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# A circle is inscribed in equilateral triangle ABC such that point D

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Math Expert
Joined: 02 Sep 2009
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A circle is inscribed in equilateral triangle ABC such that point D  [#permalink]

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17 Sep 2018, 02:03
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Difficulty:

55% (hard)

Question Stats:

50% (02:35) correct 50% (03:06) wrong based on 25 sessions

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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) $$9\sqrt{3}-3\pi$$

(E) $$9\sqrt{3}-2\pi$$

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Re: A circle is inscribed in equilateral triangle ABC such that point D  [#permalink]

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17 Sep 2018, 02:17
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Bunuel wrote:
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) $$9\sqrt{3}-3\pi$$

(E) $$9\sqrt{3}-2\pi$$

Posted from my mobile device
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File comment: Answer B , as solved in attached picture

20180917_121346.jpg [ 6.92 MiB | Viewed 477 times ]

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Re: A circle is inscribed in equilateral triangle ABC such that point D  [#permalink]

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17 Sep 2018, 08:20
kritisinghvi wrote:
Bunuel wrote:
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) $$9\sqrt{3}-3\pi$$

(E) $$9\sqrt{3}-2\pi$$

Posted from my mobile device

won't this give the area which is outside circle but not shaded in your image? can you kindly explain?
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A circle is inscribed in equilateral triangle ABC such that point D  [#permalink]

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17 Sep 2018, 09:12
1
thefibonacci wrote:
kritisinghvi wrote:
Bunuel wrote:
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) $$9\sqrt{3}-3\pi$$

(E) $$9\sqrt{3}-2\pi$$

Posted from my mobile device

won't this give the area which is outside circle but not shaded in your image? can you kindly explain?

Let us say the area outside the circle is K = A(ABC)- A(Circle)

We get K = $$9\sqrt{3}-3\pi$$

Since ABC is an equilateral triangle,

the area of created by AB, AC and minor arc DE = the area of created by AB, BC and minor arc EF = the area of created by BC, AC and minor arc DF.

Thus, to get the area of the shaded region we divide K by 3 (as done in the equation of the image) which gives us $$3\sqrt{3}- \pi$$
A circle is inscribed in equilateral triangle ABC such that point D   [#permalink] 17 Sep 2018, 09:12
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