The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum be 8 cm, find the whole surface area of the frustum. (Use = 3.14)
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Let R and r be the radii of the circular ends of the frustum. (R> r)
2R = 207.24
R = 207.24/ (2 X 3.14)
R = 33 cm
2r = 169.56 cm
r = 169.56 / (2 X 3.14)
r = 27 cm
1² = h² + (R-r)²
= 8² + (33-27)²
l = 10 cm
Whole surface area of the frustum
= (R2 + r2 + (R+r)l)
= 3.14
((33)2 + (27)2 + (33+27)10)
= 3.14 (1089 +729 + 600)
= 3.14 X 2418 cm²
= 7592.52 cm²
2R = 207.24
R = 207.24/ (2 X 3.14)
R = 33 cm
2r = 169.56 cm
r = 169.56 / (2 X 3.14)
r = 27 cm
1² = h² + (R-r)²
= 8² + (33-27)²
l = 10 cm
Whole surface area of the frustum
= (R2 + r2 + (R+r)l)
= 3.14
((33)2 + (27)2 + (33+27)10)
= 3.14 (1089 +729 + 600)
= 3.14 X 2418 cm²
= 7592.52 cm²
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