.
The area of the pupils of his eyes dilates to 3 times their normal area to allow more light into his eyes.
Diameter = √3 or 1.7321

Question 12.
Two similar cups are filled with water. The volume of water in the big cup is 80 cubic centimeters and the volume of water in the small cup is 10 cubic centimeters. If the height of the big cup is 8 centimeters, what is the height of the small cup?
Answer:
Given,
The volume of water in the big cup is 80 cubic centimeters and
the volume of water in the small cup is 10 cubic centimeters.
height of the big cup is 8 centimeters
Let the height of the small cup be x.
80/10 = 8/x
8 = 8/x
x = 8/8
x = 1
Therefore the height of the small cup is 1 cm.

Question 13.
Edward was blowing a circular bubble whose volume grew to 27 times its original size. It then drifted 8 units to the right and 11 units up before it burst.
a) Describe the sequence of transformations that the bubble went through.
Answer:
Given,
Edward was blowing a circular bubble whose volume grew to 27 times its original size.
It then drifted 8 units to the right and 11 units up before it burst.
V = 27 = 3³
So, the scale factor is 3
The sequence of transformations that the bubble went through is a dilation with the scale factor 3 and then a translation of 8 units to the right and 11 units up.

b) The original radius of the bubble was 2 centimeters. Find the final radius of the bubble.
Answer:
Given,
The original radius of the bubble was 2 centimeters.
Scale factor = 3
We know that
scale factor = scale/actual
3 = x/2
3 × 2 = x
x = 6 centimeters
Thus the final radius of the bubble is 6 centimeters.

Question 14.
An airplane at an airport terminal is cleared to take off by the control station after it moves from P to Q. The airplane’s path is marked by the thick arrows in the diagram. Describe the sequence of transformations that the airplane undergoes from P to Q.

Answer:

Brain @ Work

Question 1.
A bridge is strung between two big trees on opposite sides of a river at P and Q. The bridge is to be removed and a new bridge is to be built across the river. The new bridge starts at P and spans the shortest distance across the river. Briefly describe how you would find the length of the new bridge using congruent triangles.

Answer:
Given data,
A bridge is strung between two big trees on opposite sides of a river at P and Q.
The bridge is to be removed and a new bridge is to be built across the river.
The new bridge starts at P and spans the shortest distance across the river.

Draw a line from Q to R so that P, Q, R lie on the same straight line. Measure the distance TR such that it is perpendicular to the river bank. And TR is the length of new bridge using congruent triangles.

Question 2.
A souvenir in the shape of a rectangular prism is 20 centimeters by 15.5 centimeters by 13 centimeters. A pattern for a rectangular box, 29.2 centimeters by 18.6 centimeters by 16.9 centimeters, needs to be scaled down to fit the souvenir better. What is the minimum size of the box? Recommend a suitable set of measurements to the nearest 0.1 centimeter.

Answer:

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 9 Lesson 9.2 Understanding and Applying Similar Figures detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 9 Lesson 9.2 Answer Key Understanding and Applying Similar Figures

Math in Focus Grade 8 Chapter 9 Lesson 9.2 Guided Practice Answer Key

Identify the figures that seem similar. Explain why.

Question 1.

Answer: A and E are the similar triangle because all the sides are equal. The shape of A and E are the same but the size is different.

Hands-On Activity

Materials

• ruler
• scissors

EXPLORE ANGLE MEASURES IN SIMILAR TRIANGLES

STEP 1: Draw an acute-angled triangle. Name the angle at each vertex A, B, and C as shown below. Draw a line segment parallel to \(\overline{B C}\). Name the line segment \(\overline{D E}\).

STEP 2: Cut △ABC along its edges. Then cut the triangle along \(\overline{D E}\). You should have one small triangle, △ADE, and a trapezoid, BDEC.

STEP 3: Place ∠B on top of the angles in A ADE, one by one. Which angle has the same measure as ∠B?

STEP 4: Place ∠C on top of the angles of △ADE, one by one. Which angle has the same measure as ∠C?

STEP 5: In the diagram in STEP 1, \(\overline{D E}\) and \(\overline{B C}\) are parallel and intersected by \(\overline{A B}\) and \(\overline{A C}\). What do you know about angle measures formed when a line intersects two parallel lines? How does this fact support what you saw in STEP 3 and STEP 4?

Math Journal Triangles ADE and ABC are similar because they have the same shape but not the same size. In these triangles, △ADE and AABC are called corresponding angles. Two other pairs of corresponding angles are angles AED and ACB and angles DAE and BAC. What did you observe about the measures of each pair of corresponding angles?
Answer: ∠B and ∠D are the corresponding angles.
∠E and ∠C are the corresponding angles.

Solve.

Question 2.
Triangle ABC and triangle DBF are similar triangles. Find the value of x.

Answer:

Question 3.
An engineer wants to make a bridge across a river at \(\overline{P R}\). The diagram shows the known measurements. \(\overline{S P}\) and \(\overline{T R}\) are straight lines and triangle RPQ is similar to triangle TSQ. Find the width x of the river.

