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The shaded region in the figure above represents a rectangul

Expert replies
by gmatter2012 » Wed Jul 11, 2012 3:45 am
The shaded region in the figure below represents a rectangular frame with length 18 inches and width 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 lenght and width of the frame, what is the length of the picture, in inches?


(A) 9*sqrt(2)
(B) 3/2
(C) 9/sqrt(2)
(D) 15(1-(1/2))
(E) 9/2
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Source: — Problem Solving |

by GMATGuruNY » Wed Jul 11, 2012 4:15 am
gmatter2012 wrote:The shaded region in the figure below represents a rectangular frame with length 18 inches and width 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 lenght and width of the frame, what is the length of the picture, in inches?


(A) 9*sqrt(2)
(B) 3/2
(C) 9/sqrt(2)
(D) 15(1-(1/2))
(E) 9/2
Total area = 18*15 = 270.
Since the frame and the picture have equal areas, the area of the picture = (1/2)270 = 135.

In the frame, height : width = 18:15 = 6:5.
Since the dimensions of the picture are in the same ratio, let the length of the picture = 6x and the width of the picture = 5x.
Thus:
(6x)(5x) = 135
x² = 9/2
x = 3/√2.

The length of the picture = 6x = 6(3/√2) = 18/√2 * √2/√2 = (18√2)/2 = 9√2.

The correct answer is A.
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by Halimah_O » Sat Dec 10, 2016 1:21 am
GMATGuruNY wrote:
gmatter2012 wrote:The shaded region in the figure below represents a rectangular frame with length 18 inches and width 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 lenght and width of the frame, what is the length of the picture, in inches?


(A) 9*sqrt(2)
(B) 3/2
(C) 9/sqrt(2)
(D) 15(1-(1/2))
(E) 9/2
Total area = 18*15 = 270.
Since the frame and the picture have equal areas, the area of the picture = (1/2)270 = 135.

In the frame, height : width = 18:15 = 6:5.
Since the dimensions of the picture are in the same ratio, let the length of the picture = 6x and the width of the picture = 5x.
Thus:
(6x)(5x) = 135
x² = 9/2
x = 3/√2.

The length of the picture = 6x = 6(3/√2) = 18/√2 * √2/√2 = (18√2)/2 = 9√2.

The correct answer is A.
Dear Mitch,

I don't understand how you arrived at this "Total area = 18*15 = 270.
Since the frame and the picture have equal areas, the area of the picture = (1/2)270 = 135."

If the two rectangles have equal areas why did you divide by 2? Please help. Thank you.
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by regor60 » Sat Dec 10, 2016 11:49 am
Halimah_O wrote:
GMATGuruNY wrote:
gmatter2012 wrote:The shaded region in the figure below represents a rectangular frame with length 18 inches and width 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 lenght and width of the frame, what is the length of the picture, in inches?


(A) 9*sqrt(2)
(B) 3/2
(C) 9/sqrt(2)
(D) 15(1-(1/2))
(E) 9/2
Total area = 18*15 = 270.
Since the frame and the picture have equal areas, the area of the picture = (1/2)270 = 135.

In the frame, height : width = 18:15 = 6:5.
Since the dimensions of the picture are in the same ratio, let the length of the picture = 6x and the width of the picture = 5x.
Thus:
(6x)(5x) = 135
x² = 9/2
x = 3/√2.

The length of the picture = 6x = 6(3/√2) = 18/√2 * √2/√2 = (18√2)/2 = 9√2.

The correct answer is A.
Dear Mitch,

I don't understand how you arrived at this "Total area = 18*15 = 270.
Since the frame and the picture have equal areas, the area of the picture = (1/2)270 = 135."

If the two rectangles have equal areas why did you divide by 2? Please help. Thank you.
You're told that the frame encompasses an area that includes the area of the frame itself and the picture. So, the total area encompassed equals the sum of the two. You're also told that both those areas are equal. Meaning that each of them is therefore half of the total area encompassed.
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by [email protected] » Sat Dec 10, 2016 12:56 pm
Hi Halimah_O,

If you want to think of this algebraically, here's how to do it:

A = Area of the 'white' rectangle
B = Area of the 'frame'

We're told that the area of the entire shape is 18 inches by 15 inches, so...
(A) + (B) = (18)(15)

We're also told that the area of the rectangle is equal to the area of the frame...

A = B

So... A + B = 270 and A = B. Using substitution, we have...

A + B = 270
A + A = 270
2A = 270
A = 135

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Rich
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