AIMGMAT770 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.
Good question - Here are my thoughts on it:
"A circle is drawn within the interior of a rectangle" - this implies that the circle isn't inscribed in the rectangle. It lies within the interior so it could be as small as almost a dot. So any dimensions of rectangle will never tell us that the circle occupies more than half the rectangle's area. All we could prove from the rectangle's dimensions is that under no conditions can the circle occupy more than half the area of the rectangle. Why is it that the circle may not occupy even half of the rectangle's area? That would depend on the kind of rectangle we have.
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In a square, a circle can take up most of the area (but not all). The more the difference between the length and width of the rectangle, the less area a circle can occupy. Of course, the circle can be much smaller but we are considering here the maximum area it can occupy. We would try to establish that the maximum the circle can occupy is less than half the area of the rectangle.
(1) The rectangle’s length is more than twice its width.
This is our case 2 in the figure above. If length is twice of width, it means the rectangle has two squares. We can put a circle in one square - it will occupy most of that area but the other square will be empty. Hence area of rectangle occupied will be less than half. 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.
If this statement is to be sufficient, we need to prove that even if the rectangle is actually a square, if the circle drawn fits into a square with (3/4) dimensions, then the circle occupies less than half the area of the square.
If length of square is s, area of square \(= s^2\)
If length of square is 3s/4, area of circle inscribed \(= \pi*(3s/8)^2 = 28s^2/64\)
Even if the rectangle is a square, its area would be s^2 but the area of the circle can be maximum 28s^2/64 - less than half the area of the square. Hence in no case can the circle occupy more than half of the area of the rectangle. Sufficient
Answer (D)