Q2 of 46 Page 1

In Fig. 1, the sides AB, BC and CA of a triangle ABC touch a circle at P, Q, and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC (in cm).

Since, AP and AR are tangents to the circle from the same point, by the property of circles they are equal. That is AP = AR.

AP = 4 cm [given]


Then AR = 4 cm [∵, AP = AR]


Also, RC = AC – AR


⇒ RC = 11 – 4 = 7 cm [∵, AC = 11 cm and AR = 4cm]


Now, we have PB = 3 cm (given) and RC = 7 cm.


Also, PB = BQ = 3 cm and RC = CQ = 7 cm [∵, PB and BQ are tangents ⇒ PB = BQ and RC and CQ are tangents ⇒ RC = CQ]


BC = BQ + CQ


⇒ BC = 3 + 7 = 10 cm


Hence, the length of BC is 10 cm.


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