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: 12

Answer

(a.) $4520 \;cm^3$ (b.) $1380 \;cm^2$.

Work Step by Step

From the given figure we have Diameter of the cone is $d=20.2\;cm$. Radius $r=\frac{d}{2}$ $r=\frac{20.2}{2}\; cm$ $r=10.1\; cm$ Height of the cone is $h=42.3\;cm$. (a.) Formula of the volume of the cone is $V=\frac{1}{3}\pi r^2 h$ Plug all values. $V=\frac{1}{3}\pi (10.1\;cm)^2 (42.3\;cm)$ Simplify (rounded value). $V=4520 \;cm^3$ (b.) Formula for the slant height is $s=\sqrt{r^2+h^2}$ plug all values. $s=\sqrt{(10.1\;cm)^2+(42.3\;cm)^2}$ Simplify. $s=43.49 \; cm$ Formula for the lateral surface area is $A=\pi r s$ Plug all values. $A=\pi (10.1\;cm) (43.49\;cm)$ Simplify (rounded value). $A=1380 \;cm^2$.
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