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10 May 2022, 02:05
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
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Question Stats:
53% (03:31) correct
47%
(02:57)
wrong
based on 34
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The diagram below shows a shaded region and an equilateral triangle of side length 2. Each vertex is the center of the semicircle shapes. Find the difference between the area of the shaded region and the area of the triangle
A. π/3
B. π/4
C. π/9
D. π/16
E. π
Are You Up For the Challenge: 700 Level Questions
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10 May 2022, 10:55
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Option E is the answer.
The solution is below.
Let the shaded area at each corner of the triangle is y and shaded area at centre is z
Area of Shaded region = 5(pi)/6 *3 +3y + z
Area of Triangle = (pi)/2 *3 + 3y + z
Subtracting, we get the area = Pi
The solution is below.
Let the shaded area at each corner of the triangle is y and shaded area at centre is z
Area of Shaded region = 5(pi)/6 *3 +3y + z
Area of Triangle = (pi)/2 *3 + 3y + z
Subtracting, we get the area = Pi
Foreheadson
Joined: 22 Jun 2020
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Location: Georgia
Concentration: Finance, General Management
Schools: Booth '25 (WL) Darden '25
GMAT 1: 720 Q51 V38
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Updated on: 30 Jun 2022, 23:30
Originally posted by Foreheadson on 30 Jun 2022, 12:34.
Last edited by Foreheadson on 30 Jun 2022, 23:30, edited 1 time in total.
Last edited by Foreheadson on 30 Jun 2022, 23:30, edited 1 time in total.
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I solved like this:
lets cut the triangle in 4 equilateral triangles on the midpoints of sides.
Area of the little triangle = \(\frac{\sqrt{3}}{4}\) (since side equals 1).
Let's calculate shaded are within, let's say, leftmost little triangle. \(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{6}\)*2
the latter being the sixth of the circle, multiplied by 2, because 2 such sectors fall into leftmost little triangle.
Then we have 2 more such vertex triangles, so 3 in total and 1 in the middle. Multiply the above expression by 3 for vertex little triangles and for the middle: \(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{2}\), since there are 3 sixths sectors falling in the middle little triangle.
So finally we have to sum these and add the outside circle sectors, which is π\(r^2\)*\(\frac{5}{6}\)*3.
So: (\(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{6}\)*2)*3+\(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{2}\)+π\(r^2\)*\(\frac{5}{6}\)*3=\(\sqrt{3}\)+\(πr^2\)
insert r=0.5, \(\sqrt{3}\)+\(0.25π\)
What am I missing?
Edit: the question is asking the difference between triangle area and the one calculacted above. so \(\sqrt{3}\)+\(0.25π\) - \(\sqrt{3}\)=\(0.25π\), which is B.
lets cut the triangle in 4 equilateral triangles on the midpoints of sides.
Area of the little triangle = \(\frac{\sqrt{3}}{4}\) (since side equals 1).
Let's calculate shaded are within, let's say, leftmost little triangle. \(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{6}\)*2
the latter being the sixth of the circle, multiplied by 2, because 2 such sectors fall into leftmost little triangle.
Then we have 2 more such vertex triangles, so 3 in total and 1 in the middle. Multiply the above expression by 3 for vertex little triangles and for the middle: \(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{2}\), since there are 3 sixths sectors falling in the middle little triangle.
So finally we have to sum these and add the outside circle sectors, which is π\(r^2\)*\(\frac{5}{6}\)*3.
So: (\(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{6}\)*2)*3+\(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{2}\)+π\(r^2\)*\(\frac{5}{6}\)*3=\(\sqrt{3}\)+\(πr^2\)
insert r=0.5, \(\sqrt{3}\)+\(0.25π\)
What am I missing?
Edit: the question is asking the difference between triangle area and the one calculacted above. so \(\sqrt{3}\)+\(0.25π\) - \(\sqrt{3}\)=\(0.25π\), which is B.
