Prasanna

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 8 Quiz Answer Key.

Texas Go Math Grade 8 Module 8 Quiz Answer Key

Texas Go Math Grade 8 Module 8 Ready to Go On? Answer Key

8.1 The Pythagorean Theorem

Find the length of the missing side.

Question 1.

Answer:
Let b = 21 and c = 35 Using the Pythagorean Theorem, we have:
a² + b² = c²
a² + 21² = 35²
a² + 441 = 1225
a² + 441 – 441 = 1225 – 441
a² = 784
a = \(\sqrt {281}\)
The length of the missing side is 28 m.

The Pythagorean Theorem Iready Quiz Answers Question 2.

Answer:
Let a = 16 and b = 30 Using the Pythagorean Theorem, we have:
a² + b² = c²
16² + 30² = c²
256 + 900 = c²
1156 = c²
c = \(\sqrt {1156}\)
c = 34
The length of the missing side is 34 m.

8.2 Converse of the Pythagorean Theorem

Tell whether each triangle with the given side lengths Is a right triangle.

Question 3.
11, 60, 61 _____________
Answer:
Let a = 11, b = 60 and c = 61. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
11² + 60² = 61²
121 + 3600 = 3721
3721 = 3721
True
Since 11² + 60² = 61², the triangle is a right triangle.

Question 4.
9, 37, 40 _____________
Answer:
Let a = 9, b 37 and c = 40. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
9² + 37² = 40²
81 + 1369 = 1600
1450 = 1600
False
Since 9² + 37² ≠ 40², the triangle is not a right triangle.

Module 8 Test Quiz Pythagorean Theorem Answer Key Question 5.
15, 35, 38 _____________
Answer:
Let a = 15,b = 35, and c = 38.
a² + b² = c² …………… (1)
15² + 35² = 38² (Substitute into the formula) ……………. (2)
225 + 1225 = 1444 (Simplify) …………. (3)
1450 ≠ 1444 (Add) …………. (4)
Since 15² + 35² ≠ 38², the triangle is not a right triangle by the converse of the Pythagorean Theorem.

Question 6.
28, 45, 53 _____________
Answer:
Let a = 28, b = 45 and c = 53. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
28² + 45² = 53²
784 + 2025 = 2809
2809 = 2809
True
Since 28² + 45² = 53² the triangle is a right triangle.

Question 7.
Keelie has a triangular-shaped card. The lengths of its sides are 4.5 cm, 6 cm, and 7.5 cm. Is the card a right triangle?
Answer:
Let a = 4.5, b = 6 and c = 7.5. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
4.5² + 6² = 7.5²
20.25 + 36 = 56.25
56.25 = 56.25
True
Since 4.5² + 6² = 7.5² , the card is a right triangle.

8.3 Distance Between Two Points

Find the distance between the given points. Round to the nearest tenth.

Math Quiz for Grade 8 Pythagorean Theorem Question 8.
A and B _____________
Answer:
Using the Distance Formula, the distance d between the points A(-2, 3) and B(4, 6) is:
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
d = \(\sqrt{(4-(-2))^{2}+(6-3)^{2}}\)
d = \(\sqrt{(6)^{2}+(3)^{2}}\)
d = \(\sqrt{36+9}\)
d = \(\sqrt{45}\)
Rounding the answer to the nearest tenth:
d ≈ 6.7
The distance between points A and B is approximately 6.7 units.

Question 9.
B and C. _____________
Answer:
Using the Distance Formula, the distance d between the points B(4, 6) and C(3, -1) is:
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
d = \(\sqrt{(3-4)^{2}+(-1-6)^{2}}\)
d = \(\sqrt{(-1)^{2}+(-7)^{2}}\)
d = \(\sqrt{1+49}\)
d = \(\sqrt{50}\)
Rounding the answer to the nearest tenth:
d ≈ 7.1
The distance between points B and C is approximately 7.1 units.

Question 10.
A and C _____________
Answer:
Using the Distance Formula, the distance d between the points A(-2, 3) and C(3, -1) is:
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
d = \(\sqrt{(3-(-2))^{2}+(-1-3)^{2}}\)
d = \(\sqrt{(5)^{2}+(-4)^{2}}\)
d = \(\sqrt{25+16}\)
d = \(\sqrt{41}\)
Rounding the answer to the nearest tenth:
d ≈ 6.4
The distance between points A and C is approximately 6.4 units.

Essential Question

Grade 8 Pythagorean Theorem Quiz Answer Key Question 11.
How can you use the Pythagorean Theorem to solve real-world problems?
Answer:
We can use the Pythagorean Theorem to find the length of a side of a right triangle when we know the lengths of the other two sides. This application is usually used in architecture or other physical construction projects. For example, it can be used to find the length of a ladder, if we know the height of the wall and the distance on the ground from the wall of the ladder.

Texas Go Math Grade 8 Module 8 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
What is the missing length of the side?

(A) 9 ft
(B) 30 ft
(C) 39 ft
(D) 120 ft
Answer:
By using the Pythagorean theorem,
a² + b² = c²
80² + b² = 89²
b² = 89² – 80²
b = √89² – 80²
b = √9²
b = 9 ft
Option A is the correct answer.

Question 2.
Which relation does not represent a function?
(A) (0, 8), (3, 8), (1, 6)
(B) (4, 2), (6, 1), (8, 9)
(C) (1, 20), (2, 23), (9, 26)
(D) (0, 3), (2, 3), (2, 0)
Answer: D is the relation that does not represent a function.

Pythagorean Theorem Module 8 Quiz Answers Question 3.
Two sides of a right triangle have lengths of 72 cm and 97 cm. The third side is not the hypotenuse. How long is the third side?
(A) 25 cm
(B) 45 cm
(C) 65 cm
(D) 121 cm
Answer:
(C) 65 cm

Explanation:
Let b = 72 and c = 97, Using the Pythagorean Theorem, we have:
a² + b² = c²
a² + 72² = 97²
a² + 5184 = 9409
a² + 5184 – 5184 = 9409 – 5184
a² = 4225
a = 65

Question 4.
What is the distance between point F and point G?

(A) 4.5 units
(B) 5 units
(C) 7.3 units
(D) 20 units
Answer:
(A) 4.5 units

Explanation:
Write the coordinates of the points F(-1, 6) and G(3, 4), and find the distance between F and G. Using the Distance Formula

Pythagorean Theorem Quiz Questions and Answers Pdf Question 5.
A flagpole is 53 feet tall. A rope is tied to the top of the flagpole and secured to the ground 28 feet from the base of the flagpole. What is the length of the rope?
(A) 25 feet
(B) 45 feet
(C) 53 feet
(D) 60 feet
Answer:
(D) 60 feet

Explanation:
A flagpole, rope and the length of secure point from the flagpole are forming the right triangle, where the rope is a hypotenuse Let a = 53 ft and b = 28 ft Use Pythagorean Formula to find length of the rope c.
a² + b² = c² ……………… (1)
c² = 53² + 28² (Substitute into formula) …………….. (2)
c² = 2809 + 784 (Simplify) …………… (3)
c² = 3593 (Add) …………. (4)
c = 59.94163 (Take the square root from both sides) …………. (5)
c ≈ 60 ft (Round to the nearest whole number) …………. (6)

Question 6.
Which set of lengths are not the side lengths of a right triangle?
(A) 36, 77, 85
(B) 20, 99, 101
(C) 27, 120, 123
(D) 24, 33, 42
Answer:
(D) 24, 33, 42

Explanation:
Check if side lengths in option A form a right triangle.
Let a = 36, b = 77 and c = 85. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
36² + 77² = 85²
1296 + 5929 = 7225
7225 = 7225
True
Since 36² + 77² = 85², the triangle is a right triangle.

Check if side lengths in option B form a right triangle.
Let a = 20, b = 99 and c = 101. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
20² + 99² = 101²
400 + 9801 = 10201
10201 = 10201
True
Since 20² + 99² = 101², the triangle is a right triangle

Check if side Lengths in option C form a right triangle
Let a = 27, b = 120 and c = 123. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
27² + 120² = 123²
729 + 14400 = 15129
15129 = 15129
True
Since 27² + 120² = 123², the triangle is a right triangle.

Check if side lengths in option D form a right triangle
Let a = 24, b = 33 and c = 42. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
24² + 33² = 42²
576 + 1089 = 1764
1665 = 1764
False
Since 24² + 33² ≠ 42², the triangle is not a right triangle.

Math Quiz for 8th Grade Geometry Module 8 Test Question 7.
Which is an irrational number?
(A) 5.4
(B) \(\sqrt {7}\)
(C) -13
(D) \(\frac{2}{3}\)
Answer: (D) \(\frac{2}{3}\)
Explanation:
\(\frac{2}{3}\) = 0.66666…
0.66666… is a recurring number.
So, Option D is the correct answer.

Question 8.
A triangle has one right angle. What could the measures of the other two angles be?
(A) 25° and 65°
(B) 30° and 15°
(C) 55° and 125°
(D) 90° and 100°
Answer:
A triangle has one right angle
Sum of three angles = 180°
25° + 65° + 90° = 180
So, option A is the correct answer.

Gridded Response

Module 8 Test Answers Math 8th Grade Question 9.
A right triangle has legs that measure 1.5 centimeters and 2 centimeters. What is the length of the hypotenuse in centimeters?

Answer:
A right triangle has legs that measure 1.5 centimeters and 2 centimeters.
Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
1.5² + 2² = c²
c² = 1.5² + 2²
c² = 2.25 + 4
c² = 6.25
c = 2.5
Therefore, the length of the hypotenuse is 2.5 cm.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 9 Answer Key Volume.

Texas Go Math Grade 8 Module 9 Answer Key Volume

Texas Go Math Grade 8 Module 9 Are You Ready? Answer Key

Evaluate each exponential expression

Question 1.
11² ___________
Answer:
Multiply the base (11) by itself the number of times indicated by the exponent (2). Then find the product of all the terms.
11² = 11 × 11 = 121

Question 2.
2⁵ ____________
Answer:
2⁵ = 2 × 2 × 2 × 2 × 2 = 32

Question 3.
(\(\frac{1}{5}\))³ ___________
Answer:
(\(\frac{1}{5}\))³ = \(\frac{1}{5}\) × \(\frac{1}{5}\) × \(\frac{1}{5}\) = (\(\frac{1}{125}\))

Go Math Grade 8 Module 9 Answer Key Pdf Question 4.
(0.3)² ___________
Answer:
(0.3)² = \(\frac{3}{10}\) × \(\frac{3}{10}\) = \(\frac{9}{100}\) = 0.09

Question 5.
2.1³ ____________
Answer:
2.1³ = 2.1 × 2.1 × 2.1 = 9.261

Question 6.
0.1³ ____________
Answer:
(0.1)³ = \(\frac{1}{10}\) × \(\frac{1}{10}\) × \(\frac{1}{10}\) = \(\frac{1}{1000}\) = 0.001

Question 7.
(\(\frac{9.6}{3}\))² ___________
Answer:
(\(\frac{9.6}{3}\))² = (3.2)² = 3.2 × 3.2 = 10.24

Question 8.
100³ ____________
Answer:
(100)³ = 100 × 100 × 100 = 1000000

Round to the underlined place.

Question 9.
2.374 _____
Answer:
2.374 → 2.37

Question 10.
126.399 _____
Answer:
126.399 → 126

Question 11.
13.9577 _____
Answer:
13.9577 → 14.0

Go Math Module 9 Answer Key Grade 8 Question 12.
42.690 _________
Answer:
42.690 → 42.69

Question 13.
134.95 _____
Answer:
134.95 → 135.0

Question 14.
2.0486 ______
Answer:
2.0486 → 2.0

Question 15.
63.6352 _______
Answer:
63.6352 → 63.64

Question 16.
98.9499 ________
Answer:
98.9499
The digit to be rounded is 9. The digit to its right is 4, which is less than 5, so the rounded number is 98.9.
98.9499 → 98.9

Simplify each expression.

Question 17.
3.14 (5)² (10) _______
Answer:
To simplify the expression, we simplify the exponent and we multiply from left to right.
3.14 ∙ (5)² ∙ 10 = 3.14 ∙ 25 ∙ 10
= 78.5 ∙ 10
= 785

Grade 8 Math Module 9 Answer Key Go Math Question 18.
\(\frac{1}{3}\)(3.14) (3)² (5)
Answer:
(\(\frac{1}{3}\)) × (3.14) × (3)² × (5) = 3 × 3.14 × 5 = 47.1

Question 19.
\(\frac{4}{3}\) (3.14) (3)³ __________
Answer:
(\(\frac{4}{3}\)) × (3.14) × 27 = 4 × 3.14 × 9 = 113.04

Question 20.
\(\frac{4}{3}\) (3.14) (6)³ _______
Answer:
(\(\frac{4}{3}\)) × (3.14) × (6³) = (\(\frac{4}{3}\)) × 3.14 × 216 = 904.32

Question 21.
3.14 (4)² (9) _______
Answer:
3.14 × (4²) × 9 = 3.14 × 16 × 9 = 452.16

Question 22.
\(\frac{1}{3}\)(3.14) (9)² (\(\frac{2}{3}\))
Answer:
(\(\frac{1}{3}\)) × (\(\frac{2}{3}\)) × (9²) × 3.14 = (\(\frac{1}{3}\)) × (\(\frac{2}{3}\)) × 81 × 3.14 = 2 × 3.14 × 9 = 56.52

Texas Go Math Grade 8 Module 9 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the empty columns in the chart. You may use words more than once.

