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Q. ABC is a right angled triangle wit $ \angle B=90{}^\circ , $ $ a=6\text{ }cm $ . If the radius of the circumcircle is 5 cm, then the area of $ \Delta ABC $ is

KEAMKEAM 2007

Solution:

Let D be the centre of circumcircle.

$ \therefore $ $ BD=5\text{ }cm $
In $ \Delta ABC, $ $ A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}} $
$ \Rightarrow $ $ 100=A{{B}^{2}}+B{{C}^{2}} $
$ \Rightarrow $ $ A{{B}^{2}}=64 $
$ \Rightarrow $ $ AB=8 $
$ \therefore $ Area of $ \Delta ABC=\frac{1}{2}\times AB\times BC $
$=\frac{1}{2}\times 8\times 6=24\,c{{m}^{2}} $

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