## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Chapter 9 - Section 9.2 - Pyramids, Area, and Volume - Exercises - Page 409: 42

#### Answer

$V = 48~in^3$

#### Work Step by Step

We can find the length $L$ of each triangular section: $L^2 = (\sqrt{34})^2-(3^2)$ $L = \sqrt{(\sqrt{34})^2-(3^2)}$ $L = \sqrt{34-9}$ $L = 5~in$ We can find the altitude when each triangular section is folded up: $altitude = \sqrt{(5~in)^2-(3~in)^2} = 4~in$ We can find the total volume of the pyramid: $V = \frac{1}{3}~(base~area)(altitude)$ $V = \frac{1}{3}~(36~in^2)(4~in)$ $V = 48~in^3$

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