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TAGS: Intern Joined: 15 Sep 2017
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A circle is inscribed in equilateral triangle ABC [#permalink]
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Question Stats: 61% (02:27) correct 38% (05:41) wrong based on 31 sessions
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]
explain it pls.?? Retired Moderator Joined: 07 Jun 2014
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GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Re: A circle is inscribed in equilateral triangle ABC [#permalink]
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Expert's post
We have been given this problem

Attachment: draw1.jpg [ 28.08 KiB | Viewed 1423 times ]

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

Attachment: draw2.jpg [ 41.29 KiB | Viewed 1425 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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Intern Joined: 27 Oct 2018
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Re: A circle is inscribed in equilateral triangle ABC [#permalink]
60-30-90 triangles make things easier!
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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