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

Question

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.

KarishmaB · Accepted Answer

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. Ques3.jpg 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)

mikemcgarry · Answer

Dear Punyata , This is a great question, and I am happy to help. Statement #1 : Suppose length is exactly twice the width. That rectangle would essentially be two side-by-side squares, with the "separator" between the two squares erased. Well, think about it: a circle fits neatly into a single square. Therefore, the circle would take up part of the area of one half, and none of the area of the other half. Now, if we make the rectangle even longer than two squares, longer than twice the width, then there will be even more area outside of the square containing the circle. Therefore, the area of the circle is much less than half the area of the rectangle. This question, alone and by itself, is sufficient. Statement #2 : For this one, see: https://magoosh.com/gmat/2012/scale-fact ... decreases/ Suppose the original rectangle is A = L*W. If length is decreased by 25%, it goes down to 3L/4. Similarly, width goes down to 3W/4. The new area is (3L/4)*(3W/4) = 9LW/16 = 9A/16. This is a new rectangle with just more than half the area of the original rectangle. In order to fit into that smaller rectangle, a circle will occupy a good fraction less than the whole --- think about a circle of radius 1 in a square of side two: the ratio of (circle are):(square area) is (pi)/4, and if we multiply 9/16 by (pi)/4, it will be less than a half. Therefore, the area of the circle is much less than half the area of the original rectangle. This question, alone and by itself, is sufficient. Answer = (D) Does all this make sense? Mike :-)

WholeLottaLove · Answer

Tricky question.

I assumed (without thinking it through...blasted time constraints!) that because we are given no indication of the size of the circle, there is no way of knowing when in reality we do know because the circle, even if it was as large as possible, will never have an area more than 1/2 of the rectangles area.

HarveyS · Answer

I am not clear with this question.

No where it says that circle touches the rectangle.... so there can be very small circle within rectangle also?

WholeLottaLove · Answer

But for #2, wouldn't it be possible to construct a rectangle that is just ever so slightly longer than it is wide? In other words, the reduction of the length and sides makes it so the width of the rectangle is touching the circle (maximizing the circles width and therefore, area) where the length is just slightly greater than it's width? If the approximate ratio of a circle to a square is 3/4, can't you construct a rectangle that is just slightly larger than that?

I see where you got your formula from and that makes sense, but a rectangle could be L=10 and W=10.1 If the circle before had wiggle room but now touches the sides of the width and is almost as wide as the width of the rectangle, it would surely take up more than 1/2 of the circle. That's why I would say this isn't sufficient.

mikemcgarry · Answer

Dear WholeLottaLove , I love your Led Zeppelin screenname. I think you are underestimating the difference in area between the two rectangles. If each side decreases by 25% to 3/4 of its original length, then the area decreases to (3/4)^2 = 9/16 = 0.5625 of the original. The smaller rectangle is only slightly larger than half the larger rectangle, so even if the circle takes up a good chunk of the smaller rectangle, it's still a long way off from the larger triangle. The tightest rectangle to fit around a circle would be a square. Remember that a square is a special case of a rectangle. cirlce in square in square.JPG All the actual sizes are shown. As you see, the ratio of the side of the smaller square to the side of the larger square if 0.75, which means each length in the larger square would have to be decreased by 25% to make the smaller square. The ratio of the squares, as predicted, is 9/16 = 0.5625, and the area, which takes up a good chunk of the smaller square ---- (pi)/4 = 0.78539816339745 approximately ---- but that's still less than half the larger square, as the final ratio indicates. Here, I am using exact values, but in the above post, I show how to arrive at this conclusion via approximations. Does all this make sense? Mike :-)

WholeLottaLove · Answer

Alright. I think I got it. You're right, I didn't fully comprehend the difference in area that reduction would make though I guess it makes sense. A 2 inch square that doubles it's length and width has an area four times larger. In other words, the ratio of the circle to the square * the ratio of the square to the larger rectangle?

mikemcgarry · Answer

Yes, that product of ratios is a good way to think of it --- ratio of circle to little square is (pi)/4, which is a bit more than 3/4 = 0.75 --- say about 0.8 = 4/5; and the ratio of little square to big square is (3/4)^2 = 9/16. (4/5)(9/16) = (1/5)(9/4) = 9/20 ---- less than 1/2 Does all this make sense? Mike :-)

MartyTargetTestPrep · Answer

Yup. So we know that the area of circle can be less than half that of the rectangle. The circle can be tiny compared to the rectangle.

