Sum

The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its curved surface area

Advertisement Remove all ads

#### Solution

Here, r_{1} = 14 cm, r_{2} = 6 cm and h = 6 cm.

Slant height of the frustum, l = \[\sqrt{h^2 + \left( r_2 - r_1 \right)^2} = \sqrt{6^2 + \left( 14 - 6 \right)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}\]= 10 cm

Curved surface area of frustrum

\[= \pi\left( r_1 + r_2 \right)l\]

\[ = 3 . 14 \times \left( 14 + 6 \right) \times 10\]

\[ = 3 . 14 \times 20 \times 10\]

\[ = 628 {cm}^2\]

∴ the curved surface area of frustum is 628 sq. cm

Concept: Frustum of a Cone

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads