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In the figure shown, J, K, and L are the centers of the thre
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29 Jul 2020, 10:39
29 Jul 2020, 10:39
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81% (03:05) correct
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GRE In the figure shown, J, K, and L are the centers of the thre.jpg [ 32.38 KiB | Viewed 2038 times ]
In the figure shown, J, K, and L are the centers of the three circles. The radius of the circle with center J is four times the radius of the circle with center L, and the radius of the circle with center J is two times the radius of the circle with center K. If the sum of the areas of the three circles is \(525\pi\) square units, what is the measure, in units, of JL?
A. 35
B. 45
C. 50
D. 65
E. 70
Kudos for the right answer and explanation
Re: In the figure shown, J, K, and L are the centers of the thre
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29 Jul 2020, 10:45
29 Jul 2020, 10:45
1
Let the radius of the circle with center L be x.
=> the radius of the circle with center K = 2x
and, the radius of the circle with center J = 4x
Now, the sum of the areas of the three circles is 525π
=> π*x*x + π*(2x)*(2x) + π*(4x)*(4x) = 525*π
=> x^2 + 4*x^2 + 16*x^2 = 525
=> 21*x^2 = 525
=> x^2 = 21
=> x = 5
=> JL = 5 + 10 + 10 + 20 = 45
So, the answer is B.
=> the radius of the circle with center K = 2x
and, the radius of the circle with center J = 4x
Now, the sum of the areas of the three circles is 525π
=> π*x*x + π*(2x)*(2x) + π*(4x)*(4x) = 525*π
=> x^2 + 4*x^2 + 16*x^2 = 525
=> 21*x^2 = 525
=> x^2 = 21
=> x = 5
=> JL = 5 + 10 + 10 + 20 = 45
So, the answer is B.
Re: In the figure shown, J, K, and L are the centers of the thre
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22 Jun 2024, 13:00
22 Jun 2024, 13:00
JL = 5 + 10 + 10 + 20 = 45 please let me know how this value are selected.
