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Q.
The sides of a right angled triangle are integers. The length of one of the sides is $12$. The largest possible radius of the incircle of such a triangle is
KVPYKVPY 2015
Solution:
In right angle $\Delta\,ABC$
Let $AB=12$
$BC=x +r$
$AC=12+x-r$
$AC^{2}=AB^{2}+BC^{2}$
$\Rightarrow (12+x-r)^{2}=(12)^{2}+(x+r)^{2}$
$\Rightarrow 144+24(x-r)+(x-r)^{2}$
$=144+(x+r)^{2}$
$\Rightarrow 24(x-r)=(x +r)^{2}-(x-r)^{2}$
$\Rightarrow 24(x-r)=4xr$
$\Rightarrow 6(x-r)=xr$
$\Rightarrow r=\frac{6x}{6+x}$
At $x=6,\, r=3$
At $x=12,\, r=4$
At $x=30,\, r=5$
$\therefore $ Maximum radius $= 5$
