Spectrum Math Grade 7 Chapter 5 Lesson 12 Answer Key Volume: Pyramids

This handy Spectrum Math Grade 7 Answer Key Chapter 5 Lesson 5.12 Volume: Pyramids provides detailed answers for the workbook questions

Spectrum Math Grade 7 Chapter 5 Lesson 5.12 Volume: Pyramids Answers Key

Volume is the amount of space a solid figure occupies. The volume of a pyramid is calculated as \(\frac{1}{3}\) base × height. This is because a pyramid occupies \(\frac{1}{3}\) of the volume of a rectangular prism of the same height. Because the base of a square pyramid is square, B = s².

So, V = \(\frac{1}{3}\)Bh or \(\frac{1}{3}\)s²h. Volume is given in cubic units, or units³.
If s = 10 cm and h = 9 cm, what is the volume?

V = \(\frac{1}{3}\)s²h V = \(\frac{1}{3}\)10² × 9 V = \(\frac{900}{3}\) V = 300 cm³
If you do not know the height but you do know the slant height or length of a triangle, you can use the Pythagorean Theorem to find the height, a = \(\frac{1}{2}\) of the side length, b = the height of the pyramid, c = length
If s = 6 m and l = 5 m, what is h? a² + b² = c² 3² + b² = 25 m b² = 16 b = 4 m

Find the volume of each pyramid. Round answers to the nearest hundredth.

Question 1.
a.

V = _____ cm³
Answer:
V =256 cm³

Explanation:
V = \(\frac{1}{3}\)s²h
Given,
s = 8 cm, h = 12 cm
V = \(\frac{1}{3}\) 8² × 12
V = \(\frac{768}{3}\)
V = 256 cm³

b.

V = _____ ft³
Answer:
V = 825 ft³

Explanation:
V = \(\frac{1}{3}\)s²h
Given,
s = 15 ft, h = 11 ft
V = \(\frac{1}{3}\) 15² × 11
V = \(\frac{2475}{3}\)
V = 825 ft³

c.

V = ___ in.³
Answer:
V = 168.75 ft³

Explanation:
V = \(\frac{1}{3}\)s²h
Given,
s = 7.5 ft, h = 9 ft.
V = \(\frac{1}{3}\) 7.5² × 9
V = \(\frac{506.25}{3}\)
V = 168.75 ft³

Question 2.
a.

V = ____ m³
Answer:
V = 546.88 m³

Explanation:
V = \(\frac{1}{3}\)s²h
Given,
s = 12.5 m, h = 10.5 m
V = \(\frac{1}{3}\) 12.5² × 10.5
V = \(\frac{1640.625}{3}\)
V = 546.88 m³

b.

V = ____ cm³
Answer:
V = 1,296 cm³

Explanation:
V = \(\frac{1}{3}\)s²h
Given,
s = 18 cm, l = 15 cm
According to the Pythagorean Theorem to find the height,
a = \(\frac{1}{2}\) of the side length,
a = \(\frac{18}{2}\) = 9 cm
b = the height of the pyramid, c = length
a² + b² = c²
9² + b² = 15²
81 + b² = 225
b² = 225 – 81
b² = 144
b = 12 cm
V = \(\frac{1}{3}\) 18² × 12
V = \(\frac{3,888}{3}\)
V = 1,296 cm³

c.

V = ____ in.³
Answer:
V = 400 in³

Explanation:
V = \(\frac{1}{3}\)s²h
Given,
s = 10 in, l = 13 in
According to the Pythagorean Theorem to find the height,
a = \(\frac{1}{2}\) of the side length,
a = \(\frac{10}{2}\) = 5 in
b = the height of the pyramid, c = length
a² + b² = c²
5² + b² = 13²
25 + b² = 169
b² = 169 – 25
b² = 144
b = 12 in
V = \(\frac{1}{3}\) 10² × 12
V = \(\frac{1,200}{3}\)
V = 400 in³

Question 3.
a.

V = ____ ft.³
Answer:
V = 122.5 ft³

Explanation:
V = \(\frac{1}{3}\)s²h
Given,
s = 7 ft, h = 7.5 ft
V = \(\frac{1}{3}\) 7² × 7.5
V = \(\frac{367.5}{3}\)
V = 122.5 ft³

b.

V = ____ m³
Answer:
V = 0.72 m³

Explanation:
V = \(\frac{1}{3}\)s²h
Given,
s = 1.2 m, h = 1.5 m
V = \(\frac{1}{3}\) 1.2² × 1.5
V = \(\frac{2.16}{3}\)
V = 0.72 m³

c.

V = ____ cm³
Answer:
V = 11,200 cm³

Explanation:
Given,
s = 40 cm, l = 29 cm
According to the Pythagorean Theorem to find the height,
a = \(\frac{1}{2}\) of the side length,
a = \(\frac{40}{2}\) = 20 cm
b = the height of the pyramid, c = length
a² + b² = c²
20² + b² = 29²
400 + b² = 841
b² = 841 – 400
b² = 441
b = 21 cm
V = \(\frac{1}{3}\)s²h
Given,
s = 40 cm, h = 21 cm
V = \(\frac{1}{3}\) 40² × 21
V = \(\frac{33,600}{3}\)
V = 11,200 cm³