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Q. The circum-radius of the triangle whose sides are $13, 12$ and $5$ is

KCETKCET 2005Straight Lines

Solution:

Let sides be $a=13, b=12, c=5$.
Now, $ a^{2}=b^{2}+c^{2}$
$\Rightarrow \,(13)^{2}=(12)^{2}+5^{2} $
$\Rightarrow 169=169$
$\Rightarrow \angle A=90^{\circ}$
We know, $R=\frac{a}{2 \sin A}$
$=\frac{13}{2 \cdot \sin 90^{\circ}}=\frac{13}{2}$