Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 9 - Section 9.2 - Pyramids, Area, and Volume - Exercises - Page 407: 24c


$V=363.623$ m$^3$

Work Step by Step

The volume of a pyramid is: $V=\frac{1}{3}Bh$ So, $V=\frac{1}{3}100h$ To find h, substitute the slant height and half of the base's length into: $a^2+b^2=c^2$ In this case, we are solving for the leg of the triangle given that the hypotenuse is 12, and the other leg is 5. $a^2+5^2=12^2$ $a=10.90$ $V=\frac{1}{3}100(10.90)$ $V=363.623$ m$^3$
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