CBSE Class 10 Answered
Pls help
Asked by | 26 Aug, 2008, 10:09: AM
Expert Answer
We know that the tangent segments drawn from an external point to a circle are equal in length.
So, in this case,
AQ=AR
BP=BQ
CP=CR
Now,
Perimeter(triangle ABC)
=AB+BC+CA
=AB+(BP+PC)+CA
=AB+(BQ+CR)+CA ..(as BP=BQ and PC=CR)
=(AB+BQ)+(CR+CA)
=AQ+AR
=2AQ( as AQ=AR)
Thus,
Perimeter(triangle ABC)=2AQ
So,
AQ=1/2(Perimeter triangle ABC)
Hence proved.
Answered by | 22 Jan, 2009, 02:55: PM
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