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 × 10^{8} 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 × 10^{8} years ago. In standard form

10^{8} = 100, ooo, 000 (Ten with eight zeros) (1)

1.75 × 10^{8} = 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