Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Module 10 Quiz Answer Key.

## Texas Go Math Grade 7 Module 10 Quiz Answer Key

**Texas Go Math Grade 7 Module 10 Ready to Go On? Answer Key**

**10.1 Volume of Rectangular Prisms and Pyramids**

**Find the volume of each figure.**

Question 1.

Answer:

l = 11 yd length of base

w = 9 yd Width of base

h = 7 yd Height of a prism

Find the area of the base B. A prism has a base that is rectangular, so use the formula:

B = l.w = 11.9 = 99 yd^{2}

Use the formula for the volume of a rectangular prism.

V = B.h = 99 . 7 = 693 yd^{3}

The volume of a rectangular prism is 693 yd^{3}.

**Grade 7 Math Module 10 Answer Key Question 2.**

Answer:

l = 12 ft length of base

w = 16 ft Width of base

h = 18 ft Height of a pyramid

Find the area of the base B.

B = l . w = 12 . 16 = 192 ft^{2}

Use the formula for the volume of a rectangular pyramid.

The volume of a rectangular pyramid is 1152 ft^{3}.

The volume of a rectangular pyramid is 1152 ft^{3}.

10.2 Volume of Triangular Prisms and Pyramids

**Find the volume of each figure.**

Question 3.

Answer:

A prism has a base that is right triangle.

a = 8 cm length of one side of the triangle

b = 17 cm length of other side of the triangle

Find the area of the base B. Use the formula:

B = 68 cm^{2} Base of prism

h = 40 cm Height of a prism

Use the formula.

V = B . h = 68 . 40 = 2720 cm^{3}

The volume of a triangular prism is 2720 cm^{3}.

**Module 10 Form B Module Test Answer Key Question 4.**

Answer:

b = 5 m length of base

h_{B} = 15 m height of base

h = 22 m Height of a pyramid

Find the area of the base B. Use the formula:

B = \(\frac{1}{2}\) . b . h_{B} = \(\frac{1}{2}\) . 5 . 15 = 37.5 m^{2}

Use the formula for the voLume of a triangular pyramid:

V = \(\frac{1}{3}\)B . h = \(\frac{1}{3}\) . (37.5) . 22 = 275 m^{3}

The volume of a triangular pyramid is 275 m^{3}.

**10.3 Lateral and Total Surface Area**

**Find the lateral and total surface area of each figure using its net.**

Question 5.

Answer:

Find the Lateral area of the triangular pyramid.

We have three triangles with base of 19.2 m. and height of 40 m.

b = 19.2 m length of base

h_{B} = 40 m Height of base

Lateral area:

For the area of the base:

b = 24 m length of base

h_{b} = 15 m Height of base

The area of the base is

The total surface area is 1152 + 180 = 1332 m^{2}

The Lateral area is 1152 m^{2}.

The total surface area is 1332 m^{2}.

**Module 10 Quiz Ready To Go On Answers 7th Grade Question 6.**

Answer:

Find the Lateral area of the rectangular prism.

We have:

Two 12 in. by 6 in. rectangle;

Two 10 in. by 6 in. rectangle;

2 . (12 . 6) = 144

2 . (10 . 6) = 120

Lateral area: 144 + 120 = 264 cm^{2}

Each base is 12 in. by 10 in.

2 . (12 . 10) = 2 . 120 = 240 cm^{2}

The total surface area is 264 + 240 = 504 cm^{2}.

The Lateral area is 264 cm.

The total surface area is 504 cm^{2}.

**Essential Question**

Question 7.

How can you use volume and surface area to solve real-world problems?

Answer:

Volume and surface area can be used in many ways to solve the real-world problems such as:

For example we are making a new thing of copper and we know the volume of that thing so we can know the exact amount of copper required for making that thing by using the density = \(\begin{gathered}

\text { Mass } \\

\hline \text { Volume }

\end{gathered}\)of copper

In another example suppose a wall is to be painted and the cost of paint one square meter of wall is given then we can easily calculate the total cost for painting wall by using the total surface area of wall.

Volume can be use for calculating amount of material required for making something.

**Texas Go Math Grade 7 Module 10 Mixed Review Texas Test Prep Answer Key **

**Selected Response**

Question 1.

The volume of a triangular pyramid is 232 cubic units. The area of the base of the pyramid is 29 square units. What is the height of the pyramid?

(A) 8 units

(B) 12 units

(C) 16 units

(D) 24 units

Answer:

(D) 24 units

Explanation:

Use formula for the volume of a triangular pyramid.

B = 29 square units Base of pyramid

h = ? Height of a pyramid

V = 232 cubic units The volume of a pyramid

Use the formula for the volume of a pyramid to find a height of the pyramid.

