#### Question

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. find the curved surface area of the frustum.

#### Solution 1

Perimeter of upper circular end of frustum = 18 cm

2π*r*_{1} =18

r_{1 =} 9/π

Perimeter of lower end of frustum = 6 cm

2π*r*_{2} = 6

r_{2 =} 3/π

Slant height (*l*) of frustum = 4 cm

CSA of frustum = π (*r*_{1} + *r*_{2}) *l*

`=pi(9/pi+3/pi)4`

= 12x4

= 48 cm^{2}

Therefore, the curved surface area of the frustum is 48 cm^{2}.

#### Solution 2

We have,

Perimeter of upper end, C=18cm,

Perimeter of lower end, c=6 cm and

Slant Height, l=4 cm

Let the radius of upper end be R and the radius of lower end be r.

As, C = 18 cm

⇒2πR = 18

`=> R = 18/(2pi)`

`=> R = 9/pi`cm

Similary c = 6 cm

`=> r = 6/(2pi)`

`=> r = 3/pi` cm

Curved surface area of the frustum=π(R + r)l

`=pi(9/pi + 3/pi) xx 4`

`= pi xx 12/pi xx 4`

`= 48 cm^2`

Hence, the curved surface area of the frustum is 48 cm^{2}