Question

11. ABCD is a quadrilateral such that ZD = 90°. A circle C(0, r) touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC = 38 cm, PB = 27 cm and radius of the circle is 10 cm, then the length of CD is

(a) 11 cm

(b) 20 cm

(c) 21 cm

(d) 15 cm




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Answer 1

Step-by-step explanation:

Since tangents to a circle is perpendicular to the radius through the point.

∴∠ORD=∠OSD=90o

It is given that ∠D=90o. Also, OR=OS. Therefore, ORDS is a square.

Since tangents from an exterior point to a circle are equal in length.

∴BP=BQ CQ=CR and, DR=DS.

Now,

BP=BQ

⇒ BQ=27                          [∵BP=27 cm  (Given)]

⇒BC−CQ=27

⇒38−CQ=27

∵ BC=38cm ⇒CQ=11 cm

⇒CR=11                          [∵CR=CQ]

⇒CD−DR=11

⇒25−DR=11                 [∵CD=25cm]

⇒DR=14 cm

But, ORDS is a square. ⇒ OR=DR=14 cm.

⇒ r=14 cm.