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Question
A right-angled triangle PQR where ∠Q = 90° is rotated about QR and PQ. If QR = 16 cm and PR = 20 cm, compare the curved surface areas of the right circular cones so formed by the triangle
Solution
In the Right Triangle
QP2 = PR2 – QR2
= 202 – 162
= 400 – 256
= 144
QP = `sqrt(144)` = 12 cm
When PQ is rotated r = 12, l = 20
C.S.A of the cone = πrl sq. units
= π × 12 × 20 cm2
= 240π cm2
When QR is rotated r = 16, l = 20
C.S.A of the cone = nrl sq. units
= π × 16 × 20
= 320π cm2
C.S.A. of a cone when rotated about QR is larger.
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