Q. 64.9( 11 Votes )
A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere. The external diameter of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.
Answer :
Given.
The external diameter of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm
Formula used/Theory.
CSA of frustum = π(R + r) × L
CSA of hemisphere = 2πr2
Radii of frustum is 
and 
= 2.5cm and 1cm
As in hemisphere radius and height are same
Height of frustum = height of cock – Radius
= 7 cm – 1 cm
= 6 cm
In frustum
L2 = height2 + 
L2 = (6 cm)2 + [2.5 cm – 1 cm]2
L2 = 36 cm2 + 2.25 cm2 = 38.25 cm2
L = √(38.25 cm2) = 6.18 cm
CSA of hemisphere = 2 × 3.14 × (1 cm)2
= 6.28 cm2
CSA of frustum = 3.14 × [2.5cm + 1cm] × 6.18cm
= 3.14 × 3.5cm × 6.18cm
= 67.92 cm2
External surface area = CSA of hemisphere + CSA of frustum
= 6.28 cm2 + 67.92 cm2
= 74.2 cm2
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Gujarat Board Mathematics