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

[Mike@Magoosh]

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?
(A) 2/(pi)
(B) [(pi) - sqrt(3)]/(pi)
(C) [4 - sqrt(3)]/(pi)
(D) (2/3) - [sqrt(3)/(4pi)]
(E) (5/6) - [sqrt(3)/(2pi)]

For other challenging circle problems, as well as the OA and a detailed OE for this particular question, see
https://magoosh.com/gmat/2014/circle-pro ... -the-gmat/

Mike

[phanikpk]

Thanks mike for such a beautiful qn

Really feast for brains

E is the answer

[GMATGuruNY]

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?
(A) 2/(pi)
(B) [(pi) - sqrt(3)]/(pi)
(C) [4 - sqrt(3)]/(pi)
(D) (2/3) - [sqrt(3)/(4pi)]
(E) (5/6) - [sqrt(3)/(2pi)]

Alternate approach:



Let r=2.
Area of the circle = πr² = π(2²) = 4π.

Outside trapezoid JNOL are 3 slivers, moving clockwise from point J:
Green sliver JN, red sliver NO, and green sliver OL.
Sum of the 3 slivers = (1/2 circle) - (trapezoid JNOL)

1/2 of the circle = (1/2)(4Ï€) = 2Ï€
Area of JNOL = (b� + b₂)/2 * h = (JL + NO)/2 * NP = (4 + 2)/2 * √3 = 3√3.
Thus:
Sum of the 3 slivers = 2π - 3√3.

Since green slivers JN and OL constitute only 2 of the 3 slivers, we get:
Sliver JN + Sliver OL = (2/3)(2π - 3√3) = (4π - 6√3)/3.

Adding the green semicircle to green slivers JN and OL, we get the entire green portion:
2π + (4π - 6√3)/3 = (6π + 4π - 6√3)/3 = (10π - 6√3)/3.

Thus:
(total green portion)/(total circle) = [ (10π - 6√3)/3 ] / 4π = (10π - 6√3)/12π = 5/6 - √3/(2π).

The correct answer is E.

[GMATinsight]

Join points A and B with centre of Circle M



Suppose Radius of Circle = 2

Then, AMJ and MBL will be Equilateral Triangles with Side 2

Area of these two triangle = 2 x (Sqrt3 / 4 x 2^2) = 2 Sqrt3

Area of Unshaded sector MAB = (60/360) x Pi x 2^2 = (2/3) Pi

Total Shaded area = Pi(2^2) - ((2/3) Pi + 2 Sqrt3) = (10/3) Pi + 2 Sqrt3

Area of Circle = 4 Pi

Required Fraction = [(10/3) Pi + 2 Sqrt3]/4 Pi

= (5/6) - [sqrt(3)/(2pi)]

Answer: Option E