swerve wrote:A circle is drawn within the interior of a rectangle. Does the circle occupy more than one-half of the rectangle's area?

1) The rectangle's length is more than twice its width.
2) If the rectangle's length and width were each reduced by 25% and the circle unchanged, the circle would still fit into the interior of the new rectangle.

The OA is D

Source: Manhattan Prep

Let's take each statement one by one.

1) The rectangle's length is more than twice its width.

Note that if a circle is inscribed in a square and it touches the sides of the square, the area of the circle would be less than that of the square. Given that the rectangle's length is more than twice its width, we see that there are two squares and a small rectangle. It is clear that since the circle is drawn within the interior of the rectangle, its area would be less than half that of the rectangle. Sufficient.

2) If the rectangle's length and width were each reduced by 25% and the circle unchanged, the circle would still fit into the interior of the new rectangle.

Say the length and the width of the rectangle be a and b, respectively. Thus, its area = ab

The length and the width of the rectangle after reducing their lengths by 25%, we get new length = 0.75a and 0.75b, respectively.

Maximum possible radius of the circle such that it remains within the rectangle = 0.75b/2 = 3b/8

Thus, its area = π*3b/8*3b/8 = 9πb^2/64

Ratio of the area of the circle to the area of the rectangle = (9Ï€b^2/64) / (ab) = (9Ï€/64)*(b/a)

Note that half of 64 is 32 and 9π << 32; thus, 9π/64 < 1/2; similarly, a ≥ b; thus, (9π/64)*(b/a) < 1/2. Sufficient

The correct answer: D

Hope this helps!

-Jay
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