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17 Sep 2018, 22:02
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Difficulty:
75%
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Question Stats:
57% (04:08) correct
43%
(04:00)
wrong
based on 14
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The shaded region in the figure above represents a rectangular frame with length 12 inches and width 8 inches. The frame encloses a rectangular picture that has same area as the frame itself. If the diagonal of the picture represents 1/2 of the diagonal distance from one outer corner of the frame to the opposite corner, what is the perimeter of the picture, in inches?
A. 30
B. 20√2
C. 2√52
D. 109
E. √52
Attachment:
image024.gif
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27 Sep 2018, 01:10
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Bunuel Can you please provide an explanation?
27 Sep 2018, 03:58
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Indeed, I don't find a result that fits with your possible solutions.
Let l = length of picture and w = width of picture
area of the frame = area of the big rectangle - area of the picture = 12*8 - l*w
We're told that the picture has the same area as the frame:
l*w = 12*8 - l*w
2*l*w = 12*8
l*w = 6*8 = 48
Then we're told that the diagonal of the picture is half the length of the diagonal of the frame:
diagonal of frame = \(\sqrt{12^2 + 8^2} = \sqrt{208} = 2*\sqrt{52}\)
diagonal of picture = \(\sqrt{52}\)
or
l^2 + w^2 = 52
We have l*w and we have L^2 + w^2. Let's work with (l+w)^2.
(l+w)^2 = l^2 + w^2 + 2*w*l = 52 + 96 = 148
By applying a square root on both sides:
\(l+w = \sqrt{148} = \sqrt{4*37} = 2*\sqrt{37}\)
We need 2*(l+w) to find the perimeter of the picture. That would be \(4*\sqrt{37}\).
This seems too lengthy for a GMAT problem (maybe there is a shorter way?) but also doesn't match any of the possible solutions.
Let l = length of picture and w = width of picture
area of the frame = area of the big rectangle - area of the picture = 12*8 - l*w
We're told that the picture has the same area as the frame:
l*w = 12*8 - l*w
2*l*w = 12*8
l*w = 6*8 = 48
Then we're told that the diagonal of the picture is half the length of the diagonal of the frame:
diagonal of frame = \(\sqrt{12^2 + 8^2} = \sqrt{208} = 2*\sqrt{52}\)
diagonal of picture = \(\sqrt{52}\)
or
l^2 + w^2 = 52
We have l*w and we have L^2 + w^2. Let's work with (l+w)^2.
(l+w)^2 = l^2 + w^2 + 2*w*l = 52 + 96 = 148
By applying a square root on both sides:
\(l+w = \sqrt{148} = \sqrt{4*37} = 2*\sqrt{37}\)
We need 2*(l+w) to find the perimeter of the picture. That would be \(4*\sqrt{37}\).
This seems too lengthy for a GMAT problem (maybe there is a shorter way?) but also doesn't match any of the possible solutions.