What we can determine by using each of the given statements is that given the constraints they describe, the area of the circle can't be more than half that of the rectangle. While we don't know how much less than half the rectangle's area he circle's is, it's definitely less than half. That's all we need to answer the question.

Each statement achieves this on its own. So both are individually sufficient.

TheKingInTheNorth · Answer

Hi Bunuel , 
Would you please explain statement 2 .

Thanks.

mikemcgarry · Answer

Dear abhisheknandy08 , I'm happy to respond for the great Bunuel. :-)

My friend, this is a question that requires visual thinking . This is hard. It's not something you can understand with formulas. You need to look at diagrams, and even draw diagrams yourself, in order to develop a sense. Look at the diagram in my post in this thread from Dec 09, 2013 . There, I drew a square (a special case of a rectangle), then a second square with sides 0.75 the length of the larger square, and put the circle inside that. See that diagram for all the ratios. Another way to think about it is as follows. If the sides are multiplied by the scale factor k = 0.75 (the multiplier for a 25% decrease), then the area is reduced to k^2 = (0.75)^2 = 0.5625. In other words, the new rectangle's total area is only slightly more than 50% of the original rectangle, so there's no way that the circle inside the reduced rectangle will be more than 50% of the larger rectangle's area. Does all this make sense? Mike :-)

Annecee · Answer

Would it be okay to pick numbers for Statement 2? For example, since the question doesn't state that you couldn't have L = W I set them both to 12.

The area of the rectangle is 144
With a 25% reduction, the new area would be 81.
Diameter thus must be less than 81
2R=9
R=9/2
Area of circle is 81pi/4 which can roughly be estimated to be 243/4 = 60
Since half of the original rectangle's area is 144/2 = 72
Therefore sufficient.

brains · Answer

Hello Experts

I have just one doubt in this question. The Question stem says that the circle lies in the interior of the rectangle. No where in the question says that the circle touches the sides of a rectangle. It may happen that the circle is of radius 1 unit centered at the intersection of diagonals of rectangle having size 20*12 units etc.

In this scenario, the answer would be E.

Kindly clarify

chetan2u · Answer

Hi costco

(1) The rectangle’s length is more than twice its width.
Let the sides be w and l, so l>2w, and area of rectangle = w*l>2w^2
But the circle within this rectangle can have it's diameter as w at the maximum, so max area =  \pi*r^2=3.14*(\frac{r}{2})^2=0.9w^2 .
Thus area of circle is less than half of that of rectangle.
Suff

(2) If the rectangle’s length and width were each reduced by 20% and the circle unchanged, the circle would still fit into the interior of the new rectangle.
It depends on the sides of rectangle.
Say rectangle has sides 100*110, and dia of circle is 1, the answer is NO.
But if rectangle has sides 100*110, and dia of circle is 80, the answer is YES
Insuff

A

carouselambra · Answer

My thoughts, please let me know if the approach is correct.

Assume the l=300, b=100 (since its given that the length is more than 2b)
Question being asked : area of circle > lb/2  
pi.r^2 >lb/2
pi.(d/2)^>(lb/2)
pi*d^2/4>lb/2
Now substitute the values that we had assumed initially :

pi*300*300/2>300*100 (have assumed diameter to be 300 since that's the longest side in the rectangle)
Which leads me to 'Yes' , i.e. Option A is sufficient
With option B, the circle is still contained in the rectangle could mean various things-
1. Circle is as small as a tiny dot
2.Circle is occupying less than 1/2 of the area
3. Circle is occupying more than 1/2 of the area
4.Circle might occupy 3/4th of the area for all we know!

clubnataraj · Answer

Hi,
Drawn within interior, doesn't mean it's inscribed, right?

circle could have been just a teeny tiny dot.

KarishmaB · Answer

Yes, it doesn't mean inscribed. It can be almost like a dot too. What we need to see is if there is a reason why the circle's area MUST be less than half of the rectangle's area in every case. Both options tell us that even if we draw the biggest circle possible (ever so slightly smaller than inscribed circle), its area will STILL be less than half the rectangle's area. Check here: https://gmatclub.com/forum/a-circle-is- ... l#p1459436

bumpbot · Answer

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

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