Understand Vocabulary

Complete the sentences using the preview words.

Question 1.
A three-dimensional figure that has one vertex and one circular base is a __________.
Answer:
A three-dimensional figure that has one vertex and one circular base is a cone.

Go Math Answer Key Grade 8 Module 9 Test Answers Question 2.
A three-dimensional figure with all points the same distance from the center is a __________.
Answer:
A three-dimensional figure with all points the same distance from the center is a sphere.

Question 3.
A three-dimensional figure that has two congruent circular bases is a _________.
Answer:
A three-dimensional figure that has two congruent circular bases is a cylinder.

Active Reading
Three-Panel Flip Chart Before beginning the module, create a three-panel flip chart to help you organize what you learn. Label each flap with one of the lesson titles from this module. As you study each lesson, write important ideas like vocabulary, properties, and formulas under the appropriate flap.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 10 Answer Key Surface Area.

Texas Go Math Grade 8 Module 10 Answer Key Surface Area

Texas Go Math Grade 8 Module 10 Are You Ready? Answer Key

Find the area of each figure.

Question 1.
square with side lengths of 9.3 cm __________
Answer:
Given,
length of the side = 9.3 cm
We have to find the area of a square.
We know that,
Area of a square = s × s
A = 9.3 × 9.3
A = 86.49 sq. cm
Thus the area of the square is 86.49 sq. cm.

Grade 8 Math Module 10 Answer Key Question 2.
triangle with base 14 in. and height 7 in ____________
Answer:
Given,
base = 14 in.
height = 7 in.
We know that,
Area of a triangle = 1/2 × base × height
Area = 1/2 × 14 × 7
A = 7 × 7
A = 49 sq. inches
Thus the area of the triangle is 49 sq. inches.

Question 3.
rectangle with length 2\(\frac{1}{2}\) ft and width 1\(\frac{1}{2}\) ft ___________
Answer:
Given,
length = 2\(\frac{1}{2}\) ft
width = 1\(\frac{1}{2}\) ft
We know that,
Area of a rectangle =  length ×  width
A = \(\frac{5}{2}\) × \(\frac{3}{2}\)
A = \(\frac{15}{2}\)
A = 7\(\frac{1}{2}\) sq. ft
Thus the area of a rectangle is 7\(\frac{1}{2}\) sq. ft

Question 4.
triangle with base 6.4 m and height 3.8 m ___________
Answer:
Given,
base = 6.4 m
height = 3.8 m
We know that,
Area of a triangle = 1/2 × base × height
Area = 1/2 × 6.4 m × 3.8 m
A = 3.2 × 3.8
A = 12.16 sq. meters
Thus the area of the triangle is 12.16 sq. meters

Find the area of each circle. Use 3.14 for π.

Go Math Grade 8 Answer Key Pdf Module 10 Question 5.
r = 18 in. _________
Answer:
Given,
radius = 18 inches
We know that,
Area of a circle = πr²
Area = 3.14 × (18)²
Area =1017.88 sq. inches
Thus the area of the circle is 1017.88 sq. inches.

Question 6.
r = 7.5 m __________
Answer:
Given,
radius = 7.5 m
We know that,
Area of a circle = πr²
Area = 3.14 × (7.5)²
Area =176.71 sq. meters
Thus the area of the circle is 1017.88 sq. meters.

Module 10 Answer Key Grade 8 Go Math Question 7.
r = 2\(\frac{1}{2}\) ft ___________
Answer:
Given,
radius =2\(\frac{1}{2}\) ft
We know that,
Area of a circle = πr²
Area = 3.14 × (2\(\frac{1}{2}\))²
Area =19.63 sq. ft
Thus the area of the circle is 19.63 sq. ft

Texas Go Math Grade 8 Module 10 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the chart.

Understand Vocabulary

Complete the sentences using the review words.

Question 1.
A three-dimensional figure in which six faces are rectangles is a ___________.
Answer: A three-dimensional figure in which six faces are rectangles is a cuboid.

Texas Go Math Grade 8 Module 10 Surface Area Question 2.
An individual surface of a solid object is a ____________.
Answer: An individual surface of a solid object is a face.

Active Reading
Tri-Fold Before beginning the module, create a tri-fold to help you learn the concepts and vocabulary in this module. Fold the paper into three sections. Label the columns “What I Know”, “What I Need to Know,” and “What I Learned.” Complete the first two columns before you read. After studying the module, complete the third column.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 7.2 Answer Key Angle Theorems for Triangles.

Texas Go Math Grade 8 Lesson 7.2 Answer Key Angle Theorems for Triangles

Texas Go Math Grade 8 Lesson 7.2 Explore Activity Answer Key

Explore Activity 1

Sum of the Angle Measures in a Triangle

There is a special relationship between the measures of the interior angles of a triangle.

A. Draw a triangle and cut it out. Label the angles A, B, and C.
B. Tear off each “corner” of the triangle. Each corner includes the vertex of one angle of the triangle.
C. Arrange the vertices of the triangle around a point so that none of your corners overlap and there are no gaps between them.
D. What do you notice about how the angles fit together around a point?
E. What is the measure of a straight angle?
F. Describe the relationship among the measures of the angles of △ ABC.
The Triangle Sum Theorem states that for △ ABC, m∠A + m∠B + m∠C = _________.

Reflect

Question 1.
Justify Reasoning Can a triangle have two right angles? Explain.
Answer:
We already know that all angles of a triangle add up to 180°. If the first two angles add up to 180° (2 right angles, 90° each), it means that the third one must measure 0°, which does not make sense. Therefore, a triangle cannot have two right angles.

Lesson 7.2 Isosceles and Equilateral Triangles Answer Key Question 2.
Analyze Relationships Describe the relationship between the two acute angles in a right triangle. Explain your reasoning.
Answer:
A right triangle is a triangle that has one right angle and two other acute angles. Since we know that all angles of a triangle add up to 180°, and we have one angle of 90° in a right triangle, it means that the sum of the other two acute angles will be 90° (180° – 90°).

Explore Activity 2

Justifying the Triangle Sum Theorem

You can use your knowledge of parallel lines intersected by a transversal to informally justify the Triangle Sum Theorem.

Follow the steps to informally prove the Triangle Sum Theorem. You should draw each step on your own paper. The figures below are provided for you to check your work.
A. Draw a triangle and label the angles as ∠1, ∠2, and ∠3 as shown.

B. Draw line a through the base of the triangle.
C. The Parallel Postulate states that through a point not on a line 2, there is exactly one line parallel to line 2. Draw line b parallel to line o, through the vertex opposite the base of the triangle.
D. Extend each of the non-base sides of the triangle to form transversal s and transversal t. Transversals s and t intersect parallel lines a and b.
E. Label the angles formed by line b and the transversals as ∠4 and ∠5.

F. Because ∠4 and ___________ are alternate interior angles, they are ____________.
Label ∠4 with the number of the angle to which it is congruent.
G. Because ∠5 and ___________ are alternate interior angles, they are ____________.
Label ∠5 with the number of the angle to which it is congruent.
H. The three angles that lie along line b at the vertex of the triangle are ∠1, ∠4, and ∠5. Notice that these three angles lie along a line.
So, m∠1 + m∠4 + m∠5 = ______________
Because angles 2 and 4 are congruent and angles 3 and 5 are congruent, you can substitute m∠2 for m∠4 and m∠3 for m∠5 in the equation above.
So, m∠1 + m∠2 + m∠3 = ______________
This shows that the sum of the angle measures in a triangle is always _____________.

Reflect

Question 3.
Analyze Relationships How can you use the fact that m∠4 + m∠1 + m∠5 = 180° to show that m∠2 + m∠1 + m∠3 = 180°?
Answer:
m∠4 + m∠1 – m∠5 = 180° ………….. (1)
m∠2 = m∠4 since ∠2 and ∠4 are alternate interior angles
m∠3 = m∠5 since ∠3 and ∠5 are alternate interior angles
Therefore, we substitute m∠2 for m∠4 and m∠3 for m∠5 in equation (1) and we get:
m∠2 + m∠1 + m∠3 = 180° …………….. (2)

Your Turn

Find the missing angle measure.

Question 4.

Answer:
From the Triangle Sum Theorem we have:
m∠J + m∠L +m∠K = 1800
We substitute the given angle measures and we solve for m∠K.
71° + 56° + m∠K = 180°
127° + m∠K = 180°
127° – 127° + m∠K = 180° – 127°
m∠K = 53°

Angle Theorems Grade 8 Interior and Exterior Angles Question 5.

Answer:
To determine the answer here, we will use the following rule:
α + β + γ = 180°
where α, β and γ are angles in a triangle.
Therefore, we know the following:
m∠S = 29°
m∠T = 61°
m∠R =?
Using the rule given in Step 1, we can compute as follows:
m∠S + m∠T + m∠R = 180°
29° + 610 + m∠R = 180°
m∠R = 180° – 29° – 61°
m∠R = 90°
Hence, the missing angle measure is:
m∠R = 90°

Explore Activity 3

Exterior Angles and Remote Interior Angles

An interior angle of a triangle is formed by two sides of the triangle. An exterior angle is formed by one side of the triangle and the extension of an adjacent side. Each exterior angle has two remote interior angles. A remote interior angle is an interior angle that is not adjacent to the exterior angle.

• ∠1, ∠2, and ∠3 are interior angles.
• ∠4 is an exterior angle.
• ∠1 and ∠2 are remote interior angles to ∠4.

There is a special relationship between the measure of an exterior angle and the measures of its remote interior angles.

A. Extend the base of the triangle and label the exterior angle as ∠4.
B. The Triangle Sum Theorem states: m∠1 + m∠2 + m∠3 = ____________.
C. ∠3 and ∠4 form a ____________,
so m∠3 + m∠4 = _____________.
D. Use the equations in B and c to complete the following equation:
m∠1 + m∠2 + _________ = _________ + m∠4
E. Use properties of equality to simplify the equation in D: _____________
The Exterior Angle Theorem states that the measure of an _________ angle is equal to the sum of the measure of its ___________ angles.

Reflect

Question 6.
Sketch a triangle and draw all of its exterior angles. How many exterior angles does a triangle have at each vertex?
Answer:
Angles ∠1, ∠2, and ∠3 are interior angles.
∠4 and ∠5 are exterior angles for ∠1.
∠6 and ∠7 are exterior angles for ∠2.
∠8 and ∠9 are exterior angles for ∠3.
So, we have two exterior angles at each vertex.

Lesson 7.2 Angle Theorems for Triangles Question 7.
How many total exterior angles does a triangle have?
Answer:
There are two exterior angles at each of the three vertex, so we have a total of 6 exterior angles.

Your Turn

Question 8.
Find m∠M and m∠N.

m∠M = ____________
m∠N = ____________
Answer:
Using the Exterior Angle Theorem we have:
m∠M + m∠N = m∠MPQ
We substitute the given angle measures and we solve the equation for y
(5y + 3)°+(4y + 8)° = 146°
5y° + 3° + 4y° + 8° = 146°
9y° + 11° = 146°
9y° + 11° – 11° = 146° – 11°
9y° = 135°
\(\frac{9 y^{\circ}}{9^{\circ}}=\frac{135^{\circ}}{9^{\circ}}\)
y = 15
We use the value of y to find m∠M and m∠N.
m∠M = (5y + 3)° = (5 ∙ 15 + 3)° = 78°
m∠M = 78°
m∠N = (4y + 8)° = (4 ∙ 15 + 8)° = 68°
m∠N = 68°

Texas Go Math Grade 8 Lesson 7.2 Guided Practice Answer Key

Find each missing angle measure. (Explore Activity 1 and Example 1)

Question 1.

Answer:
From the Triangle Sum Theorem we have:
m∠L + m∠N + m∠M = 180°
We substitute the given angle measures and we solve for m°M.
78° + 31° + m∠M = 180°
109° + m∠M = 180°
109° – 109° + m∠M = 180° – 109°
m∠M = 71°

Question 2.

Answer:
From the Triangle Sum Theorem we have:
m∠Q + m∠S + m∠R = 180°
We substitute the given angle measures and we solve for m∠Q.
m∠Q + 24° + 126° = 180°
m∠Q + 150° = 180°
m∠Q + 150° – 150° = 180° – 150°
m∠Q = 30°

Use the Triangle Sum Theorem to find the measure of each angle in degrees. (Explore Activity 2 and Example 1)

Question 3.

Answer:
From the Triangle Sum Theorem we have:
m∠U + m∠T +m∠V = 180°
We substitute the given angle measures and we solve for x
(2x + 5)° + (7x + 4)° + (5x + 3)° = 180°
2x° + 5° + 7x° + 4° + 5x° + 3° = 180°
14x° + 12° = 180°
14x° + 12° – 12° = 180° – 12°
14x° = 168°
\(\frac{14 x^{\circ}}{14^{\circ}}=\frac{168^{\circ}}{14^{\circ}}\)
x = 12
We use the value of x to find the angles.
m∠U = (2x + 5)° = (2 ∙ 12 + 5)° = 29°
m∠U = 29°
m∠T = (7x + 4)° = (7 ∙ 12 + 4)° = 88°
m∠T = 88°
m∠V = (5x + 3)° = (5 ∙ 12 + 3)° = 63°
m∠V = 63°

Lesson 7.2 Angle Theorems for Triangles Answer Key Question 4.

