JKL is an equilateral triangle. Point M is the midpoint of

Question

equilateral triangle, circle on base.JPG In the diagram above, JKL is an equilateral triangle. Point M is the midpoint of segment JL, and M is the center of a circle that passes through points J and L. The shaded regions in the diagram indicate all the regions inside the circle that are outside the triangle. What fraction of the total area of the circle is outside the triangle ? answers.JPG For a detailed explanation for this question, as well as several other challenging questions on circles, see: https://magoosh.com/gmat/2014/circle-pro ... -the-gmat/ Mike :-)

Kconfused · Answer

Hey Mike, would you rate this a 750 question? Seemingly difficult, but rather straight forward apparently!

mikemcgarry · Answer

Dear Kconfused , To be honest, my friend, it's always hard to make such calls. With the Magoosh questions I write, sometimes I think it's hard, and many folks get it right; other times, I think it's easy, and no one can solve it. It's hard to judge as the question writer. In some sense, the person who writes a math question is least well equipped to assess its difficulty, because that is the one person who never had the opportunity to approach the question "fresh," from scratch, if you see what I mean. If you know your geometry, and are comfortable with all the basic theorems, and the distinctions of circle vs. sector vs. circular segment, then potentially, this is a very straightforward question. It's not clear to me, for example, what percent of folks looking at this diagram even will recognize that the points where the circle intersects sides JK & KL are actually the midpoints of those sides --- it might take some people more than two minutes just to realize that much! If you found this question relatively doable, I would urge you to follow the link to the Magoosh blog and try the last question there. Mike :-)

SVaidyaraman · Answer

Let the point where JK and KL meet the circle be P and Q. So we will have three equilateral triangles at the top and in total there are six such equilateral triangles inside the circle.

Area of each circular segment = (area of the circle - area of six equilateral triangles) / 6
Multiply by 2 to get the area of two circular segments.

Area of circle outside the triangle = Area of the two circular segments + area of the semicircle -- (1)

fraction of total area of the circle outside the triangle = (1) / area of the circle

We will get E as the answer.

manish2014 · Answer

I think this is a 600 level question... i mean if I cud solve it easily... then it must be easy

PareshGmat · Answer

Lets say radius of the circle = 1

Area of circle = \pi  ...................... (1)

Refer diagram below

Joined the intersection points of triangle on the circle to the centre of circle

Now 2 equilateral traingles are formed (shaded in pink) & one 60Degress sector is created

Area of both equilateral triangles  = 2 * \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}  .............. (2)

Area of 60 Degrees sector  = \frac{\pi}{6}  .............. (3)

Total Area Outside Triangle but inside circle = (1) - (2) - (3)

= \pi - \frac{\sqrt{3}}{2} - \frac{\pi}{6}

Fraction = \frac{\pi - \frac{\sqrt{3}}{2} - \frac{\pi}{6}}{\pi} = \frac{5}{6} - \frac{\sqrt{3}}{2\pi}

Answer = E

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JKL is an equilateral triangle. Point M is the midpoint of

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JKL is an equilateral triangle. Point M is the midpoint of

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