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