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A shuttlecock used for playing badminton has the shape of a frustum of a Cone mounted on a hemisphere. The external diameters of the frustum are 5 cm and 2 cm, and the height of the entire shuttlecock is 7cm. Find the external surface area.(Ans: 74.26cm2)
Ans: r1 = radius of lower end of frustum = 1 cm
r2 = radius of upper end = 2.5 cm
l = = 6.18 cm
External surface area of shuttlecock = π (r1 + r2) l + 2π r2 1
On substituting we get, = 74.26 cm2
ABCD is a parallelogram which AB and CD are divides by P and Q. Such that AP:PB=3:2 and CQ:QD=4:1. If PQ and AC are meet at R, show that AR=3/7AC.
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