the incircle of ∆ABC touches the sides AB, BC,,CA respectively at P,Q and R.If AB=14,BC=11 and CA=7 then find AP
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adi5166
In2 triangle ABC, we have
BP= BQ = 3cm
AP= AR = 4cm
( tangents drawn from an external point to the circle are equal).
So, RC = AC-AR
=11-4=7cm
Hence RC= CQ= 7 cm
Then, BC= BQ+QC
7+3=10cm
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