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## Texas Go Math Grade 7 Module 9 Quiz Answer Key

**Texas Go Math Grade 7 Module 9 Ready to Go On? Answer Key**

**9.1 Angle Relationships**

**Use the diagram to name a pair of each type of angle.**

Question 1.

Supplementary angles _____

Answer:

∠DFC and ∠BFC

Question 2.

Complementary angles. _____

Answer:

∠AFB and ∠BFC

Question 3.

Vertical angles. _____

Answer:

∠EFD and ∠BFC

**9.2, 9.3 Finding Circumference and Area of Circles**

**Find the circumference and area of each circle. Use 3.14 for π.**

Question 4.

Circumference ≈ _____

Answer:

d = 36 cm

Use the formula for the circumference of the circle when given diameter

C = π(d) Substitute 36 for d, and 3.14 for π.

C ≈ 3.14(36)

C ≈ 113.04

The circumference of the circle is 113.04 cm

C ≈ 113.04 cm

Question 5.

area ≈ ___

Answer:

Use the formula for the area of the circle.

Since the diameter is twice a radius, the formula for area of a circle, when given the diameter, is

A = π\(\left(\frac{d}{2}\right)^{2}\) Substitute 36 for d, and 3.14 for π.

A ≈ 3.14 . \(\left(\frac{36}{4}\right)^{2}\)

A ≈ 3.14 18^{2}

A ≈ 3.14 . 324

A ≈ 1017.36

The area of the circle is about 1017.36 cm^{2}.

A ≈ 1017.36 cm^{2}

Question 6.

Circumference ≈ _____

Answer:

The radius of the circle is 7 m.

Use the formula for the circumference of the circle.

C = 2π(r) Substitute 7 for r, and \(\frac{22}{7}\) for π.

The circumference is about 44 m.

C ≈ 44

Question 7.

area ≈ ___

Answer:

Use the formula for the area of the circle

A = Substitute 7 for r, and 314 for w.

A ≈ 3.14(7)^{2}

A ≈ 3.14 49

A ≈ 153.86

The area of the circle is about 153.86 m^{2}

A ≈ 153.86 m^{2}.

**9.4 Area of Composite Figures**

**Find the area of each figure. Use 3.14 for π.**

Question 8.

Answer:

Separate the figure into a triangle and one semicircle.

Area of the triangle

base = 14 m

height = 10 m

Use the formula for the area of the triangLe.

A_{1} = \(\frac{1}{2} b \cdot h\)

A_{1} = \(\frac{1}{2} 14 \cdot 10\)

A_{1} = \(\frac{70}{2}\)

A_{1} = 35

The area of the triangle is 35 m_{2}.

Area of the circle

diameter = 14 m

Use the formula for the area of the circte when given diameter

A_{c} = \(\pi\left(\frac{d}{2}\right)^{2}\) Subsitute 14 for d and 3.14 for π.

A_{c} = 3.14\(\left(\frac{14}{2}\right)^{2}\)

A_{c} = 3.14 . 7^{2}

A_{c} = 3.14 . 49

A_{c} = 153.86

Area of the semicircle is half the area of the circle.

A_{2} = \(\frac{A_{c}}{2}\) = \(\frac{153.86}{2}\) = 76.93

The area of the semicircle is 76.93 m^{2}.

Add the areas to find the total area.

A = A_{1} + A_{2} = 35 + 76.93 = 111.93

The area of the figure is 111.93 m^{2}.

Question 9.

Answer:

Separate the figure into a rectangle and a parallelogram.

Area of the rectangle

length = 20 cm

width = 5.5 cm

Use the formula for the area of the rectangle.

A_{1} = l . w

A_{1} = 20 . 5.5

A_{1} = 110

The area of the rectangle is 110 cm^{2}.

Area of the paralleogram

b = 20 cm

h = 4.5 cm

Use the formula for the area of the paralleogram.

A_{2} = b . h

A_{2} = 20 . 4.5

A_{2} = 90

The area of the parallelogram is 90 cm^{2}

Add the areas to find the total area.

A = A^{1} + A^{2} = 110 + 90 = 200

The area of the figure is 200 cm^{2}.

**Essential Question**

Question 10.

