The radii of the circular ends of a solid frustum of a cone are $33 c m$ and $27 c m$ and its slant height is $10 c m$ . Find its total surface area.

Open in App

Solution

TSA of frustum of cones $= π (R_{1} + R_{2}) l + π (R_{1}^{2} + R_{2}^{2})$
$= \frac{22}{7} [(33 + 27) \times 10 + (33^{2} + 27^{2})]$
$= 7599.42 {c m}^{2}$

Suggest Corrections

Similar questions

Q. The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm, and its slant height is 10 cm. Find its capacity and total surface area.

The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm, and its slant height is 10 cm. Find its capacity and total surface area. $[T a k e π = 22 / 7.]$

Q. The perimeters of the ends of a frustum of a right circular cone are

44 c m

and

33 c m .

If the height of the frustum is

16 c m,

find its volume, the slant surface, and the total surface.

Join BYJU'S Learning Program

The radii of the circular ends of a solid frustum of a cone are 33cm and 27cm and its slant height is 10cm. Find its total surface area.

TSA of frustum of cones =π(R1+R2)l+π(R21+R22) =227[(33+27)×10+(332+272)] =7599.42cm2

The radii of the circular ends of a solid frustum of a cone are $33 c m$ and $27 c m$ and its slant height is $10 c m$ . Find its total surface area.

TSA of frustum of cones $= π (R_{1} + R_{2}) l + π (R_{1}^{2} + R_{2}^{2})$
$= \frac{22}{7} [(33 + 27) \times 10 + (33^{2} + 27^{2})]$
$= 7599.42 {c m}^{2}$