Consider sector of circle $MAB$. $∠AMB = 120◦$.
A circle $S$ touches side $AM$, side $MB$ and arc $AB$ as shown in the figure.
Area of circle $S$ is $75π/(7 + 4√3)$ . Find $4√3$ times the area of $△AMB$.
Here , I know the area of circle , so radius can be calculated. For triangle $AMB$ , it's area is equal to $1/2*AM^2*\sin 120$(degrees). So I just require the length of $AM$, that is the radius of sector.
$AM$ and $MB$ are tangents to the circle so that may be of some help? But I'm stuck here .
Any hints are apreciated . (This is not class-homework , I'm solving sample questions for a competitive exam )
Best Answer
$$A_S=\frac{75\pi}{7+4\sqrt3}\implies R=5\sqrt{\frac3{7+4\sqrt3}}$$
Let now $\;K\;$ be the intersection point between the circle (with center $\;O\;$ ,say) and the radius $\;AM\;$, and form the $\;30-60-90\;$ straight-angle triangle $\;KOM\;$ ,so that
$$R=KO=\color{red}{\frac{\sqrt3}2}\,MO\implies MO=\color{red}{10}\sqrt{\frac1{7+4\sqrt3}}$$
so finally, the radius $\;r\;$ of the circular sector is
$$r=MO+R=\frac{5(\color{red}2+\sqrt3)}{\sqrt{7+4\sqrt3}}=\color{red}5\;\;\left(\text{because}\;\;(2+\sqrt3)^2=7+4\sqrt3\ldots\right)$$