Answer:
From the Triangle Sum Theorem we have:
m∠X + m∠Y + m∠Z = 180°
We substitute the given angle measures and we soLve for n.
n° + (\(\frac{1}{2}\)n)° + (\(\frac{1}{2}\)n)° = 180°
2n° = 180°
\(\frac{2 n^{\circ}}{2^{\circ}}=\frac{180^{\circ}}{2^{\circ}}\)
n = 90
We use the value of n to find the angles.
m∠X = n° = 90°
m∠X = 90°
m∠Y = (\(\frac{1}{2}\)n)° = (\(\frac{1}{2}\) ∙ 90)° = 45°
m∠Y = 45°
m∠Z = (\(\frac{1}{2}\)n)° = (\(\frac{1}{2}\) ∙ 90)° = 45°
m∠Z = 45°

Use the Exterior Angle Theorem to find the measure of each angle in degrees. (Explore Activity 3 and Example 2)

Question 5.

Answer:
Write the Exterior Angle Theorem as it applies to this triangle.
∠C + ∠D = ∠DEF ……………… (1)
4y° + (7y + 6)° = 116° (Substitute the given angle measures) …………. (2)
4y° + 7y° + 6° = 116° (Remove parentheses) ……………. (3)
11y° + 6° = 116° (Simplify) ………… (4)
11y° = 116° – 6° (Take 6 from both sides) …………….. (5)
11y° = 110° (Simplify) ……………. (6)
y° = \(\frac{110}{11}\) (Divide both sides by 11) ………… (7)
y° = 10° (Simplify) ………… (8)
Now, we can calculate the measure of the angles:
m∠C = 4 ∙ 10° = (4 ∙ 10)° = 40°
m∠D = (7y + 6)° (7 ∙ 10 + 6)° = (70 + 6)° = 76°
m∠DEC = 180° – 116° = 64°

Question 6.

Answer:
Using the Exterior Angle Theorem we have:
m∠M + m∠L = m∠JKM
We substitute the given angle measures and we solve the equation for z.
(5z – 3)° + (18z + 3)° = 161°
5z° – 3 + 18z° + 3° = 161°
23z° = 161°
\(\frac{23 z^{\circ}}{23^{\circ}}=\frac{161^{\circ}}{23^{\circ}}\)
z = 7
We use the value of z to find m∠M and m∠L.
m∠M = (5z – 3)° = (5 ∙ 7 – 3)° = 32°
m∠M = 32°
m∠L = (18z + 3)° = (18 ∙ 7 + 3)° = 129°
m∠L = 129°
From the Triangle Sum Theorem we have:
m∠M + m∠L + m∠LKM = 180°
We substitute the given angle measures and we solve for m∠LKM
32° + 129° + m∠LKM = 180°
161° + m∠LKM = 180°
161° – 161° + m∠LKM = 180° – 161°
m∠LKM = 19°

Essential Question Check-In

Question 7.
Describe the relationships among the measures of the angles of a triangle.
Answer:
The sum of all, measures of the interior angles of a triangle is 180°
The measure of the exterior angle of a triangle is equal to the sum of its remote interior angles.

Texas Go Math Grade 8 Lesson 7.2 Independent Practice Answer Key

Find the measure of each angle.

Question 8.

Answer:
By the triangle sum theorem, sum of the angles of the triangle is 180°
∠D + ∠E + ∠F = 180° ………………. (1)
98° + x° + X° = 180° (Substitute the measures of the angles) ……………… (2)
98° + 2x° = 180° (Simplify) ………….. (3)
2x° = 180° – 98° (Take 98° from both sides) ………….. (4)
2x° = 82° (Simplify) …………….. (5)
x° = \(\frac{82}{2}\) (Divide both sides by 2) …………….. (6)
x° = 41° (Simplify) ……………. (7)
m∠E = 41°
m∠F = 41°

Angle Theorems Grade 8 Lesson 7.2 Answer Key Question 9.

Answer:
Angles of the triangle are m∠W = 90°, m∠V = X° and m∠T = 2x°.
By the triangLe sum theorem, sum of the angles of the triangle is 180°
∠W + ∠V + ∠T = 180° ……………… (1)
90° + x° + 2x° = 180° (Substitute the measures of the angles) …………… (2)
90° + 3x° = 180° (Simplify) ………… (3)
3x° = 180° – 90° (Take 90° from both sides) …………….. (4)
3x° = 90° (Simplify) …………… (5)
x° = \(\frac{90}{3}\) (Divide both sides by 3) …………… (6)
x° = 30° (Simplify) …………. (7)
m∠T = 2x° = (2 ∙ 30)° = 60°
m∠T = 60°
m∠V = 30°

Question 10.

Answer:
Angles of the triangle are m∠G = 5x°, m∠H = 4x° and m∠J = 3x°.
By the triangle sum theorem, sum of the angles of the triangle is 180°
∠G + ∠H + ∠J = 180° ………………. (1)
5x° + 4x° + 3x° = 180° (Substitute the measures of the angles) …………… (2)
12x° = 180° (Simplify) ……………. (3)
x° = \(\frac{180}{12}\) (Divide both sides by 12) ……………. (4)
x° = 15° (Simplify) ………………. (5)
m∠G = 5x° = (5 ∙ 15)° = 75°
m∠H = 4x° = (4 ∙ 15)° = 60°
m∠U = 3x° = (3 ∙ 15)° = 45°

Question 11.

Answer:
Write the Exterior Angle Theorem as it applies to this triangle.
∠Q + ∠P =∠QRS …………….. (1)
(3y + 5)° + (2y – 7)° = 153° (Substitute the given angle measures) …………… (2)
3y° + 5° + 2y° – 7° = 153° (Remove parentheses) ……………. (3)
5y° – 2° = 153° (Combine like terms) ……………. (4)
5y° = 153° + 2° (Add 2 to both sides) …………… (5)
5y° = 155° (Simplify) …………… (6)
y° = \(\frac{155}{5}\) (Divide both sides by 5) ………………. (7)
y° = 31° (Simplify) …………… (8)
Now, we can calculate the measure of the angles:
m∠Q = (3y + 5)° = (3 ∙ 31 + 5)° = (93 + 5)° = 98°
m∠P = (2y – 7)° = (2 ∙ 31 – 7)° = (62 – 7)° = 55°
m∠QRP = 180° – 153° = 27°

Question 12.

Answer:
Using Triangle sum theorem we have
m∠A + m∠B + m∠ACB = 180° ……………. (1)
78° + 58° + m∠AGB = 180° (Substitute measures of the angle into the formula) ………….. (2)
136° + m∠ACB = 180° (Simplify) ……………. (3)
m∠ACB = 180° – 136° (Take 136° from both sides) ………… (4)
m∠ACB = 44° …………… (5)
Now, let’s find measure of the angle m∠DCE
m∠D + m∠E + m∠DCE = 180° ………….. (6)
85° + 60° + m∠DCE = 180° (Substitute measures of the angle into the formula) …………… (7)
145° + m∠DCE = 180° (Simplify) …………….. (8)
m∠DCE = 180° – 145° (Take 136° from both sides) ……………… (9)
m∠DCE = 35° ……………….. (10)
As we can see,
m∠ACB + m∠BCD + m∠DCE = 180° …………….. (11)
44° + m∠BCD + 350 = 180° (Substitute measures of the angle into the formula) ……………. (12)
79° + m∠BCD = 180° (Simplify) ……………. (13)
m∠BCD = 180° – 79° (Take 79° from both sides) ………….. (14)
m∠BCD = 101° ………………. (15)

Question 13.

Answer:
Angles of the triangle are m∠K = 2x°, m∠L = 3x° and m∠KML = 3x°.
By the triangle sum theorem, sum of the angles of the triangle is 180°.
∠K + ∠L + ∠KML = 180° …………… (1)
2x° + 3x° + x° = 180° (Substitute the measures of the angles) …………….. (2)
6x° = 180° (Simplify) ……………. (3)
x° = \(\frac{180}{6}\) (Divide both sides by 6) …………… (4)
x° = 30° (Simplify) ………… (5)
m∠K = 2x° = (2 ∙ 30)° = 60°
m∠L = 3x° = (3 ∙ 30)° = 90°
m∠KML = x° = 30°
Angle m∠LMN is supplementary with m∠KML so we have
m∠KML + m∠LMN = 180° ………………….. (6)
30° + m∠LMN = 180° (Substitute) …………… (7)
m∠MN = 180° – 30° (Take 30° from both sides) ………… (8)
m∠LMN = 150° (Simplify) …………….. (9)
m∠K = 60°
m∠L = 90°
m∠KML = 30°
m∠LMN = 150°

Question 14.
Multistep The second angle in a triangle is five times as large as the first. The third angle is two-thirds as large as the first. Find the angle measures.
Answer:
Lets name the interior angLes of the triangle ∠1, ∠2 and ∠3. As we don’t know the measure of any angle in a triangle, Lets put ∠1 = x.
Angle ∠2 is five times as large as the first, so
∠2 = 5 ∙ ∠1 (Substitute ∠1 = x)
= 5x
Angle ∠3 is two-thirds as large as the first, so
∠3 = \(\frac{2}{3}\) ∙ ∠1 (Substitute ∠1 = x)
= \(\frac{2}{3}\)x
By Triangle Sum Theorem

Question 15.
Analyze Relationships Can a triangle have two obtuse angles? Explain.
Answer:
An obtuse angle is a type of angle that measures greater than 90° but Less than 180°.

In the question “Can a triangle have two obtuse angles?”, the answer is no If one of the interior angles of a triangle is an obtuse angle, it follows that the other two remaining interior angles are both acute angles. This is because the sum of the interior angles of a triangle is 180°. So, it is impossible to have two obtuse angles in the interior angles of a triangle.

H.O.T. Focus on Higher Order Thinking

Question 16.
Critical Thinking Explain how you can use the Triangle Sum Theorem to find the measures of the angles of an equilateral triangle.
Answer:
All angles have the same measure ¡n an equilateral triangle
Using the Triangle Sum Theorem, you can name all angles
∠x
∠x + ∠x + ∠x = 180
then simplify to
3(∠x) = 180
and solve for ∠x

Go Math Grade 8 Lesson 7.2 Answer Key Question 17.
a. Draw Conclusions Find the sum of the measures of the angles in quadrilateral ABCD.
(Hint: Draw diagonal \(\overline{A C}\). How can you use the figures you have formed to find the sum?)
Sum = ____________

Answer:
Let’s find the sum of the measures of the angles in quadrilateral ABCD. If we draw diagonal \(\overline{A C}\), we form two triangles ABC and ADC. By the Triangle Sum Theorem
∠DAC + ∠D + ∠DCA = 180° ………………. (1)
∠CAB + ∠B + ∠BCA = 180° ………………. (2)
Notice that ∠A = ∠DAC + ∠CAB and ∠C = ∠DCA + ∠BCE
So, the sum of the measures of the angles in quadrilateral ABCD is
∠A + ∠B + ∠C + ∠D = ∠DAC + ∠CAB + ∠B + ∠DCA + ∠BCA + ∠D ……………….. (3) Substitute
= (∠DAC + ∠D + ∠DCA) + (∠CAB + ∠B + ∠BCA) ……………….. (4) (Combine angles triangles ABC and ADC)
= 180°+ 180° …………… (5) (By Triangle Sum Theorem)
= 360° ………………. (6) (Simplify)

b. Make a Conjecture Write a “Quadrilateral Sum Theorem.” Explain why you think it is true.
Answer:
“Quadrilateral. Sum Theorem:” The sum of the angle measures of a quadrilateral is 360°. According to a. every quadrilateral can be divided by a diagonal into two triangles, so the sum of angles is 180° + 180° = 360°.

Question 18.
Communicate Mathematical Ideas Describe two ways that an exterior angle of a triangle is related to one or more of the interior angles.
Answer:
An exterior angle is formed by one side of the triangle and the extension of an adjacent side Each exterior angle has two remote interior angles that are not adjacent to the exterior angle.

An exterior angle is related to the interior angles of a triangle First, the relationship is that the exterior angle is supplementary to its adjacent interior angle. The sum of their angles is 180°.

Another relationship is that the sum of the two remote interior angles which are not adjacent to the exterior angle is equal to the measure of the exterior angle.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Unit 2 Study Guide Review Answer Key.

Texas Go Math Grade 8 Unit 2 Study Guide Review Answer Key

Proportional Relationships

Essential Question
How can you use proportional relationships to solve real-world problems?

Exercises

Question 1.
The table represents a proportional relationship. Write an equation that describes the relationship. Then graph the relationship represented by the data. (Lessons 3.1,3.3, 3.4)


Answer:
slope = \(\frac{4-3}{8-6}\) = \(\frac{1}{2}\) = 0.5
Find the slope using given points by
Slope(m) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
where
(x₂, y₂) = (8, 4) and (x₁, y₁) = (6, 3)

y = 0.5x Substitute the slope m in the slope-intercept form: y = mx + b.
Since the relationship is proportional, b = 0

Graph the equation

Find the slope and the unit rate represented on each graph. (Lesson 3.2)

Go Math Grade 8 Answer Key Pdf Unit 2 Test Review Math Answers Question 2.

Answer:
Explanation A:
The slope formula is:
m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Substitute the values from the graph and find the value of slope (m).
m = \(\frac{4-0}{2-0}\)
m = \(\frac{4}{2}\)
m = 2

Explanation B:
Firstly, we will find the slope using the following formula:
m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

We can calculate the slope using given formula on any two points located on a given function,
m = \(\frac{8-4}{4-2}\)
m = 2

Unit rate is defined as the comparison of two values where one of them is one unit. To simplify, in this exercise, unit rate would be the number of words in one minute, so we have to find a point on function where x value is 1. On given function, unit rate is 2 words per minute.

