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A cirlce is drawn within the interior of a rectangle. Does

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A cirlce is drawn within the interior of a rectangle. Does

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Manhattan Prep

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 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.

OA D.

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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 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.

OA D.
Given: A circle is drawn within the interior of a rectangle.
Question: Does the circle occupy more than one-half of the rectangle's area?

Let's take each statement one by one.

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

Note that if the length and the width of the rectangle are equal, i.e., the rectangle becomes a square; and the area of the biggest possible circle (a circle touching all the sides of the rectangle/square) would be less than the area of the rectangle.

A rectangle with its length equal to twice its width is like two side-by-side squares. So, we have one biggest possible circle and two squares. It's obvious that the area of the biggest possible circle is less than half of the two squares.

So, if the rectangle length is more than twice its width, the area of the biggest possible circle would be less than half of the rectangle. The answer is No, the circle does NOT occupy more than one-half of the rectangle's area. 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.

This is interesting...

Since from Statement 1, we get the answer No; so if Statement 2 alone also suffices, we must get the answer No.

Needless to say that for a very small circle, the area of the circle is less than half of the area of the rectangle.

Let's see what happens when we have the biggest possible circle.

Say the rectangle is a square and after the reduction of its length and the width, the circle (the biggest possible circle) touches the four sides.

Say the diameter of the biggest possible circle is d; thus, its area = πd^2/4

Since the circle touches the four sides, side of the square (after reduction) = d

Thus, the side of the square before reduction = d/75% = 4d/3

Area of the original square (before reduction) = 4d/3 * 4d/3 = 16d^2/9

Ratio of (area of the biggest possible circle) to (area of the original square (before reduction))

= (πd^2/4) / (16d^2/9) = (9π) / 64 = ~27/64 < 1/2

Since half of 64 is 32, and the numerator is 27 (< 32), the area of the biggest possible circle is less than half of the rectangle. The answer is No, the circle does NOT occupy more than one-half of the rectangle's area. Sufficient.

Even if you find out the actual value of 9π, it will not less than 32. Note that 27 < 9π < 29.

The correct answer: D

Hope this helps!

-Jay
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