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30 Aug 2018, 05:11
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (01:52) correct
28%
(01:42)
wrong
based on 148
sessions
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Four semicircular arcs of length 2π are joined to make the figure above. What is the area of the enclosed region?
A. \(8\pi\)
B. \(8 + 8\pi\)
C. \(16 + 8\pi\)
D. \(16 + 16\pi\)
E. \(24\pi\)
Attachment:
image019.jpg [ 2.6 KiB | Viewed 4430 times ]
30 Aug 2018, 06:08
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Bunuel
If one SEMI-circle has a perimeter of 2π, then an ENTIRE circle must have a circumference of 4π
Circumference = (diameter)(π)
So, if the circumference of the circle is 4π, then the diameter of each circle is 4
This means the square in the middle of the figure must have area 16.
Now we need only find the area of 4 SEMIcircles.
If we combine 2 semicircles we get an entire circle.
So, if we combine 4 semicircles we get 2 entire circles.
Each circle has diameter 4, which means each circle has RADIUS 2
Area of a circle = π(radius²)
So, area of one circle = π(2²) = 4π
So, the area of TWO circles = 8π
So, the TOTAL area = 16 + 8π
Answer: C
Cheers,
Brent
General Discussion
30 Aug 2018, 06:08
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Bunuel
Area of enclosed region=4*Area of semicircle+Area of square (refer affixed diagram)
Given, length of semicircular arc=2π
Or, \(\frac{2πr}{2}=2π\)
Or, r=2
Area of semi-circle=\(\frac{1}{2}*π*r^2=2π\)
Area of square=4*(2r)=16
So , area of enclosed region=4*2π+16=8π+16
Ans. (C)
Attachments
semic.JPG [ 22.09 KiB | Viewed 3724 times ]
