A Shuttlecock used for playing badminton has the shape of a frustum of a cone mounted on a hemishphre as shown in the adjoining figure. the external diameters of the frustum are 5 cm and 2 cm, the height of the entire shuttlecock is 7 cm . find its external surgace area.
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Given: r1 = 1 cm, r2 = 2.5 cm, h = 6 cm
l = √h² + (r2-r1)²
= √(6)² + (2.5-1)² cm
= √36 + 2.25 cm
= √38.25 cm
= 6.18cm
So,
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
l = √h² + (r2-r1)²
= √(6)² + (2.5-1)² cm
= √36 + 2.25 cm
= √38.25 cm
= 6.18cm
So,
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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87
R1 = 1 cm, r2 = 2.5 cm, h = 6cm
L = underoot h square + (r2-r1) whole suqare = underoot (6) square + (2.5-1) square cm
= underoot 36 + 2.25 cm = underoot 38.25 cm = 6.18cm
External surface area = CSA of frustum + CSA of hemisphere
= π (r1+r2) l + 2π r 1 square
= π [(1+2.5) 6.18 + 2 x 1 square] cm square
= 22/7 [3.5x6.18+2] cm square
=22/7 [21.63+2] cm square = 22/7 x 23.63 cm square
= 519.86/7 cm square
= 74.26 cm square
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