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12 Jul 2020, 08:17
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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:
55%
(hard)
Question Stats:
68% (02:31) correct
32%
(03:28)
wrong
based on 69
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Attachment:
Screenshot 2020-07-12 at 8.47.48 PM.png [ 11.04 KiB | Viewed 2833 times ]
A. 4π
B. 6π
C. 12π
D. 16π
E. 26π
alternate to an easier version herel
12 Jul 2020, 09:08
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Kritisood
Attachment:
Screenshot 2020-07-12 at 8.47.48 PM.png
For each of the three circles shown above, the diameter of the smaller circle is equal to the radius of the larger. If the area of the largest circle is 52π square inches greater than the area of the shaded region, then what is the area of the shaded region in square inches?A. 4π
B. 6π
C. 12π
D. 16π
E. 26π
alternate to an easier version herel
Let the radius be:
R3, R2, R1 such that R3 > R2 > R1
R3 = 2*R2
R2 = 2*R1
=> R3 = 4R1
Area of shaded region = \(\pi\) * [(R2)^2 - (R1)^2] = \(\pi\) * [(2*R1)^2 - (R1)^2] = \(\pi\) * 3 * (R1)^2
Area of largest circle = \(\pi\) * (R3)^2 = \(\pi\) * 16 * (R1)^2
Difference between the areas = \(\pi\) * [16 * (R1)^2 - 3 * (R1)^2] = 52*\(\pi\)
=> 13*(R1)^2 = 52
=> R1 = 2
=> R2 = 4
=> R3 = 8
Area of shaded region = \(\pi\) * 3 * (R1)^2 = \(\pi\) * 3 * (2)^2 = \(\pi\) * 12
Answer: C
12 Jul 2020, 09:37
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Let the radius of smallest circle be X, then the radii of smaller circle will be 2X and large circle will be 4X.
Given that π*(4X)^2 = π*(2X)^2 - π*(X)^2 + 52π
=> 13*π*X^2 = 52π
=> X = 2
So, area of shaded part = π*(2X)^2 - π*(X)^2
=> 3*π*X^2
=> 12π
OA: C
Posted from my mobile device
Given that π*(4X)^2 = π*(2X)^2 - π*(X)^2 + 52π
=> 13*π*X^2 = 52π
=> X = 2
So, area of shaded part = π*(2X)^2 - π*(X)^2
=> 3*π*X^2
=> 12π
OA: C
Posted from my mobile device
