Question

A right angled triangle PQR where Q= 90° is rotated about QR and PQ. If QR=16
and PR=20 cm, compare the curved surface areas of the right circular
cones so formed by the triangle.​

Answer 1

Answer:

The curved surface area of cone formed by rotating the side QR is 251.3 sq cm more than the curved surface area of the cone formed by rotating the side PQ.

Step-by-step explanation:

Triangle PQR is right angle triangle with angle Q= 90°

QR= 16 cm.

PR = 20cm.

PQ=√((20) ^2) -((16) ^2)) =12 cm.

If the triangle is rotated about QR, then curved surface area of cone so formed is

= π×16×20=1005.3sq cm.

If the triangle is rotated about PQ, then curved surface area of cone so formed is

= π×12×20=753.98 sq cm.

Difference in the curved surface area is = (1005.3 - 753.98) sq cm. = 251.3 sq cm.

Answer 2

Answer:

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