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


  • $\begingroup$ How is the answer that you got is different from the answer that Alijah Ahmed got ( see below) ? $\endgroup$ – A Googler Mar 31 '14 at 7:47
  • $\begingroup$ @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. $\endgroup$ – DonAntonio Mar 31 '14 at 10:40
  • $\begingroup$ 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 ? $\endgroup$ – A Googler Mar 31 '14 at 10:51
  • $\begingroup$ Oh, I see @AGoogler. Well, let me check for some minutes. $\endgroup$ – DonAntonio Mar 31 '14 at 10:53
  • 1
    $\begingroup$ @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. $\endgroup$ – 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.

enter image description here

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


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


...............................................................................................................................................(30 characters limit) enter image description here

enter image description here

  • $\begingroup$ Wow ! Have you solved all the questions ? $\endgroup$ – A Googler Apr 1 '14 at 14:25
  • $\begingroup$ 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. $\endgroup$ – Ajay Apr 1 '14 at 14:31
  • $\begingroup$ 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 . $\endgroup$ – A Googler Apr 1 '14 at 15:26
  • $\begingroup$ 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? $\endgroup$ – Ajay Apr 2 '14 at 2:26
  • $\begingroup$ 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. $\endgroup$ – Ajay Apr 2 '14 at 3:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.