 QUESTION

# The slant height of the frustum of a cone is 4cm and the perimeter of its circular ends are 18cm and 6 cm. Find the area of its whole surface and volume.

Hint: Find Radius of top and bottom using given data. Then, find out height using the formula for height. Since we have all the required data after mentioned calculations, with help of TSA and Volume formulas, we get the answer.

Now given that perimeter of the top circular region is 6cm.

So $2\pi r$ = 6 hence r(top) = $\dfrac{3}{\pi }$-(Eq 1)

Now, the perimeter of the bottom circular region is 18cm.

So $2\pi R$ = 18 hence $R\left( {bottom} \right) = \dfrac{9}{\pi }$- (Eq 2)

Slant height of the cone is 4cm.

Height of frustum cone is h = $\sqrt {{l^2} - {{\left( {R - r} \right)}^2}}$

So substituting values h = $\sqrt {{4^2} - {{\left( {\dfrac{9}{\pi } - \dfrac{3}{\pi }} \right)}^2}}$

h = $\sqrt {16 - {{\left( {\dfrac{6}{\pi }} \right)}^2}}$

h = $\sqrt {16 - \left( {\dfrac{{36}}{{{\pi ^2}}}} \right)}$

h = $\sqrt {12.35224}$

So our h = 3.5146 cm

Now curved surface area of frustum of cone = $\pi \left( {R + r} \right)l$

So substituting values here we get,

= $\pi \left( {\dfrac{9}{\pi } + \dfrac{3}{\pi }} \right)4$

= $\dfrac{{12}}{\pi }\left( 4 \right)\left( \pi \right)$

So, it comes out to be $48{cm^2}$

T.S.A of frustum of cone = $48{cm^2}$

Now, total volume of a frustum of a cone = $\dfrac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr} \right)$

Using the above we have,

V =$\dfrac{{\pi (3.5146)}}{3}\left( {{{\dfrac{9}{{{\pi ^2}}}}^2} + {{\dfrac{3}{{{\pi ^2}}}}^2} + \dfrac{{9(3)}}{{{\pi ^2}}}} \right)$

V =$\dfrac{{\pi (3.5146)}}{3}\left( {\dfrac{{117}}{{{\pi ^2}}}} \right)$

V = $43.65c{m^3}$

Note: Such problems are always based on standard formulas. Having a good grasp of formula can surely help while practising such types of questions more