Pls help

Asked by  | 26th 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  | 22nd Jan, 2009, 02:55: PM

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