V = \(\frac{1}{3}\) . B . h.

232 = \(\frac{1}{3}\) . 29 . h

232 = (9.66) . h

Divide both sides by 9.66.

The height of the pyramid is 24 units

Question 2.

What is the volume of a rectangular prism that has a length of 8.5 centimeters (cm), a width of 3.2 centimeters, and a height of 6 centimeters?

(A) 19.2 cm^{3}

(B) 27.2 cm^{3}

(C) 51 cm^{3}

(D) 163.2 cm^{3}

Answer:

(D) 163.2 cm^{3}

Explanation:

h = 6 cm Height of prism

l = 8.5 cm Length of base

w = 3.2 cm Width of base

Use the formula for the volume of a rectangular prism

V = B . h = l . w . h

= (8.5) (3.2) . h

= (27.2) . 6

= 163.2 cm^{3}

The volume of a rectangular prism is V = 163.2 cm^{3}

Question 3.

What is the volume of the rectangular pyramid shown?

(A) 1,650 yd^{3}

(B) 2,200 yd^{3}

(C) 3,300 yd^{3}

(D) 6,600 yd^{3}

Answer:

h = 22 yd Height of pyramid

l = 15 yd Length of base

w = 20 yd Width of base

Use the formula for the volume of a rectangular pyramid.

V = \(\frac{1}{3}\) . B . h. = \(\frac{1}{3}\) . l . w . h.

= \(\frac{1}{3}\) . 15 . 20 . 22

= 2200 yd^{3}

The volume of the rectangular pyramid is V = 2200 yd^{3}.

**Module 10 Volume Quiz Answer Key Question 4.**

A circle has a circumference of 56π centimeters (cm). What is the radius of the circle?

(A) 28 cm

(B) 56 cm

(C) 88 cm

(D) 112 cm

Answer:

(A) 28 cm

Explanation:

Given circumference of circle in probem = 56π cm

We know that circumference of circle of radius “r” = 2π . r

2π . r = 56π (By using above formula)

r = \(\frac{56 \pi}{2 \pi}\) (Dividing both side by 2π)

r = 28 cm (Simplifying)

hence, option A is correct answer.

Question 5.

What is the volume of a triangular prism that has a height of 45 meters and has a base with an area of 20 square meters?

(A) 225 m^{3}

(B) 300 m^{3}

(C) 450 m^{3}

(D) 900 m^{3}

Answer:

(D) 900 m^{3}

Explanation:

h = 45 m Height of a prism

B = 20 m^{2} Base area

Use the formula for the volume of a triangular prism.

V = B . h

= 20 . 45

= 900 m^{3}

The volume of a triangular prism is V = 900 m^{3}.

Question 6.

What is the total surface area of the square pyramid whose net is shown?

(A) 256 in^{2}

(B) 336 in^{2}

(C) 576 in^{2}

(D) 896 in^{2}

Answer:

(C) 576 in^{2}

Explanation:

For lateral area:

There are four triangles with base of 16 in. and height of 10 in.

Lateral area: 320 in^{2}

The base has the shape of 16 in. by 16 in. rectangle.

Base area:

1 (16 . 16) = 256 in^{2}

The total surface area is 320 + 256 = 576 in^{2}

The surface area is 576 in^{2}.

**Gridded Response**

**Volume and Surface Area Quiz Answers Module 10 Question 7.**

What is the lateral area in square meters of the prism?

Answer:

Lateral area of the prism will be the sum of areas of all the surface except base in the given figure.

Length of the rectangle = 18 in

Breadth of the rectangle = 10 m

Base of the triangle = 12 in

Height of the triangle = 8 in

We know that area of the rectangle is = Length × Breadth

We know that area of the triangle is = \(\frac{1}{2}\) × Base × Height

Area of rectangle = 18 × 10 m^{2}

= 180 m^{2}

Area of triangle = \(\frac{1}{2}\) × 12 × 8 m^{2}

= 48 m^{2}

Lateral area of prism = 2 × Area of rectangle + 2 × Area of triangle

= (2 × 180) + (2 × 48)

= 360 + 96

= 456 m^{2}

Hence, lateral area of the prism is 456 m^{2}

Steps to plot the given box are:

In 1^{st} column : mark “+” sign

In 2^{nd} column : mark “0”

In 3^{rd} coumn: mark “4”

In 4^{th} column : mark “5”

In 5^{th} column : mark “6”

In 6^{th} column : mark “0”

In 7^{th} column : mark “0”

Lateral area of the prism is 456 m^{2}