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dby A shuttle cock used for playing badminton has theshape of a frustum of a cone mounted on ahemisphere the external diameters of the frustumare 5 cm and 2 cm and the height is 7 cm. Findexternal surface area of the cock. |
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Answer» r1= 1 cm,r2= 2.5 cm,h= 6 cml = √h² + (r2-r1)²= √(6)² + (2.5-1)² cm= √36 + 2.25 cm= √38.25 cm= 6.18cmSo,External Surface area = (curved surface area of frustum) + (curved surface area of hemisphere)= 22/7(r1+r2)l +2.22/7.r1= 22/7 [(1+2.5)6.18 + 2×(1)²] cm²= 22/7 (3.5 × 6.18 +2) cm²= 22/7 (21.63+2) cm²= 22/7 × 23.63 cm²= 519.86/7 cm²= 74.26 cm² |
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