Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 12 - Section 12.9 - Cylinders - Exercise - Page 434: 6

Answer

$90,1\bar{0}0\ ft^3$

Work Step by Step

The volume of a pyramid is given by $$ V=\frac{1}{3} Bh=\frac{1}{3} \times 109.4\times 83.2 \times 29.7= 90110.592\ ft^3\approx 90,1\bar{0}0\ ft^3 $$ where $B$ is the area of the base (parallelogram with area $A=bh$) and $h$ is the height of the pyramid.
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