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# $ABC$ is a triangle with $\angle A=30^{\large\circ}$ $BC=10$cm.The area of the circumcircle of the $\Delta$le is?

$\begin{array}{1 1}(a)\;100\pi sqcm&(b)\;5sqcm\\(c)\;25sqcm&(d)\;\large\frac{100\pi}{3}\normalsize sqcm\end{array}$

Can you answer this question?

In $\Delta$ABC $\angle A=30^{\large\circ}$
$BC=10$ cm
Let $O$ be the centre of the circle.
$\Rightarrow\:\angle BOC=60^{\large\circ}$
$OB$ and $OC\Rightarrow radius$
$\angle OBC=\angle OCB =60^{\large\circ}$
$\Rightarrow \Delta OBC$ is an equilateral $\Delta$le.
Radius of circle is $OB=OC=BC=10cm$
Area of the circumcircle is $\pi r^2$
$\Rightarrow \pi(10)^2$
$\Rightarrow 100\pi$sq.cm
Hence (a) is the correct answer.
answered Nov 18, 2013
edited Mar 25, 2014