# 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).

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 )

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

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

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

• How is the answer that you got is different from the answer that Alijah Ahmed got ( see below) ? – A Googler Mar 31 '14 at 7:47
• @AGoogler, I can't tell (I haven't even read the other answer) but mine was the first one...:) And even if it wasn't: I bet you know sometimes several pretty similar answers pop up almost at once. This usually happens with elementary questions, like yours. – DonAntonio Mar 31 '14 at 10:40
• No , not that. I mean that you got the radius of the sector as something irrational , while Alijah got it as equal to 5. Who's right? I can't find a flaw in either of you , so I'm confused . Can you take a look at his answer ? – A Googler Mar 31 '14 at 10:51
• Oh, I see @AGoogler. Well, let me check for some minutes. – DonAntonio Mar 31 '14 at 10:53
• @AGoogler, good you asked me that: it was a stupid mistake of mine. Thanks. I edited my answer and the edited parts are in red now. It all was that in fact $\;R=\frac{\sqrt3}2MO\;$ and not $\;R=\sqrt3 MO\;$ as I wrote. Forgot that $\;1/2\;$ factor. – DonAntonio Mar 31 '14 at 11:05

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 $5$.

...............................................................................................................................................(30 characters limit)

• Wow ! Have you solved all the questions ? – A Googler Apr 1 '14 at 14:25
• Maths 2013 - All. Maths 2012 - All. Maths 2011 - Q21 is remained as answer is not given in solutions. Q33 is wrong. Maths 2011 Sample - 45,47,52,53,54 As answers are not given. Maths 2010 - Q55 I didn't got it.Q 58 is wrong. Q67,69,70,73 ans not given. Maths 2009 - Didn't got Q83. Q 87,93,94 no ans is given. Maths 2008 - 95,97,a,b,98,99,100. Maths 2007 - 104,107,108,109,110,112. Just two left from physics I think they are wrong And all from chemistry. – Ajay Apr 1 '14 at 14:31
• how much time on an average did it take you to solve one maths question? I'm fearing that I won't be able to complete the question paper in time . – A Googler Apr 1 '14 at 15:26
• 3-4 minutes average sometimes it take 6-7 minutes to solve questions like Q49 due to long calculations needed to solve it. What about you? – Ajay Apr 2 '14 at 2:26
• Got 55 and 21. I was reading on ray BC and AB instead of in ray BC and AB. Thanks to you I tried to solve them again. – Ajay Apr 2 '14 at 3:03