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circle is drawn within the interior rectangle

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circle is drawn within the interior rectangle

by j_shreyans » Mon Jun 08, 2015 7:25 am
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.
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by [email protected] » Mon Jun 08, 2015 9:28 am
Hi j_shreyans,

I'm going to give you some hints so that you can try this question again:

1) This question can be solve by TESTing VALUES. To make everything simple, you should start with the circle - choose a radius and 'lock in' that circle so that it doesn't change.
2) Consider that that the circle is drawn WITHIN the rectangle.
3) Think about what the dimensions of the rectangle COULD be (and TEST VALUES to think about really small or really big rectangles), given the restrictions in the two Facts.

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by MartyMurray » Tue Jun 09, 2015 3:41 pm
j_shreyans 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.
Rich's method of testing values is pretty nice for this one. Here's another take as well.

Statement 1 tells us that the diameter of the circle must be at less half of the length of the rectangle. So basically the circle has to fit into half of the rectangle. A circle that fits into a rectangular shape must have a smaller area than that rectangular shape. So from Statemnent 1 we know that the circle's area is less than that of half of the rectangle, and so obviously the area of the circle is less than half that of the rectangle.

Statement 1 is sufficient.

Statement 2 tells us that the circle fits into something that is .75 x the length of the rectangle long and .75 x the width of the rectangle wide.

We could substitute in numbers, as in we could make the rectangle 1 long and 1 wide. The area of the rectangle would be 1, and the area the circle fits into would be .75 x .75 = something a little bigger than .7 x .7 = .49.

That's the way I would do it.

Alternatively one could do it algebraically.

For the full rectangle L x W = LW = Area.

For the smaller rectangle that the circle still fits in .75L x .75W = approx .56LW = .56Area

Using either method one finds that the circle fits into a smaller rectangle that has an area a little over half that of the full rectangle.

Without calculating it out, one can realize that a circle that fits in a rectangle has an area significantly less than that of the rectangle into which it fits. So the circle that fits into a smaller rectangle that has an area .56 times the size of the full rectangle must have an area less than half of or less than .50 x that of the full rectangle.

So Statement 2 is sufficient also.

Choose D.
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by nikhilgmat31 » Fri Jun 26, 2015 12:59 am
As per Statement B- circle can fit in .56Area of Rectangle.

But we don't know - circle can fit in .30 Area then it is taking less than half of rectangle.
But if it take .55 Area of Rectangle then it takes more than half of area of rectangle.

so it is not sufficient.

Answer is A . Please comment.
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by nikhilgmat31 » Fri Jun 26, 2015 1:04 am
Statement A
(1) The rectangle's length is more than twice its width.

L > 2W

To fit circle inside rectangle - radius of circle < w/2

Area of rectangle - LW ~ >2w^2

Area of circle < pie/4 * w^2

pie/4 i.e. 22/28 is almost 1 so we can ignore it.

w^2 < 2w^2 Circle is always less than half of rectangle.

A is sufficient.
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by Poisson » Tue Oct 04, 2016 7:51 am
j_shreyans 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.
Step 1. Recognize that this is a Yes/No question.

Statement 1:
Translation: L > 2W
It's important to realize that this also means that the length is more than twice the diameter of the circle. If this is an inscribed circle, the circle's diameter cannot be larger than the shortest side of a rectangle (the width). Because the width is less than 1/2 the length of the rectangle, the radius of the circle is at most 1/2 the width (could someone please confirm this reasoning)?

Using Test Cases:
W = 4 and L = 9, Area of rectangle = L*W = 36
Diameter = 4 and radius = 2, Area of circle = pi*(2)^2 = 3.14*4 = 12.56
This shows the circle occupies about 1/3 of the rectangle (not half)

I could also pick numbers that were very small for the length:
W = 4 and L = 8.0000001, Area of rectangle would be very close to 32
Diameter = 4 and radius = 2, Area of circle = pi*(2)^2 = 3.14*2 = 12.56
This shows the circle occupies about 3/8 of the circle (less than half)

Statement 1 is sufficient.

Statement 2:
Use Test Numbers that can easily be reduced by 25%.
L = 100 and W = 100 (all squares are rectangles)
In the original rectangle, the circle's area would be (100/2)^2 = 50^2 = 2500*3.14 = 7850

Reduce L and W by 25%
New area of rectangle: 75*75 = 7*8 = 5625

Here, I realized that the circle I picked doesn't work for statement 2. It's area (7850) is bigger than the area of the new rectangle (5625).

I went back and used the area of the new rectangle as the limits of the circle.
If width = 75, then diameter = 75 and radius = 75/2 = 37.5.
37.5 * 3.14 = 117.75

This is much less than half of the new rectangle (117.75/5625) = about 2%

Statement 2 is sufficient. Answer is D.

Could someone please look through my steps and help me understand the most efficient way of addressing this problem? How do I get an always yes answer for statement 2? I got stuck on statement 2 after I realized that the circle's area was bigger than that of the new rectangle.
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