A circle is inscribed in sector of another bigger circle.Given A(circle) find the A(triangle formed by the center and the endpoints of the sector).

Question

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 )

Answer 1

$$A_S=\frac{75\pi}{7+4\sqrt3}\implies R=5\sqrt{\frac3{7+4\sqrt3}}$$

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

$$r=MO+R=\frac{5(\color{red}2+\sqrt3)}{\sqrt{7+4\sqrt3}}=\color{red}5\;\;\left(\text{because}\;\;(2+\sqrt3)^2=7+4\sqrt3\ldots\right)$$

Answer 2

Denote the centre of the smaller (inscribed) circle as $C$, and let the circle touch tangent $AM$ at point $X$, as illustrated in the figure below.

We are given the area of the inscribed circle, which is $\frac{75\pi}{7+4\sqrt{3}}$. Thus we have $$\frac{75\pi}{7+4\sqrt{3}}=\pi R^2 \Rightarrow R= 5\sqrt{\frac{3}{7+4\sqrt{3}}}$$

Now the line $CM$ can be calculated using the right angled triangle $\triangle MXC$, where $\angle XMC=\frac{\pi}{3}$, $\angle CXM=\frac{\pi}{2}$ ($XM$ is tangent of circle), so that $\angle XCM=\frac{\pi}{6}$, and $CX=R$.

$$CM=\frac{CX}{\sin \frac{\pi}{3}}=\frac{2R}{\sqrt{3}}$$

The radius of the big circle which the sector is part of is $$AM=R+CM=\left(1+\frac{2}{\sqrt{3}}\right)R=\left(\frac{\sqrt{3}+2}{\sqrt{3}}\right)R$$

Substituting in the value of $R$, we have

$$AM=5\sqrt{\frac{3}{7+4\sqrt{3}}}\left(\frac{\sqrt{3}+2}{\sqrt{3}}\right)=5\sqrt{\frac{3(2+\sqrt{3})^2}{(7+4\sqrt{3})(3)}}=5\sqrt{\frac{3(7+4\sqrt{3})}{(7+4\sqrt{3})(3)}}=5$$

Thus the length of $AM$, which is the radius of the sector, is $5$.

Answer 3

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