Question

The radii of 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.

Answer 1

SOLUTION:-

GIVEN,
R= 33CM , r= 33cm , hieght (l) = 10cm

FORMULA IS ,

● slant height of frustum, l = √h²+(R - r)²

10 = √h²+(33 - 27)²
10² = h² + 36
h² = 100 - 36
h² = 64
h = 8 cm,

● volume of the frustum
= πh/3(R²+r²+Rr)
= (22/7)×(8/3){33²+27²+(33×27)}
= (22/7)×(8/3)×2709
= 476784/21
[Volume = 22704 cm³]

● Total surface area of the frustum
= π{R² + r² +l(R + r)}
= 22/7{33² + 27² + 10(33 + 27)}
= 22/7×(1089 + 729 + 600)
= 22/7×2418
= 53196/7
{Total surface area = 7599.43 cm^2}ans

Answer 2

Given :-

Radius of the bottom of frustum ( R ) :- 33cm

Radius of the top of the frustum ( r ) = 27cm

Slant height of the frustum ( l ) :- 10cm

• Total Surface are of frustum =

π [ R² + r² + l ( R + r ) ] sq.units

=> T.S.A = π [ ( 33 )² + ( 27 )² + 10 × 60 ]

=> 22/7 [ 1089 + 729 + 600 ]

=> 7599.428 cm²