$ABC is a right angled triangle with ∠ B = 90^{0}, a = 6 cm . If the radius of the circumcircle is 5 cm, then the area of Δ A B C i s$

25 cm²

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30 cm²

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36 cm²

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24 cm²

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Solution

The correct option is D
24 cm²

$\text{Let D be the centre of circumcircle.}\\

\therefore BD=5~cm\\
\text{In}~\Delta ABC,\\
AC^2=AB^2+BC^2\\
{\Rightarrow}100=AB^2+36\\
{\Rightarrow}AB^2=64{\Rightarrow}AB=8\\
\therefore \text{Area of }~\Delta ABC=\frac{1}{2}\times AB\times BC\\
=\frac{1}{2}\times8\times6=24cm^2$

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ABC is a right angled triangle with ∠B=900,a=6cm. If the radius of the circumcircle is 5 cm, then the area of ΔABC is

$ABC is a right angled triangle with ∠ B = 90^{0}, a = 6 cm . If the radius of the circumcircle is 5 cm, then the area of Δ A B C i s$