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(in $c m^{2}$ ).

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Solution

The correct option is D
48

Slant height of a frustum of a cone = l =4 cm

Perimeter of its first circular end =18 cm

Perimeter of its second circular end =6 cm
Curved Surface Area of frustum of the cone = $π . (r_{1} + r_{2}) l$
$= (π . r_{1} + π . r_{2}) l = (2 π . r_{1} + 2 π . r_{2}) \frac{l}{2} = (18 + 6) 4 / 2 = 48 c m^{2}$

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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(in cm2).

The correct option is D 48 Slant height of a frustum of a cone = l =4 cm Perimeter of its first circular end =18 cm Perimeter of its second circular end =6 cm Curved Surface Area of frustum of the cone = π.(r1+r2)l = (π.r1+π.r2)l=(2π.r1+2π.r2)l2=(18+6)4/2=48cm2

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(in $c m^{2}$ ).