Question

hiii can you please tell me the attachment fast .....
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Answer 1

Tangents drawn from one external point to the circle are equal

So,AR=AQ ,CP=CQ , BP=BR

Now, AR+AQ+CP+CQ+BP+BR= perimeter of Triangle ABC

2(AR+BP+CQ)=perimeter ofABC

AR+BP+CQ=1/2 of perimeter of ABC

Similarly, AQ+BR+CP=1/2 of perimeter of ABC

So, AQ+BR+CP=AR+BP+CQ=1/2 Of perimeter of triangle ABC

HOPE..IT HELPS U

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Answer 2

$<b>Hope it helps...$

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