As it is shown in this exercise, unit rate and slope of the graph have the same value, but in a different format. Slope is an integer and unit rate is a rate of change of a mathematical function.

Question 3.

Answer:
Slope/unitrate = \(\frac{2-0}{5-0}\) = \(\frac{2}{5}\) = 0.4 Find the slope using given points by Slope(m) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) where (x₂, y₂) = (5, 2) and (x₁, y₁) = (0, 0)
The unit rate and the slope is
0.4 ft per sec

Module 4 Nonproportional Relationships

Essential Question
How can you use nonproportional relationships to solve real-world problems?

Exercises

Complete each table. Explain whether the relationship between x and y is proportional or nonproportional and whether it is linear. (Lesson 4.1)

Question 1.
y = 10x – 4

Answer:
y = 10x – 4 Given

complete the table
The relationship is Linear but not proportional Compare with general linear equation: y = mx + b. Since, b ≠ 0, relationship is not proportional.

Question 2.
y = –\(\frac{3}{2} x\)

Answer:
Firstly, we will complete the given table by calculating the missing x and y values. To calculate the missing y values, we will multiply the corresponding x value with -1.5 and to get the missing x values we will divide the corresponding y value with -1.5.

The completed table is shown below.

Equation is linear if its graph is a straight line. Considering the fact that the graph of an equation y =  \(\frac{3}{2} x\) is a straight line, we can claim that the relationship between x and y is linear.

To determine whether the given equation is proportional, we have to check two things. The first one is if ratios between x and y values are the same- Therefore:
–\(\frac{1}{1.5}\) = –\(\frac{2}{3.0}\) = –\(\frac{3}{4.5}\)
–\(\frac{2}{3} x\) = –\(\frac{2}{3}\) = –\(\frac{2}{3}\)

The other thing is if the graph goes through the origin. Considering the fact that point (0, 0) is shown in the table, we know that the graph goes through the origin.
Hence, we can conclude that given equation is proportional.

Question 3.
Find the slope and y-intercept for the linear relationship shown in the table. Graph the line. Is the relationship proportional or nonproportional? (Lessons 4.2,4.4)


slope ____
y-intercept ____
The relationship is ______
Answer:
Firstly, we will graph the equation using given x and y values from the table. Graph is shown in the picture below.

We can calculate the slope as following using any two points from the graph:
m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
m = \(\frac{4-2}{0-(-1)}\)
m = 2
y-intercept is defined as the value of y coordinate in a point where the function intersects with the y axis. Looking
at the graph, we can conclude that the line intersects with the y axis in point (0, 4).
Therefore, b = 4.

Having calculated the values of the slope and y-intercept, we can write the equation of a line:
y = mx + b
y = 2x + 1

Given relationship is not proportional considering the fact that the graph does not pass through the origin and noticing that the ratio between x and y values is not the same.

Go Math Grade 8 Proportional Relationships Unit Study Guide Answer Key Question 4.
Tom’s Taxis charges a fixed rate of $4 per ride plus $0.50 per mile. Carla’s Cabs does not charge a fixed rate but charges $1.00 per mile. (Lessons 4.3, 4.5)
Answer:
Toms taxis charge a fixed rate of $4 per ride plus $0.50 per mile. Carlas Cabs does not charge a fixed rate but charges $ 1 .00 per tuile. To represent the cost of Toms Taxis and Carias cabs in equation, let us apply express
them using the slope—intercept form y = mx + b.

a. Write an equation that represents the cost of Tom’s Taxis. ______
Answer:
Representing the cost of Tom’s Taxis
Let y be the cost of Tom’s Taxis
x be the distance traveL by Tom’s Taxis (in mile)
m be the amount to be paid per mile
b be the amount to be paid per ride
Using y = mx + b, the representation of the cost of Tom’s Taxis where m = 0.50 and b = 4 is
y = 0.50x + 4

b. Write an equation that represents the cost of Carla’s cabs.
Answer:
Representing the cost of Carla’s cabs.
Let y be the cost of Carla’s cabs
x be the distance travel by Carla’s cabs (in mile)
m be the amount to be paid per mile
b be the amount to be paid per ride

Using y = mx + b, the representation of the cost of Carla’s cabs where m = 1 and b = 0 is
y = 1x + 0
y = x

c. Steve calculated that for the distance he needs to travel, Tom’s Taxis will charge the same amount as Carla’s Cabs. Graph both equations. How far is Steve going to travel and how much will he pay?

Answer:
To determine the distance Steve needs to travel that Tom’s Taxis will charge the same amount as Carla’s Cabs and
the amount he will pay, let us show the graph the two equations by setting the values of x and determine the corresponding values of y gives the following ordered pairs
For Tom’s Taxis y = 0.50x + 4

Plotting and connecting the points gives the graph of

Now, looking at the graph, the two lines intersect at (8, 8), meaning
Steve needs to travel 8 miles and he will pay an amount of $8.

See the explanation

Writing Linear Equations

Essential Question
How can you use linear equations to solve real-world problems?

Exercises

Question 1.
Ms. Thompson is grading math tests. She is giving everyone that took the test a 10-point bonus. Each correct answer is worth 5 points. Write an equation in slope-intercept that represents the scores on the tests. (Lesson 5.1) _______
The table shows a pay scale based on years of experience. (Lessons 5.1, 5.2)

Answer:
y = 5x + 10 Substitute m = 5 points and b = 10 point in scope-intercept form: y = mx + b. where x is the number of correct answers.

Question 2.
Find the slope for this relationship. ______
Answer:
Slope will be calculated using the fotbwing formula:
m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\),
for any two points (x₁, y₁) and (x₂, y₂) on the graph.

Therefore, we can use formula from the previous step:
m = \(\frac{19-14}{4-2}\)
m = 2.5
Hence, the calculated slope for given relationship is:
m = 2.5

Question 3.
Find the y-intercept.
Answer:
9 = 2.5(0) + b Substituting the value of slope (m) and (x, y) in the slope intercept form to find y intercept (b): y = mx + b
9 = 0 + b
b = 9

Question 4.
Write an equation in slope-intercept form that represents this relationship. ____
Answer:
y = 2.5x + 9
Substituting the value of slope (m) and y-intercept (b) in slope-intercept form: y = mx + b

Unit 2 Proportional Relationships Answer Key Grade 8 Go Math Question 5.
Graph the equation, and use it to predict the hourly pay of someone with 10 years of experience.

Answer:
To graph the equation, we will find given points on the coordinate system and connect them to draw the graph.

In order to predict the hourly pay for someone with 10 years of experience, we will have to find a point on function where x = 10 and value of y coordinate in that points is the hourly pay as shown in the picture below.

Hence, the hourly pay for someone with 10 years of experience is 34 dollars.

Does each of the following graphs represent a linear relationship? Why or why not? (Lesson 5.3)

Question 6.

Answer:
To determine the answer here, we will review the definition of a linear function.

Equation is a linear function only if its graph is a straight line. Considering the fact that given graph is a straight line, we can conclude that the relationship is linear.

Question 7.

Answer:
Non-linear relationship Graph of a linear equation is a straight line

Functions

Essential Question
How can you use functions to solve real-world problems?

Exercises

Determine whether each relationship is a function. (Lesson 6.1)

Question 1.

Answer:
Explanation A:
To determine the answer here, we will have to review on how to recognize a function.
A certain relationship represents a function only if each input value has one output value. To clarify, every X value
on the graph can have only one y value.
Considering the fact that in a given graph there is no repetition of values, we can claim that the given relationship is a function.

Explanation B:
Based on the graph, it is a function since each input is paired with exactly one output

8th Grade Math Unit 2 Test Answer Key Question 2.

Answer:
Function Each input is paired with exactly one output

Tell whether the function is linear or nonlinear. (Lesson 6.2)

Question 3.
y = 5x + \(\frac{1}{2}\): ____
Answer:
y = 5x + \(\frac{1}{2}\) Compare with general linear equation: y = mx + b
Function is linear

Question 4.
y = x² + 3 : ______
Answer:
y = x² + 3 Compare with general linear equation: y = mx + b
Function is nonlinear

Question 5.
Elaine has a choice of two health club memberships. The first membership option is to pay $500 now and then pay $150 per month. The second option is shown in the table. Elaine plans to go to the club for 12 months. Which option is cheaper? Explain.

Answer:
To determine which is cheaper between Elaine’s two health club membership, where she plans to go for 12 months. Let us apply the slope-intercept form y = mx + b.

The first membership option is pay now $500 and then pay $150 per month.
Let us represent the following variables.
Let, y be the total membership fee for 12 months, m be the payment for every month, x be the number of months she plans to go the club, and b be the down Payment.

Now, to determine the total membership fee y for the first option, substitute the values of m = 150, x = 12, and
b = 500.
y = mx + b
y = 150(12) + 500
y = 1800 + 500
y = 2300
The total health club membership fee for the first option is
$2300

The second option is shown in the table below.
| Months. x|1 |2 |3 |
|—— |——| ——|———|
|Total paid($),y |215| 430| 645 |

Before we can find the total membership fee of the second option, Let us first determine the slope in using the first
two months of the payment plan, where x₁ = 1, x₂ = 2, y₁ = 215, and y₂ = 430

Next, Let us get the value of b, using x = 1 and y = 215, where m = 215.
y = mx + b Formula.
215 = 215(1) + b Substitute.
215 = 215 + b Simplify.
b = 215 – 215
b = 0

Now, to determine the total membership fee y for the second option, let us substitute the values of m = 215, x = 12, and b = 0.
y = mx + b
y = 215(12) + 0
y = 2580 + 0
y = 2580
The total health club membership fee for the second option is
$2580

Since the total health club membership fee for the first option $2300 is less than the total health club membership fee for the second option $2380, then
**the first option is cheaper that the second option**.

Texas Go Math Grade 8 Unit 2 Performance Tasks Answer Key

Question 1.
CAREERS IN MATH Cost Estimator To make MP3 players, a cost estimator determined it costs a company $1500 per week for overhead and $45 for each MP3 player made.
a. Define a variable to represent the number of players made. Then write an equation to represent the company’s total cost per week, c.
Answer:
Let n be the number of players made in a week Identify the variable
c = 45n + 1500 Company’s total cost

b. One week, the company spent $5460 making MP3 players. How many players were made that week? Show your work.
Answer:
c = 45n + 1500 Substitute c = $5460 in the equation
5460 = 45n + 1500
45n = 5460 – 1500
45n = 3960
n = \(\frac{3960}{45}\) = 88
The number of MP3 prayers made were 88

c. If the company sells MP3 players for $ 120, how much profit would it make if it sold 80 players in one week? Explain how you found your answer.
Answer:
c = 45n + 1500 Find the cost of making 80 players by substituting n = 80
c = 45(80) + 1500
c = $5100
r = 120 * 80 = $9600 each Find the revenue made by selling 80 players at $120 each
P = 9600 – 5100 = $4500 Profit = Cost – Revenue

Grade 8 Math Module Unit 2 Answer Key Question 2.
A train from Portland, Oregon, to Los Angeles, California, travels at an average speed of 60 miles per hour and covers a distance of 963 miles. Susanna is taking the train from Portland to Los Angeles to see her aunt. She needs to arrive at her aunt’s house by 8 p.m. It takes 30 minutes to get from the train station to her aunt’s house.

a. By what time does the train need to leave Portland for Susanna to arrive by 8 p.m.? Explain how you got your answer. As part of your explanation, write a function that you used in your work.
Answer:
Given that, a train traveling from Portland, Oregon to Los Angeles, California travels at an average speed of
60 miles per hour and covers a distance of 963 miles.
Time = Distance/Speed
Time = 963/60
Time = 16.05 hours
Subtract 16.05 hours from 8.00 P.M.
8 – 16.05 hours = 4.55 A.M

b. Susanna does not want to leave Portland later than 10 p.m. or earlier than 6 a.m. Does the train in part a meet her requirements? If not, give a new departure time that would allow her to still get to her aunt’s house on time, and find the arrival time of that train.
Answer:
Susanna is traveling to her aunt’s house from Portland to Los Angeles. It will take 30 min to get from the station to her aunt’s house and she needs to arrive aunt’s house by 8 pm.
The departure time of the train should be 4.55 am to meet her requirements.
Susanna does not want to leave Portland later than 10 pm or earlier than 6 am. The arrival time of the train will be 9 pm.

Texas Go Math Grade 8 Unit 2 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
Rickie earns $7 an hour babysitting. Which table represents this proportional relationship?

Answer:

Explanation:
y = 7x where y represents the total earning and x represents the number of hour
y = 7(4) = 28 Substitute x = 4, 6, 8
y = 7(6) = 42
y = 7(8) = 56
Option A represents this proportional relationship

Question 2.
Which of the relationships function?
(A) (6, 3), (5, 2), (6, 8), (0, 7)
(B) (8, 2), (1, 7), (-1, 2), (1, 9)
(C) (4, 3), (3, 0), (-1, 3), (2, 7)
(D) (7, 1), (0, 0), (6, 2), (0, 4)
Answer:
(C) (4, 3), (3, 0), (-1, 3), (2, 7)

Explanation:
Each input has exactly one output. In other options, there is an input paired with more than one output

Question 3.
Which set best describes the numbers used on the scale for a standard thermometer?
(A) whole numbers
(B) rational numbers
(C) real numbers
(D) integers
Answer:
(B) rational numbers

Explanation:
The standard thermometer is measured nearest to 0.1 degree

Question 4.
Which term refers to slope?
(A) rate of change
(B) equation
(C) y-intercept
(D) coordinate
Answer:
(A) rate of change

Explanation:
Rate of change
Slope and rate of change gives the change in second quantity as compared to unit of first quantity.

