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.(take $π = 3.14$ )

Open in App

Solution

Here,
$r_{1} = 33 c m$
$r_{2} = 27 c m$
$l = 10 c m$

T.S.A. = $π [(r_{1} + r_{2}) l + {r_{1}}^{2} + {r_{2}}^{2}]$

= $3.14 [(33 + 27) 10 + 1089 + 729]$
= $3.14 [600 + 1089 + 729]$
= $3.14 \times 2418 = 7592.52 c m^{2}$

Suggest Corrections

Similar questions

Q. 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.

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. $[U s e π = 3.14.]$

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 33 cm and 27 cm and its slant height is 10 cm. Find its total surface area.(take π=3.14)

Here,r1=33 cmr2=27 cml=10 cmT.S.A. = π[(r1+r2)l+r12+r22] =3.14[(33+27)10+1089+729] = 3.14[600+1089+729] = 3.14×2418=7592.52 cm2

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.(take $π = 3.14$ )

Here,
$r_{1} = 33 c m$
$r_{2} = 27 c m$
$l = 10 c m$

T.S.A. = $π [(r_{1} + r_{2}) l + {r_{1}}^{2} + {r_{2}}^{2}]$

= $3.14 [(33 + 27) 10 + 1089 + 729]$
= $3.14 [600 + 1089 + 729]$
= $3.14 \times 2418 = 7592.52 c m^{2}$