Question

The radii of the ends of a frustum of a cone 45cm high are 28 cm and 7cm. Find its volume, the curved surface area, and the total surface area $(take\pi =\frac{22}{7})$.

Answer 1

The frustum can be viewed as a difference of two right circular cones OAB and OCD . let the height of the cone OAB be $h_1$ and its slant height $l_1$

i.e., OP=$h_1$ and OA=OB=$l_1$.

Let h2 be the height of cone OCD and l2 its Slant height .

We have $r_1$=28cm , $r_2$=7cm and the height of frustum (h) =45cm

$h_1=45+h_2$ ----(i)

Since the triangles OPB and OQD are similar , we have

$\frac{h_1}{h_2}=\frac{28}{7}=\frac{4}{1}-----\left(ii\right)$

From (i) and (ii) $h_2$=15cm and $h_1$= 60 cm.

now, the volume of the frustum = volume of the cone OAB – volume of the cone OCD

$=\frac{1}{3}\pi (r_1{)}^2{h}_1-\frac{1}{3}\pi (r_2{)}^2{h}_2$
$=\left[\frac{1}{3}\right(\frac{22}{7}\left)\right(28{)}^2\left(60\right)-\frac{1}{3}\left(\frac{22}{7}\right)\left(7{)}^2\right(15\left)\right]$
$=48510c{m}^2$

The respective slant height $l_2$and $l_1$ of the cones OCD and OAB are given by

$l_2=\sqrt{(7{)}^2+(15{)}^2}=16.55cm$
$l_1=\sqrt{(28{)}^2+(60{)}^2}=4\sqrt{(7{)}^2+(15{)}^2}=66.20cm$

Thus, the curved surface area of the frustum = $\pi r_1{l}_1-\pi r_2{l}_2$

$=\frac{22}{7}\times 7\times 16.55(16-1)$
$=5461.5c{m}^2$

Now, the total surface of the frustum = curved surface area + $\pi (r_1{)}^2+\pi (r_2{)}^2$

$=5461.5+\frac{22}{7}\left(28\right)\left(28\right)+\frac{22}{7}\left(7\right)\left(7\right)$

= 5461.5+2464+154

$=8079.5c{m}^2$