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In Fig. 3, the radius of incircle of $\bigtriangleup ABC$ of area $84\;cm^2$ is 4 cm and the lengths of the segments AP and BP into which side AB is divided by the point of contact P are 6 cm and 8 cm. Find the lengths of the sides AC and BC.

Answers (2)

Given : $\text{area of }\bigtriangleup ABC=84\; cm^2$

$\text{radius of circle}=4\; cm$

$AB=AP+PB$

$=6+8$

$AB=14\; cm$

Now, we know tangents drawn from external point to the circle are equal, hence,

$AP=AR=6\; cm$

$B\!P=BQ=8\; cm$

$RC=QC=x\; cm$

$\text{area of }\bigtriangleup ABC=\text{area of }(\bigtriangleup O\!AB + \bigtriangleup O\!AC + \bigtriangleup O\!BC)$

$84= \frac{1}{2} \times O\!P \times AB + \frac{1}{2} \times AC \times O\!R + \frac{1}{2} \times BC \times OQ$

$84= \frac{1}{2} \times 4 \times (AP+BP) + \frac{1}{2} \times (AR+RC) \times 4 + \frac{1}{2} \times (BQ+QC) \times 4$

$84= \frac{1}{2} \times 4 [AP+BP+AR+RC+BQ+QC]$

$\frac{84}{2}= [6+8+6+x+8+x]$

$48= 28+2x$

$2x=48-28$

$\Rightarrow 2x=20$

$\Rightarrow x=10 \; cm$

Length :

$AB=AP+BP=14\; cm$

$AC=AR+RC=6+10=16\; cm$

$BC=BQ+QC=8+10=18\; cm$

Posted by

Safeer PP

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84/2 =6+8+6+x+8+x

= 42= 28+2x

=2x=42-28

=X=14/2

=X=7

Length

AC= 6+x

=6+7=13

BC= 8+x

=8+7=15

Posted by

Anand

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