Question 5.
The graph of which equation is shown below?

(A) y = 4x + 3
(B) y = -4x – 0.75
(C) y = -4x + 3
(D) y = 4x – 0.75
Answer:
(A) y = 4x + 3

Explanation:
Option B and D are rejected. The graph has a y intercept of 3

Option C is rejected The slope of the graph would be positive (as the graph rises from left to right)
Hence, the equation of the garph is y = 4x + 3
Option A

Question 6.
Which equation represents a nonproportional relationship?
(A) y = 5x
(B) y = -5x
(C) y = 5x + 3
(D) y = –\(\frac{1}{5} x\)
Answer:
(C) y = 5x + 3

Explanation:
y = 5x + 3 Comparing with linear equation: y = mx + b. Since b ≠ 0, the equation is non-proportional represents a non-proportional relationship

Question 7.
Which describes the solution of a system of linear equations for two lines with the same slope and the same y-intercepts?
(A) one nonzero solution
(B) infinitely many solutions
(C) no solution
(D) solution of 0
Answer:
(B) infinitely many solutions

Question 8.
Which is 7.0362 × 10^-4 written in standard notation?
(A) 0.000070362
(B) 0.00070362
(C) 7.0362
(D) 7036.2
Answer:
(B) 0.00070362

Explanation:
7.0362 × 10^-4 Given
4 use me exponent the power of 10 to see how many places to move the decimal point.
places
0.00070362 Place the decimal point. Since you are going to write a number less than 7.0362, move the decimal to the left Add placeholder zeros if necessary.

Question 9.
Which term does not correctly describe the function shown in the table?

(A) relationship
(B) linear
(C) proportional
(D) nonproportional
Answer:
(D) nonproportional

Explanation:
Non-proportional The relationship is a function as each input has exactly one output. It is linear as the rate of change is constant. It is proportional as the table contains (0, 0). Hence, a relationship that is proportional cannot be non
proportional.

8th Grade Math Unit 2 Test Answers Question 10.
As part of a science experiment, Greta measured the amount of water flowing from Container A to Container B. Container B had half a gallon of water in it to start the experiment. Greta found that the water was flowing at a rate of two gallons per hour. Which equation represents the amount of water in Container B?
(A) y = 2x
(B) y = 0.5x
(C) y = 2x + 0.5
(D) y = 0.5x + 2
Answer:
(C) y = 2x + 0.5

Explanation:
y = 2x + 0.5
where y is the number of gallons after x hours. The slope or rate of change is 2 gallons per hr and the initial number of gallons in Container B (y-intercept).
Use the slope-intercept form: y = mx + b

Gridded Response

Question 11.
A factory produces gaskets at a constant rate. After 6 hours, 4620 gaskets have been produced. How many gaskets does the factory produce per hour?

Answer:
Given that,
The factory produces gaskets at a constant rate.
After 6 hours there are 4620 gaskets.
Per 1 hour = 4620/6 = 770
The gaskets per 1 hour = 770

Question 12.
The table below represents a linear relationship.

What is the y-intercept?

Answer:

The Y-intercept is (0,8)

Hot Tip!
Estimate your answer before solving the question. Use your estimate to check the reasonableness of your answer.

Question 13.
The table below shows some input and output values of a function.

What number is missing?

Answer:
31.5 – 27 = 4.5
18+4.5 = 22.5

Texas Go Math Grade 8 Unit 2 Vocabulary Preview Answer Key

Use the puzzle to preview key vocabulary from this unit. Unscramble the circled letters within the found words to answer the riddle at the bottom of the page.

• The y-coordinate of the point where the graph crosses the y-axis.   (Lesson 4-2)
• A rule that assigns exactly one output to each input.   (Lesson 6-1)
• The result after applying the function machines rule.   (Lesson 6-1)
• A relationship is written as y = kx, where the data increases or decreases together at a constant rate (2 words).   (Lesson 3-4)
• The ratio of change in rise to the corresponding change in run on a graph.  (Lesson 3-2)
• A set of data that is made up of two paired variables.  (Lesson 5-3)
• An equation whose solutions form a straight line on a coordinate plane.  (Lesson 4-1)

Question.
How much of the money earned does a professional sports team pay its star athlete?
Answer:
An ____ ____ ____ ____ ____ ____ – ____ ____ ____ ____ ____ ____ ____!

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 8 Answer Key The Pythagorean Theorem.

Texas Go Math Grade 8 Module 8 Answer Key The Pythagorean Theorem

Texas Go Math Grade 8 Module 8 Are You Ready? Answer Key

Find the square of each number.

Question 1.
5 _________
Answer:
To determine the square of a number x, denoted by x² just multiply the given number x by itself.
Now, let us get the square of 5 or the equivalent of 5², just multiply 5 by itself, that is
5² = 5 × 5
= 25
or

Grade 8 Math Module 8 Answer Key Question 2.
16 __________
Answer:
When solving for the square of a number, we need to multiply the certain number by itself.
To find the square of 16, we simply multiply it by itself
16² = 16 × 16
= 256

Question 3.
-11 _________
Answer:
To determine the square of a number x, denoted by x² just multiply the given number x by itself.
Now, let us get the square of -11 or the equivalent of (-11)², just multiply 11 by itself, that is
(-11)² = -11(-11)
= 121

Question 4.
\(\frac{2}{7}\) __________
Answer:
To solve for the square of \(\frac{2}{7}\), we simply multiply it by itself. Take note that in multiplying fractions, the product is equal to the product of all numerators divided by the product of all denominators.
(\(\frac{2}{7}\))² = \(\frac{2}{7}\) × \(\frac{2}{7}\) = \(\frac{4}{49}\)

Evaluate each expression.

Go Math Grade 8 Answer Key Pdf Module 8 Question 5.
\(\sqrt{(6+2)^{2}+(3+3)^{2}}\) __________
Answer:
\(\sqrt{(6+2)^{2}+(3+3)^{2}}=\sqrt{8^{2}+6^{2}}\) (First, operate within parentheses.) …… (1)
= \(\sqrt{64+36}\) (Next, simplify exponents.) ………. (2)
= \(\sqrt{100}\) (Then add and subtract left to right.) ……….. (3)
= 10 (Finally, take the square root.) ……… (4)
= 10

Question 6.
\(\sqrt{(9-4)^{2}+(5+7)^{2}}\) __________
Answer:
\(\sqrt{(9-4)^{2}+(5+7)^{2}}=\sqrt{5^{2}+12^{2}}\) (First, operate within parentheses.) …… (1)
= \(\sqrt{25+144}\) (Next, simplify exponents.) ………. (2)
= \(\sqrt{169}\) (Then add and subtract left to right.) ……….. (3)
= 13 (Finally, take the square root.) ……… (4)
= 13

Question 7.
\(\sqrt{(10-6)^{2}+(15-12)^{2}}\) __________
Answer:
\(\sqrt{(10-6)^{2}+(15-12)^{2}}=\sqrt{4^{2}+3^{2}}\) (First, operate within parentheses.) …… (1)
= \(\sqrt{16+9}\) (Next, simplify exponents.) ………. (2)
= \(\sqrt{25}\) (Then add and subtract left to right.) ……….. (3)
= 5 (Finally, take the square root.) ……… (4)
= 5

Question 8.
\(\sqrt{(6+9)^{2}+(10-2)^{2}}\) __________
Answer:
\(\sqrt{(6+9)^{2}+(10-2)^{2}}=\sqrt{15^{2}+8^{2}}\) (First, operate within parentheses.) …… (1)
= \(\sqrt{225+64}\) (Next, simplify exponents.) ………. (2)
= \(\sqrt{289}\) (Then add and subtract left to right.) ……….. (3)
= 17 (Finally, take the square root.) ……… (4)
= 17

Simplify each expression.

Module 8 The Pythagorean Theorem Answer Key Question 9.
5(8)(10) ___________
Answer:
5(8)(10) = (40)(10) (Multiply from left to right)
= 400

Question 10.
\(\frac{1}{2}\)(6)(12) ___________
Answer:
\(\frac{1}{2}\)(6)(12) = (3)(12) (Multiply from left to right)
= 36

Question 11.
\(\frac{1}{3}\)(3)(12) ___________
Answer:
\(\frac{1}{3}\)(3)(12) = (1)(12) (Multiply from left to right)
= 12

Question 12.
\(\frac{1}{2}\)(8²)(4) ___________
Answer:
Simplify expression
\(\frac{1}{2}\)(8²)(4) = \(\frac{1}{2}\)(64)(4) (Simplify the exponent) …………. (1)
= (32)(4) (Multiply from left to right) ……………… (2)
= 128

Question 13.
\(\frac{1}{4}\)(10)²(15) ___________
Answer:
Simplify expression
\(\frac{1}{4}\)(10²)(15) = \(\frac{1}{4}\)(100)(15) (Simplify the exponent) …………. (1)
= (25)(15) (Multiply from left to right) ……………… (2)
= 375

Texas Go Math Grade 8 Module 8 Answer Key Pdf Question 14.
\(\frac{1}{3}\)(9)²(6) ___________
Answer:
Simplify expression
\(\frac{1}{3}\)(9²)(6) = \(\frac{1}{3}\)(81)(6) (Simplify the exponent) …………. (1)
= (27)(6) (Multiply from left to right) ……………… (2)
= 162

Texas Go Math Grade 8 Module 8 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the graphic.

Understand Vocabulary

Match the term on the left to the correct expression on the right.

Answer:
1 – C. Hypotenuse: The side opposite the right angle in a right triangle
2 – A. Theorem: An idea that has been demonstrated as true.
3 – B. Legs: The two sides that form the right angle of a right triangle.

Active Reading
Booklet Before beginning the module, create a booklet to help you learn about the Pythagorean Theorem. Write the main idea of each lesson on each page of the booklet. As you study each lesson, write important details that support the main idea, such as vocabulary and formulas. Refer to your finished booklet as you work on assignments and study for tests.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 7 Quiz Answer Key.

Texas Go Math Grade 8 Module 7 Quiz Answer Key

Texas Go Math Grade 8 Module 7 Ready to Go On? Answer Key

7.1 Parallel Lines Cut b a Transversal

In the figure, line P|| line q. Find the measure of each angle if m∠8 = 115°.

Question 1.
m∠7 = ______________
Answer:
According to the Exterior Angle Theorem, we have:
m∠7 + m∠8 = 180°
m∠7 + 115° = 180°
m∠7 + 115° – 115° = 180° – 115°
m∠7 = 65°

Go Math Grade 8 Module 7 Answer Key Question 2.
m∠6 = ______________
Answer:

In the given figure, line P is parallel to line q. So, the angles given in line P are equal to the angles in line q.(corresponding angles)
so, m∠8 is parallel to m∠6 or m∠8 = m∠6 = 115°

Question 3.
m∠1 = ______________
Answer:

this is the given figure of the question

taking a section of a figure, we have

In the previous question, we got m∠6 = 115°
Figure, ∠1 and ∠6 are alternate exterior angles
so, m∠1 = m∠6 = 115°

7.2 Angle Theorems for Triangles

Find the measure of each angle.

Question 4.
m∠A = ______________
Answer:

From the triangle sum theorem
the sum of the angles of the triangle is 180°
m∠A + m∠B + m∠C = 180°
4y + (3y + 22) + 74° = 180°
7y = 180 – 96 = 84
y = 12°

m∠A = 4y = 4 × 12 = 48°
and m∠B = 3y + 22 = 3 × 12 + 22
= 58°

Module 7 Quiz Ready To Go On Grade 8 Question 5.
m∠B = ________
Answer:
Using the Exterior Angle Theorem we have:
m∠A + m∠B = m∠BCD
We substitute the given angle measures and we solve the equation for y.
(4y)° + (3y + 22)° = 106°
4y° + 3y° + 22° = 106°
7y° + 22° = 106°
7y° + 22° – 22° = 106° – 22°
7y° = 84°
\(\frac{7 y^{\circ}}{7^{\circ}}\) = \(\frac{84^{\circ}}{7^{\circ}}\)
y = 12
We use the value of y to find m∠B.
m∠B = (3y + 22)° = (3 . 12 + 22)° = 58°
m∠B = 58°

Question 6.
m∠BCA = ________
Answer:

In the given triangle,
m∠A = 4y m∠B = 3y + 22°
and m∠BCD = 106°
using exterior angle theorem

m∠BCD + m∠BCA = 180°
m∠BCA = 180° – 106°
= 74°
So, m∠BCA = 74°

∠BCA = 74 degrees

7.3 Angle-Angle Similarity

Triangle FEG is similar to triangle IHJ. Find the missing values.

