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13 Feb 2020, 07:38
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In the figure attached, the length of the large rectangle is 12 inches and the width of the large rectangle is 10 inches. The area of the unshaded rectangle is equal to the area of the shaded region. If the ratio of the length to the width of the unshaded rectangle is equal to the ratio of the length to the width of the large rectangle, what is the length of the unshaded rectangle, in inches?
A) 5
B) 6
C) \(5\sqrt{2}\)
D) \(6\sqrt{2}\)
E) 10
Attachment:
217826.18.GMC74351img01.gif [ 5.56 KiB | Viewed 4265 times ]
13 Feb 2020, 08:30
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We're dividing the big rectangle of area 120 into two regions of equal area. So the area of those two regions must each be 60. So we have a small rectangle of area 60, and it is similar to the large rectangle. If you know that when similar shapes have areas in a 2 to 1 ratio, their sides will be in a √2 to 1 ratio, you can then answer the question almost immediately. But that principle might not be familiar to everyone, and you can also solve algebraically.
The small rectangle's sides are in a 12 to 10, or 6 to 5 ratio. So the width of this rectangle is 5/6 the length, and if L is its length, then 5L/6 is the width. Since its area is 60, we have
L * (5L/6) = 60
L^2 = 360/5 = 72
L = √72 = √36 * √2 = 6 √2
The small rectangle's sides are in a 12 to 10, or 6 to 5 ratio. So the width of this rectangle is 5/6 the length, and if L is its length, then 5L/6 is the width. Since its area is 60, we have
L * (5L/6) = 60
L^2 = 360/5 = 72
L = √72 = √36 * √2 = 6 √2
13 Feb 2020, 08:40
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Bunuel
Let the sides of smaller rectangle be 12k & 10k where 0<k<1
12*10 - 12k*10k = 12k*10k
1 = 2k^2
k = 1/sqrt(2)
Length of smaller rectangle = 12/sqrt(2) = 6*sqrt(2)
IMO D
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