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In the figure above, the radius of the larger circle is twice the rad
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18 Feb 2025, 00:53
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66% (01:02) correct
33% (00:39) wrong based on 3 sessions
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In the figure above, the radius of the larger circle is twice the radius of the smaller circle. How many times is the area of the shaded region more than the area of the unshaded region?
Re: In the figure above, the radius of the larger circle is twice the rad
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18 Feb 2025, 15:49
1
Let x = the radius of the smaller circle The area of the smaller circle is πx^2; this is also the area of the unshaded region.
Given that the radius of the larger circle is twice the radius of the smaller circle, the larger circle's radius = 2x The area of the larger circle is π(2x)^2 = 4πx^2
The area of the shaded region is the area of the larger circle minus the area of the smaller circle = 4πx^2 - πx^2 = 3πx^2
Since the ratio of the area shaded region to the area of the unshaded region is 3πx^2 : πx^2, the area of the shaded region is 3 times more than the area of the unshaded region.
Re: In the figure above, the radius of the larger circle is twice the rad
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12 May 2025, 01:03
1
The area of bigger circle is pi*r^2 The area of the shaded region is pi*r^2 - pi*(r/2)^2 Now, we need to find how many time the area of smaller circle is equal to the shaded region. Hence, this area is equal to x*pi*(r/2)^2 Equating both will be x=3 hence, C
gmatclubot
Re: In the figure above, the radius of the larger circle is twice the rad [#permalink]