Question

In the following figure, $\overline{KL}$ is a chord of the circle centered at $O$, with $\overline{KL}\perp \overline{MO}$. If $KL=12$ and $MO=6$, what is the area of the circle?

• A. $36\pi$
• B. $48\pi$
• C. $72\pi$
• D. $96\pi$
• E. $144\pi$

Answer 1

The correct option is C $72\pi$

Given, $KL=12$
Since $M$ is the midpoint $KM=6$ and $ML=6$
$\therefore OM=6$
If we join line $OL$ we will get a triangle $OML$
In $△OML$, $OM=$ perpendicular, $ML=$base and $OL=$ hypotenuse as well as radius of circle
As per Pythagoras theorem,
$h^2=p^2+b^2$
$\Rightarrow h=\sqrt{p^2+b^2}$
$\Rightarrow h=\sqrt{6^2+6^2}$
$\Rightarrow h=\sqrt{72}$
Hence, we also get the radius of circle $=$ $\sqrt{72}$
Area of circle is $\pi r^2$
$\Rightarrow \pi {\left(\sqrt{72}\right)}^2$
$\Rightarrow 72\pi$