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29 Jan 2016, 03:23
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C
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In the figure given below, ABC and CDE are two identical semi-circles of radius 2 units. B and D are the mid points of the arc ABC and CDE respectively. What is the area of the shaded region?
A. 4π - 1
B. 3π - 1
C. 2π - 4
D. ½(3π - 1)
E. 2π - 2
Untitled.png [ 25.32 KiB | Viewed 14496 times ]
A. 4π - 1
B. 3π - 1
C. 2π - 4
D. ½(3π - 1)
E. 2π - 2
Attachment:
Untitled.png [ 25.32 KiB | Viewed 14496 times ]
29 Jan 2016, 09:12
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Bunuel
Hi,
the figure may look complicated but can be actually taken as one circle divided into two half..
We can work on one semicircle and then multiply the area of shaded region by 2..
join point B with th center of the diameter say, O..
so OBC is a right angle triangle, as B is the midpoint of arc AC...
what will be the area of shaded region..
=area of sector BC - area of triangle OBC..
sector BC is 1/4 of the circle of radius 2, so area = π *2^2/4= π ..
area of OBC= 1/2* (OB*OC)= 2^2/2=2..
so area of one shaded region= π - 2..
area of total shaded region= 2(π - 2)
C
29 Jan 2016, 09:54
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I took a similar approach to Chetan4u, but worked with a square inside the circle instead of a triangle.
If you inscribe a square inside a circle, then there will be 4 small regions inside the circle but outside the square. The question is asking us to find the area of two of these regions. Pictures are so much easier:
Square in circle.png [ 4.56 KiB | Viewed 12988 times ]
Now, the area of the circle is \(\pi r^2 = 4\pi\)
The length of one side of the square is \(\sqrt{2^2+2^2} = \sqrt{8}\)
So the area of the square is \((\sqrt{8})^2 = 8\)
And the area of the shaded region is half the area outside the square
\(\frac{1}{2}(4\pi-8) = 2\pi-4\)
Answer: C
*For some reason the math editor is adding a vertical line at the end of each expression... it is not part of the expression and should be ignored.
If you inscribe a square inside a circle, then there will be 4 small regions inside the circle but outside the square. The question is asking us to find the area of two of these regions. Pictures are so much easier:
Attachment:
Square in circle.png [ 4.56 KiB | Viewed 12988 times ]
Now, the area of the circle is \(\pi r^2 = 4\pi\)
The length of one side of the square is \(\sqrt{2^2+2^2} = \sqrt{8}\)
So the area of the square is \((\sqrt{8})^2 = 8\)
And the area of the shaded region is half the area outside the square
\(\frac{1}{2}(4\pi-8) = 2\pi-4\)
Answer: C
*For some reason the math editor is adding a vertical line at the end of each expression... it is not part of the expression and should be ignored.
