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Question 84
A circle is inscribed in a quadrilateral ABCD touching AB, BC, CD and AD at the points P, Q, R and S, respectively, and $$\angle B = 90^\circ$$. If AD = 24 cm, AB = 27 cm and DR = 6 cm, then what is the circumference of the circle?
Solution
DR = DS = 6 cm
AS = AD - DS = 24 - 6 = 18 cm
AS = AP = 18 cm
PB = AB - AP = 27 - 18 = 9 cm
PB = r = 9 cm
Circumference of the circle = $$2\pi r = 2 \times 9 \pi = 18 \pi$$
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