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area of the circle is pi*r^2
so percentage increase equal to (1.06)^2 = 12.36 %
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Bunuel
If the radius of a circle is increased by 6%, then the area of the circle is increased by

(A) .36%
(B) 3.6%
(C) 6%
(D) 12.36%
(E) 36%

We can let the original area = πr^2; thus, the new area is π(1.06r)^2 = π(1.1236)r^2.

We use the percent increase formula: (New - Old)/Old x 100. Thus, the area is increased by:

[π(1.1236)r^2 - πr^2]/πr^2 x 100

[0.01236πr^2/]/πr^2 x 100 = 12.36%

Answer: D
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Bunuel
If the radius of a circle is increased by 6%, then the area of the circle is increased by

(A) .36%
(B) 3.6%
(C) 6%
(D) 12.36%
(E) 36%

Asked: If the radius of a circle is increased by 6%, then the area of the circle is increased by

Area of a circle = \pi r^2 ; where r is the radius of the circle

New area = \(\pi (1.06r)^2 = \pi 1.1236r^2 = (1 + .1236) \pi r^2\)

IMO D
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