PQR is a right angled triangle with $P Q = 12 c m$ and $Q R = 5 c m$ . A circle with centre O and radius x is inscribed in $Δ P Q R$ . Find the value of x.

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Solution

In $Δ P Q R$
$(P R)^{2} = (P Q)^{2} + (Q R)^{2}$
$(P R)^{2} = (12)^{2} + (5)^{2}$
$(P R)^{2} = 169$
$P R = 13$
Now AOBQ is a square
So, $Q B = x$
Then, $B R = 5 - x$
Similarly $A Q = x$
Then $A P = 12 - x$
Also, $C R = B R$
$C R = 5 - x$
[ $∵$ Length of tangents drawn from external point are equal]
And, $C P = A P = 12 - x$
[ $∵$ Length of tangents drawn from external point are equal]
$P R = P C + C R$
$13 = 5 - x + 12 - x$
$2 x = 4 \Rightarrow x = 2$ cm.

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