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Length of a Rectangle

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Length of a Rectangle

by pareekbharat86 » Wed Nov 13, 2013 8:02 am
A rectangular frame has a length of 18 inches and width of 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?

(A) 9 * sqrt2
(B) 3/2
(C) 9/sqrt2
(D) 15*(1-1/sqrt2)
(E) 9/2

OA is A.

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by Brent@GMATPrepNow » Wed Nov 13, 2013 8:28 am
pareekbharat86 wrote:A rectangular frame has a length of 18 inches and width of 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?

(A) 9√2
(B) 3/2
(C) 9/√2
(D) 15(1- 1/√2)
(E) 9/2
So, we have something like this
Image

The frame encloses a rectangular picture that has the same area as the frame itself.
The area of the 18 by 15 rectangle = (18)(15) = 270
Half of this area is frame and half is picture.
So, the area of the blue frame = 135, and the area of the green rectangle = 135
In other words, xy = 135


The length and width of the picture have the same ratio as the length and width of the frame
In other words, x/y = 18/15
We can rewrite this as: y = 15x/18

We now have two equations:
xy = 135
y = 15x/18

Take the first equation and replace y with 15x/18 to get...
(x)(15x/18) = 135
Simplify: 15x²/18 = 135
Divide both sides by 15 to get: x²/18 = 9
Multiply both sides by 18 to get: x² = 162
Find square root of both sides to get: x = √162
Simplify: x = (√81)(√2)
Simplify: x = [spoiler]9√2 = A[/spoiler]

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by GMATGuruNY » Wed Nov 13, 2013 10:39 am
pareekbharat86 wrote:A rectangular frame has a length of 18 inches and width of 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?

(A) 9 * sqrt2
(B) 3/2
(C) 9/sqrt2
(D) 15*(1-1/sqrt2)
(E) 9/2
The frame encloses a rectangular picture that has the same area as the frame itself.
Thus, the picture's area is equal to half of the TOTAL area:
(1/2) * 18 * 15 = 135.

The length and width of the picture have the same ratio as the length and width of the frame.
L:W = 18:15 = 6:5 = 12:10.
If L=12 and W=10, then the area of the picture = 12*10 = 120.
Since the actual area is 135 -- a bit more than 120 -- the actual value of L must be a bit more than 12.

Only A works:
9√2 ≈ 9 * 1.4 = 12.6

The correct answer is A.
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by fifafreak » Thu Nov 14, 2013 5:35 am
GMATGuruNY wrote:
pareekbharat86 wrote:A rectangular frame has a length of 18 inches and width of 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?

(A) 9 * sqrt2
(B) 3/2
(C) 9/sqrt2
(D) 15*(1-1/sqrt2)
(E) 9/2
The frame encloses a rectangular picture that has the same area as the frame itself.
Thus, the picture's area is equal to half of the TOTAL area:
(1/2) * 18 * 15 = 135.

The length and width of the picture have the same ratio as the length and width of the frame.
L:W = 18:15 = 6:5 = 12:10.
If L=12 and W=10, then the area of the picture = 12*10 = 120.
Since the actual area is 135 -- a bit more than 120 -- the actual value of L must be a bit more than 12.

Only A works:
9√2 ≈ 9 * 1.4 = 12.6

The correct answer is A.
Isn't that an incorrect way of framing a question. It says the frame has L =18 W = 15. So, area of frame has to be 18*15. hence the term "area of frame = that of picture" doesn't stand. My question is - In GMAT will questions be asked on similar lines, or it will be specified that area of left over area in the frame = area of the picture ... [confused]
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by [email protected] » Thu Nov 14, 2013 1:23 pm
Hi fifafreak,

While the frame has dimensions of 18 x 15, that DOES not mean that its area is 18 x 15. It's a frame, so there's a "hole" in it where the picture would go, thus the area of the frame is something LESS THAN 18 x 15.

We're told that the area of the picture = area of the frame. This tells us that the frame area + picture area = 18 x 15.

These types of geometry questions are always specifically worded, so you have to pay careful attention to the details (drawing a picture usually helps), but you won't see that many on the actual GMAT (maybe 1, possibly 0).

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by pareekbharat86 » Fri Nov 15, 2013 8:19 pm
GMATGuruNY wrote:
pareekbharat86 wrote:A rectangular frame has a length of 18 inches and width of 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?

(A) 9 * sqrt2
(B) 3/2
(C) 9/sqrt2
(D) 15*(1-1/sqrt2)
(E) 9/2
The frame encloses a rectangular picture that has the same area as the frame itself.
Thus, the picture's area is equal to half of the TOTAL area:
(1/2) * 18 * 15 = 135.

The length and width of the picture have the same ratio as the length and width of the frame.
L:W = 18:15 = 6:5 = 12:10.
If L=12 and W=10, then the area of the picture = 12*10 = 120.
Since the actual area is 135 -- a bit more than 120 -- the actual value of L must be a bit more than 12.

Only A works:
9√2 ≈ 9 * 1.4 = 12.6

The correct answer is A.
I still don't get it. Why are we halving the area?
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by [email protected] » Sat Nov 16, 2013 12:27 am
Hi Bharat,

The question tells us that we have a frame and a picture within the frame. The total area of EVERYTHING is 18in x 15in = 270 sq. inches. We're also told that the frame has the SAME AREA as the picture. In math terms:

Area of Frame = Area of Picture
Area of Frame + Area of Picture = 270

Using substitution, you'll see that each Area = 135. This is HALF of the total (270).

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by pareekbharat86 » Sat Nov 16, 2013 1:12 am
[email protected] wrote:Hi Bharat,

The question tells us that we have a frame and a picture within the frame. The total area of EVERYTHING is 18in x 15in = 270 sq. inches. We're also told that the frame has the SAME AREA as the picture. In math terms:

Area of Frame = Area of Picture
Area of Frame + Area of Picture = 270

Using substitution, you'll see that each Area = 135. This is HALF of the total (270).

GMAT assassins aren't born, they're made,
Rich
Hi Rich.C,

Thanks for the explanation. I could have never been able to interpret that Area of Frame+Area of Picture= 270. Its certainly not a straight forward question. Thanks again.
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by Zach.J.Dragone » Sat Nov 30, 2013 1:55 pm
I know this may be stupid but:

We now have two equations:
xy = 135
y = 15x/18

Why is that? In other words, why do we set up the equation with those numbers? I see where they come from but I don't know why we use them. Why can't we set up the frame equation as 6x*5y=135?
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