The width of the river is feet.
Answer:

The width of the river is 32 feet.

Question 4.
△ABC and △DEF are similar triangles. Find m∠D, m∠E, and m∠F.

The corresponding angles of similar triangles have measures.
So m∠D = m∠ = ,
m∠E = m∠ = ,
and m∠F = 180° – – ∠ sum of triangle
= .
Answer:
The corresponding angles of similar triangles have same or equal measures.
So m∠D = m∠A = 62°,
m∠E = m∠B = 51°,
and m∠F = 180° – 62° – 51° ∠ sum of triangle
m∠F = 67°

Question 5.
The area of △AXY is 12 square centimeters. Find the area of △ABC.
Use the ratio of corresponding lengths to find the ratio of the areas.

Let the area of △ABC be x square centimeters.
Use the ratio of the areas to find x.
\(\frac{x}{12}\) = k² The ratio of the areas equals k².
\(\frac{x}{12}\) = \(\frac{?}{?}\) Substitute for k²,
\(\frac{x}{12}\) • 12 = \(\frac{?}{?}\) • Multiply both sides by .
x = Simplify.
The area of △ABC is square centimeters.
Answer:

\(\frac{x}{12}\) = k² The ratio of the areas equals k².
\(\frac{x}{12}\) = \(\frac{81}{16}\) Substitute 81/16 for k²,
\(\frac{x}{12}\) • 12 = \(\frac{81}{16}\) • 12 Multiply both sides by 12.
x = 60.75 Simplify.
The area of △ABC is 60.75 square centimeters.

Hands-On Activity

Materials:

• two similar triangles
• protractor
• ruler

EXPLORE A MINIMUM CONDITION FOR TWO SIMILAR TRIANGLES

STEP 1: Measure the side lengths of the triangles. Find out if △ABC is similar to △PQR by finding the ratios of the corresponding side lengths.

STEP 2: Measure ∠A, ∠B, ∠P, and ∠Q. Complete the following:
m∠A = m∠ and m∠B = m∠.

Math Journal Without measuring ∠C and ∠R, do you know whether they have the same measure? Explain. State a minimum condition for two triangles to be similar.
Answer:
m∠A = m∠p and m∠B = m∠Q.

Identify whether △ABC is similar to △DEF. Explain with a test for similar triangles.

Question 6.

All three pairs of corresponding side lengths have the ratio.
So, △ABC △DEF.
Answer:

All three pairs of corresponding side lengths have the same ratio.
So, △ABC ≈ △DEF.

Question 7.

Two pairs of corresponding side lengths have ratios.
So, △ABC △DEF.
Answer:

Two pairs of corresponding side lengths have different ratios.
So, △ABC are not similar to △DEF.

Question 8.

m∠BAC ≠ m∠
Two pairs of corresponding side lengths have the ratio and the included angles have measures.
So, △ABC △DEF.
Answer:

m∠BAC ≠ m∠DEF
Two pairs of corresponding side lengths have the different ratio and the included angles have different measures.
So, △ABC is not similar to △DEF.

Math in Focus Course 3B Practice 9.2 Answer Key

Identify the figures that seem similar. Explain why.

Question 1.

Answer:
A and D are similar figures with the same shape.

Question 2.

Answer: B, C and D are similar figures with a similar shape.

Triangle ABC is similar to triangle POR. Find the scale factor by which △ABC is enlarged to △PQR.

Question 3.

Answer:
Scale factor = scale/actual length
Scale factor = 4.5/3 = 1.5

Question 4.

Answer:
Scale factor = scale/actual length
Scale factor = 12/4 = 3

Question 5.

Answer:
Scale factor = scale/actual length
Scale factor = 20/15 = 4/3 = 1 1/3

Each pair of figures are similar. Find the value of each variable.

Question 6.
ABCD ~ EFGH

Answer:
By seeing the above figures we can say that it is a parallelogram.
We know that the opposite sides of the parallelogram are equal.
AB = 3EF
AD = 3EH
AB = 6 in.
EF = 3 × 6 in = 18 in.
AD = 9 in.
EF = 3x = 3 × 9 = 27 in.
Thus the value of x is 6 in.

Question 7.
△ABC ~ △DEF

Answer:
As the given triangles are similar the angles will be same.
∠B = 64°
∠C = 78°
Sum of angles = 180°
∠x° + 64° + 78° = 180°
∠x° = 180° – 64° – 78°
x° = 38°
x° = y° = 38°

Question 8.
ABCD ~ EFGH

Answer:
90/6 = 10/x
9/6 = 1/x
x × 9 = 6
x = 6/9
x = 2/3

Question 9.
△ABC ~ △DEF

Answer:
AB = ED
BC = EF
AC = DF
BC/EF = 4.5/3 = 1.5
Scale factor = 1.5
AD = 5.2 cm
AB = 5.2 × 1.5 = 7.8
z° = 35°

Explain, with a test, why the two triangles in each figure are similar. Find the unknown lengths.

Question 10.

