Chapter 12.4, Problem 19PPS

To determine

The height of the cone whose volume and diameter are $306.464f{t}^3,8ft$ respectively.

Answer to Problem 19PPS

The height of the cone is $57.46ft$

Explanation of Solution

Given:

  $\begin{array}{l}V=306.464f{t}^3\\ d=8ft\end{array}$

Concept Used:

The volume of the cone is given by the following expression:

  $V=\pi r^2\frac{h}{3}$

Where,

Radius of cone: $r$

Height of cone: $h$

Calculation:

The volume of the cone is given by the following expression:

  $V=\pi r^2\frac{h}{3}$

Put the given data in eq.

  $\begin{array}{l}V=\pi r^2\frac{h}{3}\\ 306.464=\pi {\left(\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2\frac{h}{3}\\ 306.464=\pi {\left(\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2\frac{h}{3}\\ h=\frac{306.464\times 3}{16}\\ h=57.46ft\end{array}$

Conclusion:

Hence, the height of the cone is $57.46ft$