Elementary Technical Mathematics

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

Chapter 1 - Section 1.3 - Area and Volume - Exercise - Page 18: 24

Answer

$5728\;ft^3$.

Work Step by Step

Divide the figure into three sections. First rectangular section:- Height $=20\;ft$ Length $=8\;ft$ Width $=8\;ft$ Volume $=$ Height $\times$ Length $\times$ Width $\Rightarrow V_1=(20\;ft)\times(8\;ft)\times(8\;ft) $ $\Rightarrow V_1=1280\;ft^3 $ Second rectangular section:- Height $=20-12=8\;ft$ Length $=60-20-8=32\;ft$ Width $=8\;ft$ Volume $=$ Height $\times$ Length $\times$ Width $\Rightarrow V_2=(8\;ft)\times(32\;ft)\times(8\;ft) $ $\Rightarrow V_2=2048\;ft^3 $ Third rectangular section:- Height $=15\;ft$ Length $=20\;ft$ Width $=8\;ft$ Volume $=$ Height $\times$ Length $\times$ Width $\Rightarrow V_3=(15\;ft)\times(20\;ft)\times(8\;ft) $ $\Rightarrow V_3=2400\;ft^3 $ Total volume $\Rightarrow V=V_1+V_2+V_3$ $\Rightarrow V=1280\;ft^3+2048\;ft^3+2400\;ft^3$ $\Rightarrow V=5728\;ft^3$.
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