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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