Answer to Question #149315 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

Let $r_1,r_2$ and $h$ be the radius of the upper base,lower base and height of the frustum of cone.

Then volume of the frustum of cone is

$=(1/3)\pi h(r_1^2+r_1r_2+r_2^2)$ .........(1)

According to the problem, diameter of one base is twice that of other, therefore radius of one base is twice that of other also.

So,let us take $r_1=x$ m

$r_2=2x$ m

and given $h=3$ m

Also given that ,slant height of the of the frustum is inclined to the greater base at an angle $45^○$ .

Then we have, $h/(r_2-r_1)=tan45^○$

$\implies h/(2x-x)=1$

$\implies h/x=1$

$\implies x=h$

$\implies x=3$

Therefore, $r_1= 3$ m

$r_2=6$ m

So required volume of the frustum is

$=(1/3)\pi h(r_1^2+r_1r_2+r_2^2)$ $=(1/3)\pi ×3×[3^2+(3×6)+6^2]$$m^3$

$=(1/3)\pi ×3×63$ $m^3$

$=63\pi$ $m^3$