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28 Dec 2020, 09:00
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D
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Difficulty:
15%
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Question Stats:
83% (02:27) correct
17%
(02:10)
wrong
based on 36
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In the figure below, the centers of all three circles lie on the same line. The medium-sized circle has a radius twice the size of the radius of the smallest circle, and the smallest circle has a radius whose length is 2. What is the area of the shaded region?
(A) 3π
(B) 4π
(C) 6π
(D) 8π
(E) 10π
2020-12-21_21-18-40.png [ 26.61 KiB | Viewed 2024 times ]
(A) 3π
(B) 4π
(C) 6π
(D) 8π
(E) 10π
Attachment:
2020-12-21_21-18-40.png [ 26.61 KiB | Viewed 2024 times ]
28 Dec 2020, 11:19
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Given, r1= 2, r2= 4
2*R= 2*r1 + 2*r2= 2*2 +2*4, R= 6
Now, we can see area leftover(Inside Bigger circle) is symmetric in GREY and WHITE color, so area in GREY color
= {Pie (6^2) - Pie (4^2)- Pie (2^2)}/2 = 16*Pie /2 = 8*Pie
So, I think D.
2*R= 2*r1 + 2*r2= 2*2 +2*4, R= 6
Now, we can see area leftover(Inside Bigger circle) is symmetric in GREY and WHITE color, so area in GREY color
= {Pie (6^2) - Pie (4^2)- Pie (2^2)}/2 = 16*Pie /2 = 8*Pie
So, I think D.
28 Dec 2020, 21:54
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\(R_s=2; R_m=4; \)
\(D_l = 2*R_s+2*R_m=2*2+4*2=12\)
\(R_l=\frac{12}{2}=6\)
By removing the smaller and medium circle area from the larger circle will help us in finding the remaining area.
Remaining \(Area = πR_l^2-(πR_m^2+πR_s^2)\)
To find the area shaded region, which is half of the remaining area. Therefore,
Shaded \(Area = \frac{πR_l^2-(πR_m^2+πR_s^2)}{2}\)
\(=π\frac{(6^2-(4^2+2^2)}{2}=π \frac{(36-20)}{2}=π\frac{16}{2}=8π\)
Ans D
\(D_l = 2*R_s+2*R_m=2*2+4*2=12\)
\(R_l=\frac{12}{2}=6\)
By removing the smaller and medium circle area from the larger circle will help us in finding the remaining area.
Remaining \(Area = πR_l^2-(πR_m^2+πR_s^2)\)
To find the area shaded region, which is half of the remaining area. Therefore,
Shaded \(Area = \frac{πR_l^2-(πR_m^2+πR_s^2)}{2}\)
\(=π\frac{(6^2-(4^2+2^2)}{2}=π \frac{(36-20)}{2}=π\frac{16}{2}=8π\)
Ans D
Attachments
Circle.jpg [ 29.27 KiB | Viewed 1801 times ]
