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01 May 2021, 04:30
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D
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Difficulty:
45%
(medium)
Question Stats:
68% (02:24) correct
32%
(02:37)
wrong
based on 38
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Arcs DE, EF, FD are centered at C, B and A, in equilateral triangle ABC, as shown above. If the area of the triangle is 6, what is the area of the shaded region formed by the intersection of the arcs and the triangle?
(A) \(6 - \frac{\sqrt{3}}{3}\)\(\pi\)
(B) \(6 - \frac{\sqrt{3}}{2}\)\(\pi\)
(C) \(\frac{\pi}{\sqrt{3}}\)
(D) \(6 - \sqrt{3}\pi\)
(E) \(\frac{\pi}{6}\)
Attachment:
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09 May 2021, 00:00
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Bunuel
The area of an equilateral triangle is \(area=a^2*\frac{\sqrt{3}}{4}\), where \(a\) is the length of a side. Thus \(a^2*\frac{\sqrt{3}}{4}=6\) --> \(a^2=8*\sqrt{3}\) --> radius of the circles (which make arcs) is a/2, so \(r^2=\frac{a^2}{4}\). Now, as the triangle is equilateral each arc is 60 degrees, so the area of white region (non-shaded) is \(\frac{60*3}{360}*\pi{r^2}=\frac{1}{2}*\pi{\frac{8*\sqrt{3}}{4}}=\pi{\sqrt{3}}\) (three 60 degrees arcs).
The area of the shaded region is \(6-\pi{\sqrt{3}}\).
Answer: D.
General Discussion
01 May 2021, 06:45
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As it is an equilateral triangle, each angle = 60' (60*3=180), and thus we are dealing with a half a circle. To get the area of the shaded area we must subtract the area of the semi circle from the area of the triangle.
Formula for area equilateral triangle is: a^2 * \(\sqrt{3}/4\) where a is the side.
a^2 * \(\sqrt{3}/4\) = 6
a^2 = 24/\(\sqrt{3}\)
= \(\sqrt{576}/\sqrt{3}\)
= \(\sqrt{192}\sqrt{3}/\sqrt{3}\)
= \(\sqrt{192}\)
= \(\sqrt{64}*\sqrt{3}\)
= \(8\sqrt{3}\)
Therefore a = \(\sqrt{ 8 √3 }\)
The radius of the semi-circle will be half of a side of the equilateral triangle.
Radius = ( \(\sqrt{ 8 √3}\) )/ 2
Change this into \(\sqrt{4 * 2 √3 }\) / \(\sqrt{4}\)
the two √4 cancel leaving you with \(\sqrt{ 2 √3}\)
Formula for area of half a circle is: πr^2
(π\(\sqrt{2√3}\))^2 /2
(π*2√3)/2
√3π
Therefore 6 - √3π
Answer D
This obviously isn't ideal for the GMAT setting, so it's probably best to estimate. If one divides the triangle into 4 smaller equilateral triangles (with the shaded region within the central one) one sees that the shaded region is around 1/2 of the area of one of these smaller triangles.
So the area of each smaller triangles is around 6/4 which means that the shaded region has an area of around 3/4 (or 0.75).
Looking at the answer choices C and E can be eliminated as we need to be subtracting something with pi.
A and B will give answers much bigger than what we want.
Looking at D: 6-(3.1*1.7) will give us 6 - 5.27 = 0.73
Answer D
Formula for area equilateral triangle is: a^2 * \(\sqrt{3}/4\) where a is the side.
a^2 * \(\sqrt{3}/4\) = 6
a^2 = 24/\(\sqrt{3}\)
= \(\sqrt{576}/\sqrt{3}\)
= \(\sqrt{192}\sqrt{3}/\sqrt{3}\)
= \(\sqrt{192}\)
= \(\sqrt{64}*\sqrt{3}\)
= \(8\sqrt{3}\)
Therefore a = \(\sqrt{ 8 √3 }\)
The radius of the semi-circle will be half of a side of the equilateral triangle.
Radius = ( \(\sqrt{ 8 √3}\) )/ 2
Change this into \(\sqrt{4 * 2 √3 }\) / \(\sqrt{4}\)
the two √4 cancel leaving you with \(\sqrt{ 2 √3}\)
Formula for area of half a circle is: πr^2
(π\(\sqrt{2√3}\))^2 /2
(π*2√3)/2
√3π
Therefore 6 - √3π
Answer D
This obviously isn't ideal for the GMAT setting, so it's probably best to estimate. If one divides the triangle into 4 smaller equilateral triangles (with the shaded region within the central one) one sees that the shaded region is around 1/2 of the area of one of these smaller triangles.
So the area of each smaller triangles is around 6/4 which means that the shaded region has an area of around 3/4 (or 0.75).
Looking at the answer choices C and E can be eliminated as we need to be subtracting something with pi.
A and B will give answers much bigger than what we want.
Looking at D: 6-(3.1*1.7) will give us 6 - 5.27 = 0.73
Answer D