How can you use geometric formulas in real-world situations?

Answer:

There are times that we need to determine the area or perimeter or volume of a certain object or even a place and

the only given is the dimensions, the geometric formuLas are very helpful in such a way that we can calculate it by ourselves.

The geometric formulas are very helpful.

**Texas Go Math Grade 7 Module 9 Mixed Review Texas Test Prep Answer Key **

**Selected Response**

**Use the diagram for Exercises 1-3.**

Question 1.

What is the measure of ∠BFC?

(A) 18°

(B) 72°

(C) 108°

(D) 144°

Answer:

(C) 108°

Explanation:

∠AFB and ∠BFC are supplement angles, hence

∠AFB + ∠BFC = 180° Substitute 72° for ∠AFB

72° + ∠BFC = 180° Subtract 72° from both sides.

∠BFC = 180° – 72°

∠BFC = 108°

Question 2.

Which describes the relationship between ∠BFA and ∠CFD?

(A) adjacent angles

(B) complementary angles

(C) supplementary angles

(D) vertical angles

Answer:

(D) vertical angles

Question 3.

Which information would allow you to identify ∠BFA and ∠AFE as complementary angles?

(A) m∠AFE = 108°

(B) ∠DFE is a right angle.

(C) ∠BFA and ∠BFC are supplementary angles.

(D) ∠BFA and ∠BFC are adjacent angles.

Answer:

(B) ∠DFE is a right angle.

Explanation:

∠DFE is a right angle, as we see in the diagram.

Question 4.

David pays $7 per day to park his car. He uses a debit card each time. By what amount does his bank account change due to parking charges over a 40-day period?

(A) -$280

(B) -$47

(C) $47

(D) $280

Answer:

(A) -$280

Explanation:

David pays $7 each day for 40 days, so amount on his bank account decreases for $7 . 40 = $280.

Question 5.

What is the circumference of the circle? Use 3.14 for π.

(A) 34.54 m

(B) 69.08 m

(C) 379.94 m

(D) 1,519.76 m

Answer:

(B) 69.08 m

Explanation:

r = 11 m

Use the formula for circumference.

C = 2πr(r) Substitute 11 for r and 314 for π

C ≈ 2 3.14 . 11

C ≈ 69.08

The circumference of the circle is 69.08 m.

Question 6.

What is the area of the circle? Use 3.14 for π.

(A) 23.55 m^{2}

(B) 176.625 m^{2}

(C) 47.1 m^{2}

(D) 706.5 m^{2}

Answer:

(B) 176.625 m^{2}

Explanation:

d = 15m

Use the formula for the area of the circle.

A = πr^{2}

Since the diameter is twice a radius, the formula for area of a circle, when given the diameter, is

A = π(r)^{2} Substitute \(\frac{d}{2}\) for r.

A = \(\pi\left(\frac{d}{2}\right)^{2}\) Substitute 15 for d, and 3.14 for π.

A ≈ 3.14 . \(\left(\frac{15}{2}\right)^{2}\)

A ≈ 3.14 . (7.5)^{2}

A ≈ 3.14 . 56.25

A ≈ 176.62

The area of the circle is about 176.62 m^{2}.

**Gridded Response**

Question 7.

Find the area in square meters of the figure below? Use 3.14 for π.

Answer:

Separate the figure into square and a quater of circle.

Area of the square

side = 6 m

Use the formula for the area of the square

A_{1} = 8^{2}

A_{1} = 6^{2}

A_{1} = 36

The area of the sqaure is 36 m^{2}.

To find the area of the quarter of a circle, find the area of the circle and divide it by 4.

Area of the circle

radius = 6 m

Use the formula for the area of the circle when given radius.

A_{c} = πr^{2} Subsitute 10 for r and 314 for w.

A_{c} = 3.14(6)^{2}

A_{c} = 3.14 . 36

A_{c} = 113.04

A_{2} = \(\frac{A_{c}}{4}\) = \(\frac{113}{404}\) = 28.26

The area of the quarter of the circle is 28.26 m^{2}.

Add the areas to find the total area.

A = A_{1} + A_{2} = 36 + 28.26 = 64.26

The area of the figure is 64.26 m^{2}.