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

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Solution

Let radii of two end be $r_{1}$ and $r_{2}$ . Then,
$2 π r_{1} = 18 ⟹ π r_{1} = 9$
And, $2 π r_{2} = 6$
$π r_{2} = 3$
Therefore,
$π (r_{1} + r_{2}) = 9 + 3 = 12$
Curved surface area = $π (r_{1} + r_{2}) l = 12 \times 4 = 48$ $c m^{2}$

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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.

Q. The slant height of the frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface of the frustum.

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

Let radii of two end be r1 and r2. Then,2πr1=18⟹πr1=9And, 2πr2=6πr2=3Therefore,π(r1+r2)=9+3=12Curved surface area = π(r1+r2)l=12×4=48 cm2

Let radii of two end be $r_{1}$ and $r_{2}$ . Then,
$2 π r_{1} = 18 ⟹ π r_{1} = 9$
And, $2 π r_{2} = 6$
$π r_{2} = 3$
Therefore,
$π (r_{1} + r_{2}) = 9 + 3 = 12$
Curved surface area = $π (r_{1} + r_{2}) l = 12 \times 4 = 48$ $c m^{2}$