Question 7
x = ________
Answer:
In similar triangles, corresponding side lengths are proportional
\(\frac{H J}{E G}\) = \(\frac{I J}{F G}\)
\(\frac{x+12}{42}\) = \(\frac{40}{60}\)
\(\frac{x+12}{42}\) = \(\frac{4}{6}\)
6 . 42 . \(\frac{x+12}{42}\) = \(\frac{4}{6}\) . 6 . 42
6(x + 12) = 168
6x + 72 = 168
6x + 72 – 72 = 168 – 72
6x = 96
\(\frac{6 x}{6}\) = \(\frac{96}{6}\)
x = 16

Grade 8 Math Module 7 Answer Key Pdf Question 8.
y = ________
Answer:
In similar triangles, corresponding angles are congruent
m∠HJI = m∠EGF
(5y + 7)° = 52°
5y + 7° = 52°
5y° + 7° – 7° = 52° – 7°
5y° = 45
\(\frac{5 y^{\circ}}{5^{\circ}}\) = \(\frac{45^{\circ}}{5^{\circ}}\)
y = 9

Question 9.
m∠H = ___
Answer:
From the Triangle Sum Theorem we have:
m∠E + m∠F + m∠G = 180°
We substitute the given angle measures and we solve for m∠E.
m∠E + 36° + 52° = 180°
m∠E + 88° = 180°
m∠E + 88° – 88° = 180° – 88°
m∠E = 92°
In similar triangles, corresponding angles are congruent. Therefore,
m∠H = m∠E
m∠H = 92°

Essential Question

Grade 8 Math Module 7 Answer Key Question 10.
How can you use similar triangles to solve real-world problems?
Answer:
We know that if two triangles are similar, then their corresponding angles are congruent and the lengths of their
corresponding sides are proportional. We can use this to determine values that we cannot measure directly. For example, we can calculate the length of the tree if we measure its shadow and our shadow in a sunny day.

Texas Go Math Grade 8 Module 7 Mixed Review Texas Test Prep Answer Key

Selected Response

Use the figure for Exercises 1 and 2.

Question 1.
Which angle pair is a pair of alternate exterior angles?
(A) ∠5 and ∠6
(B) ∠6 and ∠7
(C) ∠5 and ∠4
(D) ∠5 and ∠2
Answer:
(C) ∠5 and ∠4

Explanation:
∠5 and ∠4 are alternate exterior angles

Module 7 Form A Module Test Answer Key Question 2.
Which of the following angles is not congruent to ∠3?
(A) ∠1
(B) ∠2
(C) ∠6
(D) ∠8
Answer:
(B) ∠2

Explanation:
∠2 and ∠3 are same-side interior angles. They are not congruent; instead their sum is equal to 180°

Question 3.
The measures of the three angles of a triangle are given by 2x + 1, 3x – 3, and 9x. What is the measure of the smallest angle?
(A) 13°
(B) 27°
(C) 36°
(D) 117°
Answer:
(B) 27°

Explanation:
From the Triangle Sum Theorem we have:
m∠1 + m∠2 + m∠3 = 180°
We substitute the given angle measures and we solve for X.
(2x + 1)° + (3x – 3)° + (9x)° = 180°
2x° + 1 + 3x° – 3° + 9x° = 180°
14x° – 2° = 180°
14x° – 2° + 2° = 180° + 2°
14x° = 178°
\(\frac{14 x^{\circ}}{14^{\circ}}\) = \(\frac{182^{\circ}}{14^{\circ}}\)
x = 13
We use the value of x to find m∠1, m∠2 and m∠3.
m∠1 = (2x + 1)° = (2 . 13 + 1)° = 27°
m∠1 = 27°
m∠2 = (3x – 3)° = (3 . 13 – 3)° = 36°
m∠2 = 36°
m∠3 = (9x)° = (9 . 13)° = 117°
m∠3 = 117°
The smallest angle is 27°

Question 4.
Which is a possible measure of ∠DCA in the triangle below?
(A) 36°
(B) 38°
(C) 40°
(D) 70°
Answer:
(D) 70°

Explanation:
Using the Exterior Angle Theorem we have:
m∠A + m∠B = m∠DCA
m∠A + 40° = m∠DCA
As we can see, m∠DCA will be greater than 40°. The only suitable option is D, 70°.

Texas Go Math Grade 8 Module 7 Quiz Answers Question 5.
Kaylee wrote in her dinosaur report that the Jurassic period was 1.75 × 10⁸ years ago. According to Kaylee’s report, how many years ago was the Jurassic period?
(A) 1,750,000
(B) 17,500,000
(C) 175,000,000
(D) 17,500,000,000
Answer:
(C) 175,000,000

Explanation:
The Jurassic period was 1.75 × 10⁸ years ago. In standard form
10⁸ = 100, ooo, 000 (Ten with eight zeros) (1)
1.75 × 10⁸ = 1.75 × 100, 000,000 (Substitute) (2)
= 175,000,000 (3)

The correct answer is C 175, 000, 000

Question 6.
Given that y varies directly with x, what is the equation of direct variation if y is 16 when x is 20?
(A) y = 1\(\frac{1}{5}\)x
(B) y = \(\frac{5}{4}\)x
(C) y = \(\frac{4}{5}\)x
(D) y = 0.6x
Answer:
(C) y = \(\frac{4}{5}\)x

Explanation:
Given that y is proportional to x, and if y is 16, and x is 20 we can write proportion
\(\frac{y}{x}\) = \(\frac{16}{20}\) (1)
\(\frac{y}{x}\) = \(\frac{4}{5}\)x (simplify the fraction) (2)
x × \(\frac{4}{5}\) = \(\frac{4}{5}\) × x (Use properties of equality to get y by itself.) (3)
y = \(\frac{4}{5}\)x (4)

The correct answer is C

Gridded Response

Module 7 Properties of Triangles Module Quiz B Answer Key Question 7.
What is the value of h in the triangle below?

Answer:
In similar triangles the corresponding sides and lengths are proportional.
4/10 = h/12
4 × 12 = h × 10
48/10 = h
h = 4.8

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 7 Answer Key Angle Relationships in Parallel Lines and Triangles.

Texas Go Math Grade 8 Module 7 Answer Key Angle Relationships in Parallel Lines and Triangles

Texas Go Math Grade 8 Module 7 Are You Ready? Answer Key

Solve for x.

Question 1.
6x + 10 = 46
Answer:
Following the example on page 190, first, we subtract 10 from both sides Then, we simplify and divide both sides by 6.
6x + 10 = 46
6x + 10 – 10 = 46 – 10
6x = 36
\(\frac{6x}{6}\) = \(\frac{36}{6}\)
x = 6

Grade 8 Module 7 Answer Key Angle Relationships Unit Test Question 2.
7x – 6 = 36
Answer:
Following the example on page 190, first, we add 6 to both sides Then, we simplify and divide both sides by 7.
7x – 6 = 36
7x – 6 + 6 = 36 + 6
7x = 42
\(\frac{7x}{7}\) = \(\frac{42}{7}\)
x = 6

Question 3.
3x + 26 = 59
Answer:
Following the example on page 190, first we subtract 26 from both sides. Then, we simplify and we divide both sides by3.
3x + 26 = 59
3x + 26 – 26 = 59 – 26
3x = 33
\(\frac{3x}{3}\) = \(\frac{33}{3}\)
x = 11

Question 4.
2x + 5 = -25
Answer:
Following the example on page 190, first we subtract 5 from both sides. Then, we simplify and we divide both sides by 2.
2x + 5 = 25
2x + 5 – 5 = -25 – 5
2x = 30
\(\frac{2x}{2}\) = \(\frac{30}{2}\)
x = -15

Question 5.
6x – 7 = 41
Answer:
6x – 7 = 41 (Write the equation) ……….. (1)
6x = 41 + 7 (Add 7 to both sides) ………… (2)
6x = 48 (Simplify) ………….. (3)
x = \(\frac{48}{6}\) (Divide both sides by 6) ………. (4)
x = 8 (Simplify) ………….. (5)

Module 7 Angle Relationships in Parallel Lines and Triangles Answer Key Question 6.
\(\frac{1}{2}\)x + 9 = 30
Answer:
\(\frac{1}{2}\)x + 9 = 30 (Write the equation.) ……………. (1)
\(\frac{1}{2}\)x = 30 – 9 (Subtract 9 from both sides) ……………. (2)
\(\frac{1}{2}\)x = 21 (Simplify) ………………. (3)
x = 21 ∙ 2 (Multiply both sides by 2) …………… (4)
x = 42 (Simplify) …………… (5)

Question 7.
\(\frac{1}{3}\)x – 7 = 15
Answer:
\(\frac{1}{3}\)x + 7 = 15 (Write the equation.) ……………. (1)
\(\frac{1}{3}\)x = 15 + 7 (Add 7 from both sides) ……………. (2)
\(\frac{1}{3}\)x = 22 (Simplify) ………………. (3)
x = 22 ∙ 3 (Multiply both sides by 3) …………… (4)
x = 66 (Simplify) …………… (5)

Question 8.
0.5x – 0.6 = 8.4
Answer:
0.5x – 0.6 = 8.4 (Write the equation) …………. (1)
0.5x = 8.4 + 0.6 (Subtract 9 from both sides) …………. (2)
0.5x = 9 (Simplify) ………….. (3)
x = \(\frac{9}{0.5}\) (Divide both sides with 0.5) ………. (4)
x = 18 (Simplify) …………. (5)

Give two names for the angle formed by the dashed rays.

Question 9.

Answer:
The angle formed by the dashed rays can be named as: ∠MHR or ∠RHM

Question 10.

Answer:
The angle formed by the dashed rays can be named as ∠SGK or ∠KGS

Grade 8 Module 7 Angle Relationships Unit Test Answer Key Question 11.

Answer:
The angle formed by the dashed rays can be named as: ∠BTF or ∠FTB

Texas Go Math Grade 8 Module 7 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the graphic. You can put more than one word in each section of the triangle.

Understand Vocabulary

Complete the sentences using preview words.

Question 1.
A line that intersects two or more lines is a ____________.
Answer:
A transversal line is a line that intersects two or more Lines in the same plane at different points.
By definition, a line that intersects two or more lines is a transversal.
Thus, the blank part in the statement is transversal

Geometry Module 7 Answer Key Angle Relationships in Parallel Lines and Triangles Question 2.
Figures with the same shape but not necessarily the same size are ____________.
Answer:
Figures with the same shape but not necessarily the same size are similar figures.

Question 3.
An ____________________ is an angle formed by one side of the triangle and the extension of an adjacent side.
Answer:
The angle formed by one side of the triangle and the extension of an adjacent side is called an exterior angle.
By definition, an exterior angle is an angle formed by one side of the triangle and the extension of an adjacent side.
Thus, the blank part in the statement is the exterior angle.

Active Reading
Pyramid Before beginning the module, create a pyramid to help you organize what you learn. Label each side with one of the lesson titles from this module. As you study each lesson, write important ideas like vocabulary, properties, and formulas on the appropriate side.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 8.3 Answer Key Distance Between Two Points.

Texas Go Math Grade 8 Lesson 8.3 Answer Key Distance Between Two Points

Example 1

The figure shows a right triangle. Approximate the length of the hypotenuse to the nearest tenth without using a calculator.

STEP 1: Find the length of each leg.
The length of the vertical leg is 4 units.
The length of the horizontal leg is 2 units.

STEP 2: Let a = 4 and b = 2. Let c represent the length of the hypotenuse. Use the Pythagorean Theorem to find c.

STEP 3: Approximate \(\sqrt {20}\) by finding perfect squares close to 20.
\(\sqrt {20}\) is between \(\sqrt {16}\) and \(\sqrt {25}\) or \(\sqrt {16}\) < \(\sqrt {20}\) < \(\sqrt {25}\) .
Simplifying gives 4 < \(\sqrt {20}\) < 5.
Since 20 is about halfway between 16 and 25, \(\sqrt {20}\) is about halfway between 4 and 5. So, \(\sqrt {20}\) ≈ 4.5.
The hypotenuse is about 4.5 units long.

Your Turn

Question 1.
Approximate the length of the hypotenuse to the nearest tenth without using a calculator.

Answer:
Find the length of each leg. The length of the vertical leg is 4 units. The length of the horizontal Leg is 5 units. Let a = 4 and b = 5. Let c represent the length of the hypotenuse.
Use the Pythagorean Theorem to find c
a² + b² = c² ……………. (1)
4² + 5² = c² (Substitute into the formula) …………… (2)
16 + 25 = c² (Simplify) …………. (3)
41 = c² (Add) ………….. (4)
c = \(\sqrt {41}\) (Take the square root of both sides) …………. (5)
c ≈ 6.4 (Use a calculator and round to the nearest tenth) ……………… (6)

Texas Go Math Grade 8 Lesson 8.3 Explore Activity Answer Key

Finding the Distance Between Any Two Points

The Pythagorean Theorem can be used to find the distance between any two points (x₁, y₁) and (x₂, y₂) in the coordinate plane. The resulting expression is called the Distance Formula.

Use the Pythagorean Theorem to derive the Distance Formula.
A. To find the distance between points P and Q, draw segment \(\overline{P Q}\) and label its length d. Then draw horizontal segment \(\overline{P R}\) and vertical segment \(\overline{Q R}\). Label the lengths of these segments a and b. Triangle
PQR is a ______________ triangle, with hypotenuse ____________.

