Question

A is a point at a distance of 13 centre o of a circle of radius 5cm AP and AQ are tangents to the circle at P and Q .IF tangent BC is drawn at R lie on minor Arc PQ to interested AP AT B AND AQ AT C find the perimeter of triangle ABC​

Answer 1

Solution:-

BP=BR & CR=CQ

Length of tangents drawn from an external points to circle are equal.

Perimeter of triangle ABC = AB+BR+RC+CA = AB+BP+QC+CA

= AP+QA. [ AP=QA]

= 2AP

In triangle APO, using Pythagoras theorem:

AO^2= AP^2+ PO^2

13^2 = AP^2+5^2

169 = AP^2+25

AP = √169-25

AP = √144

AP = 12.

So, Perimeter of triangle ABC = 2AP

= 2×12

= 24.

Hope it helps you..

Answer 2

Refer to the Aatchement!!!!

Hope it's help you