Answer to Question #149698 in Trigonometry for yui

Question

PROBLEM 1. The height of the frustum of a cone is 3 m, the diameter of one base is twice that of the other. The slant height of the frustum is inclined to the greater base at an angle of 450. Find the volume of the frustum.
FINAL ANSWER: _______________________________________

Answer 1

The sketch of the conical frustum is as shown in the figure below:

In the figure,

$tan(45^0)=\frac{h}{x}$

$1=\frac{h}{x}$

$x=h=3$

Given, $r_{2}=2r_{1}$ , therefore,

$r_{1}+x=2r_{1}$

$r_{1}=x=3$

So, the radius of lower base is $r_{1}=3$ and the radius of greater base is $r_{2}=6$

Use the formula for volume of frustum $V=\frac{1}{3}\pi h (r_{2}^{2}+r_{1}^{2}+r_{1}r_{2})$ to find the volume as,

$V=\frac{1}{3}\pi (3) (6^2+3^2+(6)(3))$

$=\frac{1}{3}\pi (3) (36+9+18)$

$=\frac{1}{3}\pi (3) (63)$

$=63\pi$

Therefore, the volume of the frustum is $V=63\pi$ $m^3$