Q.Two circles with centers A and B have a common tangent PO of length 16 cm, and the length of PB is 10 cm; the area of triangle \(\triangle\)POB will be— (a) 24 square cm (b) 48 square cm (c) 16 square cm (d) 12 square cm
Answer: B
A and B are joined, intersecting OP at point M. In the right-angled triangle \(\triangle\)BMP: BM\(^2\) = BP\(^2\) − MP\(^2\) = 10\(^2\) − \((\frac{16}{2})^2\) = 100 − 64 = 36 ∴ BM = \(\sqrt{36}\) = 6 cm ∴ Area of triangle \(\triangle\)POB = \(\cfrac{1}{2} \times\) OP \(\times\) BM = \(\cfrac{1}{2} \times 16 \times 6\) square cm = 48 square cm.
Similar Questions