B. Since \(\overline{P R}\) is a horizontal segment, its length, a, is the difference between its x-coordinates. Therefore, a = x₂ – ___________.
C. Since \(\overline{Q R}\) is a vertical segment, its length, b, is the difference between its y-coordinates. Therefore, b = y₂ – _____________.
D. Use the Pythagorean Theorem to find d, the length of segment \(\overline{P Q}\). Substitute the expressions from B, and C for a and b.
d² = a² + b²
d = \(\sqrt{a^{2}+b^{2}}\)

Reflect

Go Math Lesson 8.3 Distance Between Two Points Answer Key Question 2.
Why are the coordinates of point R the ordered pair (x₂, y₁)?
Answer:
Since \(\overline{Q R}\) is a vertical statement, the x coordinate of R will be the same as the x coordinate of Q. On the other hand, since \(\overline{P R}\) is a horizontal statement, the y coordinate of R will be the same as the y coordinate of Q. Therefore, the coordinates of point R is the ordered pair (x₂, y₁).

Example 2

Francesca wants to find the distance between her house on one side of a lake and the beach on the other side. She marks off a third point forming a right triangle, as shown. The distances in the diagram are measured in meters.

Use the Pythagorean Theorem to find the straight-line distance from Francesca’s house to the beach.
STEP 1: Find the length of the horizontal leg.
The length of the horizontal leg is the absolute value of the difference between the x-coordinates of the points (280, 20) and (10, 20).
|280 – 10| = 270
The length of the horizontal leg is 270 meters.

STEP 2: Find the length of the vertical leg.
The length of the vertical leg is the absolute value of the difference between they-coordinates of the points (280, 164) and (280, 20).
|164 – 20| = 144
The length of the vertical leg is 144 meters.

STEP 3: Let a = 270 and b = 144. Let c represent the length of the hypotenuse. Use the Pythagorean Theorem to find c.

The distance from Francesca’s house to the beach is 306 meters.

Reflect

Question 3.
Show how you could use the Distance Formula to find the distance from Francesca’s house to the beach.
Answer:
You can use the Distance Formula, d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\) , to find the distance between the house and the beach. Let (x₁, y₁) = (10, 20) and (x₂, y₂) = (280, 164).

Your Turn

Question 4.
Camp Sunshine is also on the lake. Use the Pythagorean Theorem to find the distance between Francesca’s house and Camp Sunshine to the nearest tenth of a meter.

Answer:
Use the Pythagorean Theorem to find the straight-line distance from Francesca’s house to the Camp Sunshine. The length of the horizontal leg is the absolute value of the difference between the x-coordinates of the points (200, 20) and (10, 20).
|200 – 10| = 190
The length of the vertical leg is the absolute value of the difference between the y-coordinates of the points (200, 120) and (200, 20).
|120 – 20| = 100
Let = 190, b = 100, and Let c represent the Length of the hypotenuse. Use the Pythagorean Theorem to find c.
a² + b² = c² …………….. (1)
190² + 100² = c² (Substitute into the formula) …………. (2)
36100 + 10000 = c² (Simplify) …………… (3)
46100 = c² (Add) ………….. (4)
c = \(\sqrt {46100}\) (Take the square root of both sides) ………… (5)
c ≈ 214.7 (Use a calculator and round to the nearest tenth) ……………. (6)
The distance from Francesca’s house and Camp Sunshine is approximately 214.7 meters.

Texas Go Math Grade 8 Lesson 8.3 Guided Practice Answer Key

Question 1.
Approximate the length of the hypotenuse of the right triangle to the nearest tenth without using a calculator. (Example 1).

Answer:
The length of the vertical leg is 3 units. The length of the horizontal Leg is 5 units. Let a = 3 and b = 5. Let c represent the length of the hypotenuse. We use the Pythagorean Theorem to find c.
a² + b² = c²
3² + 5² = c²
9 + 25 = c²
34 = c²
c = \(\sqrt {34}\)
Approximate \(\sqrt {34}\) by finding perfect squares close to 34.
25 < 34 < 36
\(\sqrt {25}\) < \(\sqrt {34}\) < \(\sqrt {36}\)
5 < \(\sqrt {34}\) < 6
\(\sqrt {34}\) ≈ 5.5
The hypotenuse is about 5.5 units long.

Go Math Lesson 8.3 8th Grade Distance between Two Points Answer Key Question 2.
Find the distance between the points (3, 7) and (15, 12) on the coordinate plane. (Explore Activity).
Answer:
Using the Distance Formula we can find the distance between points (x₁, y₁) = (3, 7) and (x₂, y₂) = (15, 12) in the coordinate plane.

The distance between the points is 13 units.

Question 3.
A plane leaves an airport and flies due north. Two minutes later, a second plane leaves the same airport flying due east. The flight plan shows the coordinates of the two planes 10 minutes later. The distances in the graph are measured in miles. Use the Pythagorean Theorem to find the distance shown between the two planes. (Example 2)

Answer:
Use the Pythagorean Theorem to find the straight-line distance between two planes.
The length of the vertical leg is distance between first plane and the starting point: the absolute value of the difference
between the y-coordinates of the points (1, 80) and (1, 1).
|80 – 1| = 79
The length of the horizontal leg is distance between second plane and the starting point: it is the absolute value of the difference between the x-coordinates of the points (68, 1) and (1, 1).
|68 – 1| = 67
Let a = 79, b = 67, and let c represent the Length of the hypotenuseUse the Pythagorean Theorem to find c.
a² + b² = c² ……………….. (1)
79² + 67² = c² (Substitute into the formula) ……………. (2)
6241 + 4489 = c² (Simplify) ……………. (3)
10730 = c² (Add) ………………. (4)
c = \(\sqrt {10730}\) (Take the square root of both sides) …………….. (5)
c ≈ 103.6 (Use a calculator and round to the nearest tenth) ……………… (6)
The distance between two planes is approx 103.6 miles.

Essential Question Check-In

Question 4.
Describe two ways to find the distance between two points on a coordinate plane.
Answer:
You can draw a right triangle whose hypotenuse is the segment connecting the two points and then use the Pythagorean Theorem to find the length of that segment. You can also use the Distance Formula to find the length of that segment.
For example, plot three points; (1, 2), (20, 2), and (20, 12)
Using the Pythagorean Theorem
The length of the horizontal leg is the absolute value of the difference between the x-coordinates of the points (1, 2) and (20, 2).
|1 – 20| = 19
The length of the horizontal leg is 10.
Let a = 19, b = 10 and Let c represent the hypotenuse. Find c.
a² + b² = c²
19² + 10² = c²
361 + 100 = c²
461 = c²
distance → 21.5 ≈ e
Using the Distance Formula
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
The length of the horizontal leg in between (1, 2) and (20, 2).

The length of the horizontal leg in between (20, 2) and (20, 12).

The length of the horizontal leg in between (1, 2) and (20, 12).

Texas Go Math Grade 8 Lesson 8.3 Independent Practice Answer Key

Question 5.
A metal worker traced a triangular piece of sheet metal on a coordinate plane, as shown. The units represent inches. What is the length of the longest side of the metal triangle? Approximate the length to the nearest tenth of an inch without using a calculator.

Answer:
(5 × 5) + (6 × 6) = c × c
25 + 36 = c × c
\(\sqrt {61}\) = c
c ≈ 7.8

Go Math Lesson 8.3 Using Midpoint and Distance Formulas Answers Question 6.
When a coordinate grid is superimposed on a map of Harrisburg, the high school is located at (17, 21) and the town park is located at (28, 13). If each unit represents 1 mile, how many miles apart are the high school and the town park? Round your answer to the nearest tenth.
Answer:
The coordinates of the high school, are said to be (17, 21), whereas the coordinates of the park are (28, 13). In a coordinate plane, the distance d between the points (17, 21) and (28, 13) is:
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
d = \(\sqrt{(28-17)^{2}+(13-21)^{2}}\)
d = \(\sqrt{(11)^{2}+(-8)^{2}}\)
d = \(\sqrt{121+64}\)
d = \(\sqrt{185}\)
Rounding the answer to the nearest tenth:
d ≈ 13.6
Taking into consideration that each unit represents 1 mile, the high school and the town park are 13.6 miles apart.

Question 7.
The coordinates of the vertices of a rectangle are given by R(-3, -4), E(-3, 4), C(4, 4), and T(4, -4). Plot these points on the coordinate plane at the right and connect them to draw the rectangle. Then connect points E and T to form diagonal \(\overline{E T}\).

a. Use the Pythagorean Theorem to find the exact length of \(\overline{E T}\).
Answer:

Taking into consideration the triangLe TRE, the Length of the vertical leg (\(\overline{E R}\)) is 8 units. The Length of the horizontal. Leg (\(\overline{R T}\)) is 7 units. Let a = 8 and b = 7. Let c represent the length of the hypotenuse, the diagonal
\(\overline{E T}\). We use the Pythagorean Theorem to find c.
a² + b² = c²
8² + 7² = c²
64 + 49 = c²
113 = c²
c = \(\sqrt {113}\)
c ≈ 10.6
The diagonal \(\overline{E T}\) is about 10.63 units long.

b. How can you use the Distance Formula to find the length of \(\overline{E T}\)? Show that the Distance Formula gives the same answer.
Answer:
Using the distance Formula in a coordinate plane, the distance d between the points E(-3, 4) and T(4, 4)
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
d = \(\sqrt{(4-(-3))^{2}+(-4-4)^{2}}\)
d = \(\sqrt{(7)^{2}+(-8)^{2}}\)
d = \(\sqrt{49+64}\)
d = \(\sqrt{113}\)
d ≈ 10.63
The diagonal \(\overline{E T}\) is about 10.63 units long.
As we can see, the answer is the same as the one we found on point a.

Question 8.
Multistep The locations of three ships are represented on a coordinate grid by the following points: P(-2, 5), Q(-7, -5), and R(2, -3). Which ships are farthest apart?
Answer:
Distance Formula: In a coordinate plane, the distance d between two points (x₁, y₁) and (x₂, y₁) is:
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
The distance d₁ between the points P(2, 5) and Q(-7, -5) is:

The distance d₂ between the points Q(-7, -5) and R(2, -3) is:

The distance d₃ between the points P(-2, 5) and R(2, -3) is:

As we can see, the greatest distance is d₁ ≈ 11.8, which means that ships P and Q are farthest apart.

Go Math 8th Grade Distance and Midpoint Map Activity Answer Key Question 9.
Make a Conjecture Find as many points as you can that are 5 units from the origin. Make a conjecture about the shape formed if all the points 5 units from the origin were connected.
Answer:
Some of the points that are 5 units away from the origin are: (0, 5), (3, 4), (4, 3), (5, 0), (4, 3), (3, -4), (0, 5) (-3, 4), (-4, 3), (- 5, 0), ( 4, 3), (-3, 4) etc. If all the points 5 units away from the origin are connected, a circle would be formed.

Question 10.
Justify Reasoning The graph shows the location of a motion detector that has a maximum range of 34 feet. A peacock at point P displays its tail feathers. Will the motion detector sense this motion? Explain.

Answer:
The coordinates of the motion detector are said to be (0, 25), whereas the coordinates of the peacock are (30, 10). In a coordinate plane, the distance d between the points (0, 25) and (30, 10) is:
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
d = \(\sqrt{(30-0)^{2}+(10-25)^{2}}\)
d = \(\sqrt{(30)^{2}+(-15)^{2}}\)
d = \(\sqrt{900+225}\)
d = \(\sqrt{1125}\)
Rounding the answer to the nearest tenth:
d ≈ 33.5
Taking into consideration that each unit represents 1 foot,
the motion detector and the peacock are 33.5 feet apart. Since the motion detector has a maximum range of 34 feet, it means that it wilt sense the motion of the peacock’s feathers.

H.O.T. Focus on Higher Order Thinking

Question 11.
Persevere in Problem-Solving One leg of an isosceles right triangle has endpoints (1, 1) and (6, 1). The other leg passes through the point (6, 2). Draw the triangle on the coordinate plane below. Then show how you can use the Distance Formula to find the length of the hypotenuse. Round your answer to the nearest tenth.

Answer:
One leg of an isosceles right triangle has endpoints (1, 1) and (6. 1), which means that that leg is 5 units long. Since the triangle is isosceles, the other leg should be 5 units long too, therefore the endpoints of the second leg that passes through point (6, 2) are (6, 1) and (6, 6).
In the coordinate plane, the length of the hypotenuse is the distance d between the points (1, 1) and (6, 6).
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
d = \(\sqrt{(6-1)^{2}+(6-1)^{2}}\)
d = \(\sqrt{(5)^{2}+(5)^{2}}\)
d = \(\sqrt{25+25}\)
d = \(\sqrt{50}\)
Rounding the answer to the nearest tenth:
d ≈ 7.1
The hypotenuse is around 7.1 units long.

Question 12.
Represent Real-World Problems The figure shows a representation of a football field. The units represent yards. A sports analyst marks the locations from where the football was thrown (point A) and where it was caught (point B). Explain how you can use the Pythagorean Theorem to find the distance the ball was thrown. Then find the distance.

Answer:
To find the distance between points A and B, we draw segment \(\overline{A B}\) and label its length d. Then we draw vertical segment \(\overline{A C}\) and horizontal segment \(\overline{C B}\). We label the lengths of these segments a and b. Triangle ACB is a right triangle with hypotenuse AB.
Since \(\overline{A C}\) is a vertical segment, its length, a, is the difference between its y-coordinates. Therefore, a = 26 – 14 = 12 units.
Since \(\overline{C B}\) is a horizontal segment, its length, b, is the difference between its x-coordinates. Therefore, b = 75 – 40 = 35 units.
We use the Pythagorean Theorem to find d, the length of segment \(\overline{A B}\).
d² = a² + b²
d² = 12² + 35²
d² = 144+ 1225
d² = 1369
d = \(\sqrt {1369}\)
d = 37
The distance between points A and B is 37 yards.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 7.3 Answer Key Angle-Angle Similarity.