Answer:
By using the SSS test we can say that the two triangles are similar because the shape of two triangles are same.
AB = DE
AC = CE
BC = CD
AB = 2 cm
DE = 4 cm
scale factor = 4/2 = 2
BC = 1 cm
CD = 1 × 2 = 2 cm
CD = y = 2 cm
CE = 3 cm
CE = 3/2 = 1.5 cm
x = 1.5 cm

Question 11.

Answer:
AB = AE
2 + 1 = 3 m
So, AE = x = 3 m
AD + DE = BC
2 + 4 = 6m
Thus x = 3 and y = 6

Solve. Show your work.

Question 12.
Mia made some copies of a drawing using a photocopier. The drawing was either enlarged or reduced. Find the value of x.
a)

Answer:
7.5/6 = 1.25
3 × 1.25 = 3.75 in.
Thus the value of x is 3.75 in.

b)

Answer:
3/2.25 = 1.3 in
x = 2.4 × 1.3 = 3.12 in
Thus the value of x is 3.12 in.

Question 13.
The slope of a wheelchair ramp is \(\frac{1}{15}\).

a) Suppose a wheelchair ramp has to rise 3 feet. Find the horizontal distance it covers.
Answer:
3/x = 1/15
3 × 15 = x
x = 45 feet
Thus the horizontal distance it covers is 45 feet.

b) Suppose there is space for a wheelchair ramp to cover at most 30 feet horizontally. How high can it rise then?
Answer:
x/30 = 1/15
x = 30/15
x = 2
Thus it rise 2 feet.

Question 14.
A circle has 9 times the area of another circle. If the radius of the larger circle is 27 meters, find the radius of the smaller circle.
Answer:
Given,
A circle has 9 times the area of another circle.
Radius of the larger circle is 27 meters.
Area of circle (A1) = 9 Area of circle2(A2)
A1 = 9A2
We know that,
Area of the circle = πr²
where r is the radius of the circle
r1 = 27 meters
r2 = ?
A1 = πr²
A1 = 3.14 × r²
A1 = 9A2
π(27)² = 9πr²
r² = (27)² /9
r = 27/3
r2 = 9 meters
Thus the radius of the smaller circle = 9 meters

Question 15.
The two ellipses shown on the right are similar. The area of the inner ellipse is 18 square inches and the shaded area is 32 square inches. Find the values for x and y.

Answer:
Given,
The area of the inner ellipse is 18 square inches.
a = 12/2 = 6 in.
18 = 6 × b
18/6 = b
b = 3
x = 3 + 3
x = 6 in.

Question 16.
Two trees on a street have heights 10 feet and 28 feet. At a certain time of a day, the shorter tree casts a shadow of length 15 feet on the ground. How far apart are the trees?

Answer:
Given,
Two trees on a street have heights 10 feet and 28 feet.
At a certain time of the day, the shorter tree casts a shadow of length 15 feet on the ground.
Let x be the distance between the two trees.
x + 15
10 : 28 = x : x + 15
10/28 = x/x + 15
10(x + 15) = x × 28
10x + 150 = 28x
150 = 28x – 10
18x = 150
x = 150/18
x = 8.3
x + 15 = 8.3 + 15 = 23.3 feet

Question 17.
A bag in the shape of two regular pentagons is shown on the right. The ratio of the shaded area to the area of the larger pentagon is 27 : 36. If the larger pentagon has sides of length 4 inches, find the length of the sides of the smaller pentagon.

Answer:
Given,
A bag in the shape of two regular pentagons is shown on the right.
The ratio of the shaded area to the area of the larger pentagon is 27 : 36.
Larger pentagon has sides of length 4 inches
Let the length of the sides of the smaller pentagon be x.
27/36 = x/4
3/4 = x/4
(3 × 4)/4 = x
x = 12/4
x = 3
Thus the length of the sides of the smaller pentagon is 3 inches.

Question 18.
A cable car travels from a village up to a resort on the top of a mountain. When the cable car has traveled 150 feet along the cable, it is 100 feet above the ground. The total distance the cable car must travel to reach the resort is 12,000 feet. How high is the mountain?

Answer:
Given that,
A cable car travels from a village up to a resort on the top of a mountain. When the cable car has traveled 150 feet along the cable, it is 100 feet above the ground.
The total distance the cable car must travel to reach the resort is 12,000 feet.
12000 ÷ 150 = 80
100 × 80 = 8000 feet
Thus the height of the mountain is 8000 feet.

Question 19.
Math journal Are similar figures ever congruent? If so, under which conditions are they congruent? Use two congruent triangles and two similar triangles in your explanation.
Answer:
Similar figures are not congruent until the scale factor is 1 or -1. The two triangles are said to be similar when all the three pairs of the figure are corresponding sides have the same ratio. If the ratio of the scale factor is 1 or -1 then the corresponding lengths are the same. When all the three pairs of corresponding lengths are the same then the triangle is said to be congruent.

Question 20.
Math journal Do you know all circles are similar? What other geometric shapes are similar?
Answer:
All circles are said to be similar because the radii are equidistant from their center.
Two shapes are said to be similar if and if it has the same shape but not the same size.