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A circle is inscribed in equilateral triangle ABC

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A circle is inscribed in equilateral triangle ABC [#permalink] New post 10 Sep 2018, 22:59
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A circle is inscribed in equilateral triangle ABC such that point D lies on the circle and on the line segment AC, point E lies on the circle and on the line segment AB, point F lies on the circle and on the line segment BC. If the line segment AB = 6, what is the area of the figure created by AB, AC and minor arc DE?


A)3\(\sqrt{3}-\frac{9}{4}pi\)


B)\(3\sqrt{3}- pi\)


C)\(6\sqrt{3}-pi\)

D)\(9\sqrt{3}-3pi\)

E)\(9\sqrt{3}-2pi\)
[Reveal] Spoiler: OA

Last edited by AchyuthReddy on 11 Sep 2018, 02:51, edited 1 time in total.
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Re: A circle is inscribed in equilateral triangle ABC [#permalink] New post 10 Sep 2018, 23:40
explain it pls.??
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Re: A circle is inscribed in equilateral triangle ABC [#permalink] New post 11 Sep 2018, 05:11
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We have been given this problem

Attachment:
draw1.jpg
draw1.jpg [ 28.08 KiB | Viewed 342 times ]


Now from here we have to calculate the are aof the shaded region below.

Attachment:
draw2.jpg
draw2.jpg [ 41.29 KiB | Viewed 342 times ]


We know that radius of the circle r shown in figure above =\(\frac{a}{2 \sqrt{3}}\) where a is the side of equilateral triangle.

For derivation of the above fromula refer to the video below.



Now area of the equilateral triangle is \(\frac{\sqrt{3}}{4} \times a^2=\frac{\sqrt{3}}{4} \times 6^2=9\sqrt{3}\).

Area of the inscribed circle =\(\pi (\frac{6}{2\sqrt{3}})^2= 3 \pi\).

Substracting the area of circle from triangle = \(9\sqrt{3}-3\pi\).

This area is 3 times the shaded area shown in figure above.

Divide the obtained area by 3 and we have the solution. Hence option B.
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Re: A circle is inscribed in equilateral triangle ABC [#permalink] New post 02 Nov 2018, 23:40
60-30-90 triangles make things easier!
radius = root(3)
symmetry tells us that answer is one-third of the area we find by subtracting area of triangle from area of circle
Re: A circle is inscribed in equilateral triangle ABC   [#permalink] 02 Nov 2018, 23:40
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