Texas Go Math Grade 8 Lesson 7.3 Answer Key Angle-Angle Similarity

Essential Question
How can you determine when two triangles are similar?

Texas Go Math Grade 8 Lesson 7.3 Explore Activity Answer Key

Explore Activity 1

Discovering Angle-Angle Similarity
Similar figures have the same shape but may have different sizes. Two triangles are similar if their corresponding angles are congruent and the lengths of their corresponding sides are proportional.

A. Use your protractor and a straightedge to draw a triangle. Make one angle measure 45° and another angle measure 60°.
B. Compare your triangle to those drawn by your
classmates. How are the triangles the same?

How are they different?

C. Use the Triangle Sum Theorem to find the measure of the third angle of your triangle.

Reflect

Question 1.
If two angles in one triangle are congruent to two angles in another triangle, what do you know about the third pair of angles?
Answer:
If two angles in one triangle are congruent to two angles in another triangle, then we can call them the corresponding angles of the two triangles since they are congruent

Also, if two pairs of corresponding angles of a triangle are congruent then the triangles are similar

By the description of the angles of the two triangles, we can say that the third pair of the angles are also congruent.

Lesson 7.3 Answer Key 8th Grade Similar Triangles Worksheet Answers Question 2.
Make a Conjecture Are two pairs of congruent angles enough information to conclude that two triangles are similar? Explain.
Answer:
If two pairs of corresponding angles in a pair of triangles are congruent, it means that the third pair must also be equal since the total sum of interior angles of a triangle is 180°. Therefore, we can conclude that the triangles are similar.

Your Turn

Question 3.
Explain whether the triangles are similar.

Answer:
180° = 70° + 58° + N° First find what angle N equals
180° = 128° + N°
52° = N°
180° = 70° + 49° + Q° Then find what angle Q equals
180° = 119° + Q°
61° = Q°
Since one angle from these triangles are the same then these are not similar triangles.

Example 2.
While playing tennis, Matt is 12 meters from the net, which is 0.9 meter high. He needs to hit the ball so that it just clears the net and lands 6 meters beyond the base of the net. At what height should Matt hit the tennis ball?

Matt should hit the ball at a height of 2.7 meters.

Reflect

Question 4.
What If? Suppose you set up a proportion so that each ratio compares parts of one triangle, as shown below.

Show that this proportion leads to the same value for h as in Example 2.
Answer:
We are watching triangles ∆ABC and ∆ABD. Both triangles contain ∠A and a right angle, so ∆ABC and ∆ADE are similar. Lets set up a proportion:

This proportion leads to the same value for h as in Example 2

Your Turn

Lesson 7.3 Solve for Unknown Angles-Transversals Question 5.
Rosie is building a wheelchair ramp that is 24 feet long and 2 feet high. She needs to install a vertical support piece 8 feet from the end of the ramp. What is the length of the support piece in inches?

Answer:
Triangles formed from wheelchair ramps, supported pieces, and distances between are forming similar triangles.
Let’s set up a proportion using known lengths
\(\frac{8}{24}\) = \(\frac{h}{2}\) (Substitute the lengths from the figure.) (1)
2 × \(\frac{8}{24}\) = \(\frac{h}{2}\) × 2 (Use properties of equality to get h by itself) (2)
\(\frac{2}{3}\) = h (Simplify) (3)
h = \(\frac{2}{3}\)ft = \(\frac{2}{3}\) . 12 inches = 8 inches (4)

h = 8 inches

Question 6.
The lower cable meets the tree at a height of 6 feet and extends out 16 feet from the base of the tree. If the triangles are similar, how tall is the tree?

Answer:
Let’s set up a proportion using known lengths:
\(\frac{h}{56}\) = \(\frac{6}{16}\) (Substitute the lengths from the figure.) (1)
56 × \(\frac{h}{56}\)=\(\frac{6}{16}\) × 56 (Use properties of equality to get h by itself) (2)
h = 21ft (Simplify) (3)

h = 21ft

Explore Activity 2
Using Similar Triangles to Explain Slope

You can use similar triangles to show that the slope of a line is constant.

A. Draw a line l that is not a horizontal line. Label four points on the line as A, B, C, and D.

B. Draw the rise and run for the slope between points A and B. Label the intersection as point E. Draw the rise and run for the slope between points Cand D. Label the intersection as point F.

C. Write expressions for the slope between A and B and between C and D.

D. Extend \(\overleftrightarrow{A E}\) and \(\overleftrightarrow{C F}\) across your drawing. \(\overleftrightarrow{A E}\) and \(\overleftrightarrow{C F}\) are both horizontal
lines, so they are parallel.
Line l is a _______________ that intersects parallel lines.
E. Complete the following statements:
∠BAE and ________ are corresponding angles and are ___.
∠BEA and ________ are right angles and are ___.
F. By Angle—Angle Similarity, ∠ABE and ___ are similar triangles.
G. Use the fact that the lengths of corresponding sides of similar triangles are proportional to complete the following ratios:
H. Recall that you can also write the proportion so that the ratios compare parts of the same triangle: .
I. The proportion you wrote in step H shows that the ratios you wrote in C, are equal. So, the slope of line l is constant.

Reflect

Question 7.
What If? Suppose that you label two other points on line l as G and H. Would the slope between these two points be different than the slope you found in the Explore Activity? Explain.
Answer:
If we Label two other points on Line as G and H and we follow the steps as in the Explore Activity 2, we will see that
the sLope between these two points will be the same as the one we found before.
First, we draw the rise and run for the slope between points G and H and we label the intersection as point M. We write the expression for the slope between G and H as \(\frac{H M}{G M}\). We extend GM across our drawing, so it is parallel with the extensions of AE and CF. We find the corresponding and the right angles and we see that the triangles that were formed are similar. Therefore, we write the proportions and we conclude that the slopes are equal.

Texas Go Math Grade 8 Lesson 7.3 Guided Practice Answer Key

Question 1.
Explain whether the triangles are similar. Label the angle measures in the figure. (Explore Activity 1 and Example 1)

∆ABC has angle measures ____________________ and ∆DEF has angle measures _________________. Because ______________ in one triangle are congruent to ________________ in the other triangle, the triangles are ___.
Answer:
∆ÁBC has angle measures 40°, 30° and 109°
∠B = 180 – A – C = 180 – 41 – 30 = 109°)

and ∆DEF has angle measures 41°, 109°, and 30°
∠D = 180 – E – F = 180 – 109 – 30 = 41°)

Because
2∠s
in the other triangle, the
triangles are similar
40°, 30°, and 109°
41°, 109°. and 30°,
similar

Angle Angle Similarity Lesson 7.3 Practice A Geometry Answers Question 2.
A flagpole casts a shadow 23.5 feet long. At the same time of day, Mrs. Gilbert, who is 5.5 feet tall, casts a shadow that is 7.5 feet long. How tall in feet is the flagpole? Round your answer to the nearest tenth. (Example 2)


Answer:
In similar triangles, corresponding side lengths are proportional.
\(\frac{5.5}{7.5}\) = \(\frac{h}{23.5}\)
23.5 . 7.5 . \(\frac{5.5}{7.5}\) = \(\frac{h}{23.5}\) . 23.5 . 7.5
23.5 . 5.5 = 7.5 . h
h = \(\frac{129.25}{7.5}\)
Rounding to the nearest tenth:
h = 17.2 feet

Question 3.
Two transversals intersect two parallel lines as shown. Explain whether ∆ABC and ∆DEC are similar. (Example

∠BAC and ∠EDC are ____________ since they are _______
∠ABC and ∠DEC are ____________ since they are _______
By ____________, ∆ABC and ∆DEC are ____.
Answer:
∠BAC and ∠EDC are congruent since they are
alt. interior ∠s
∠ABC and ∠DEC are congruent since they are
alt. interior ∠s
By AA similarity, ∆ABC and ∆DEC are similar.

AA similarity; similar (see explanation)

Essential Question Check-In

Question 4.
How can you determine when two triangles are similar?
Answer:
If 2 angles of one triangle are congruent to 2 angles of another triangle, the triangles are similar by the Angle-Angle Similarity Postulate

Angle – Angle Similarity Postulate (see explanation)

Texas Go Math Grade 8 Lesson 7.3 Independent Practice Answer Key

Use the diagrams for Exercises 5-7.

Lesson 7.3 Similar Triangles Answer Key Question 5.
Find the missing angle measures in the triangles.
Answer:
a) From the Triangle Sum Theorem we have:
m∠A + m∠B + m∠C = 180°
We substitute the given angle measures and we solve for m∠B.
85° + m∠B + 53° = 180°
138° + m∠B = 180°
138° – 138° + m∠B = 180° – 138°
m∠B = 42°

b) From the Triangle Sum Theorem we have:
m∠D + m∠E + m∠F = 180°
We substitute the given angle measures and we solve for m∠F.
64° + 47° + m∠F = 180°
111° + m∠F = 180°
111° – 111° + m∠F = 180° – 111°
m∠F = 69°

c) From the Triangle Sum Theorem we have:
m∠G + m∠H + m∠I = 180°
We substitute the given angle measures and we solve for m∠H.
47° + m∠H + 69° = 180°
116° + m∠H = 180°
116° – 116°+ m∠H = 180° – 116°
m∠H = 64°

d) From the Triangle Sum Theorem we have:
m∠J + m∠K + m∠L = 180°
We substitute the given angle measures and we solve for m∠H.
85° + m∠K + 42° = 180°
127° + m∠K = 180°
127° – 127° + m∠K = 180° – 127°
m∠K = 53°

Question 6.
Which triangles are similar?
Answer:
∆ABC’ and ∆JKL are similar because their corresponding angles are congruent Also, ∆DEF and ∆GHI are similar because their corresponding angles are congruent.

Question 7.
Analyze Relationships Determine which angles are congruent to the angles in A ABC.
Answer:
∆JKL has angle measures that are the same as those is ∆ÀBC
(∠A ≅ ∠J, ∠B ≅ ∠L, and ∠C ≅ ∠K)
Therefore, they are congruent

∆JKL ≅ ∆ABC

Lesson 7.3 Similar Triangles AA Similarity Question 8.
Multistep A tree casts a shadow that is 20 feet long. Frank is 6 feet tall, and while standing next to the tree he casts a shadow that is 4 feet long.

a. Flow tall is the tree? _____
Answer:
In similar triangles, corresponding side lengths are proportional.
\(\frac{20}{4}\) = \(\frac{h}{6}\)
6 . 4 . \(\frac{20}{4}\) = \(\frac{h}{6}\) . 6 . 4
6 . 20 = 4 . h
\(\frac{4 h}{4}\) = \(\frac{120}{4}\)
h = 30
The tree is 30 feet tall.

b. Flow much taller is the tree than Frank? ____
Answer:
30 – 6 = 24
The tree is 24 feet taller than Frank.

Question 9.
Represent Real-World Problems Sheila is climbing on a ladder that is attached against the side of a jungle gym wall. She is 5 feet off the ground and 3 feet horizontally from the base of the ladder, which is 15 feet from the wall. Draw a diagram to help you solve the problem. Flow high up the wall is the top of the ladder?
Answer:

\(\frac{3}{15}\) = \(\frac{5}{h}\)
15 × 3 = 3h
75 = 3h
25 = h
h = 25

Question 10.
Justify Reasoning Are two equilateral triangles always similar? Explain.
Answer:
yes; two equilateral triangles are always similar

Each angle of an equilateral triangle is 60°
Since both triangles are equilateral they are similar

yes (see explanation)

Triangle Similarity Worksheet Answers Question 11.
Critique Reasoning Ryan calculated the missing measure in the diagram shown. What was his mistake?

Answer:
In the first line, Ryan did not take the sum of 65 and 19.5 to get the denominator on the right

The denominator on the right should be 26 instead of 19.5

correct value for h
\(\frac{3.4}{6.5}\) = \(\frac{h}{26}\)
26 × \(\frac{3.4}{6.5}\) = \(\frac{h}{26}\) × 26
\(\frac{88.4}{6.5}\) = h
13.6 cm = h

error: 19.5 should be 26
h = 13.6
see explanation

Texas Go Math Grade 8 Lesson 7.3 H.O.T. Focus On Higher Order Thinking Answer Key

Question 12.
Communicate Mathematical Ideas For a pair of triangular earrings, how can you tell if they are similar? How can you tell if they are congruent?
Answer:
The earrings are similar if two angle measures of one are equal to two angle measures of the other

The earrings are congruent if they are similar and if the side lengths of one are equal to the side lengths of the other.

similar → 2∠’S = 2∠’S
congruent → similar and side lengths = side lengths
see explanation

Question 13.
Critical Thinking When does it make sense to use similar triangles to measure the height and length of objects in real life?
Answer:
If the item is too tall or the distance is too long to measure directly, similar triangles can help with measuring

similar triangles (see explanation)

Similar Triangles AA Similarity Answers Question 14.
Justify Reasoning Two right triangles on a coordinate plane are similar but not congruent. Each of the legs of both triangles are extended by 1 unit, creating two new right triangles. Are the resulting triangles similar? Explain using an example.
Answer:
Two triangles are similar if their corresponding angles are congruent and the lengths of their corresponding sides are proportional. If each of the legs of both triangles is extended by 1 unit, the ratio between proportional sides does not change. Therefore, the resulting triangles are similar.