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Two rectangles are drawn such that they share a center and that the sh

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New post 14 Sep 2018, 01:25
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Two rectangles are drawn such that they share a center and that the shaded rectangle is 2 inches from the edge of the larger rectangle as shown above. If the perimeter of the shaded rectangle above is 27 inches, then the area of the larger rectangle is how many square inches more than the area of the shaded one?


A. 16
B. 53
C. 54
D. 70
E. 108

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image040.jpg [ 5.48 KiB | Viewed 325 times ]

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Re: Two rectangles are drawn such that they share a center and that the sh  [#permalink]

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New post 14 Sep 2018, 01:52
Bunuel wrote:
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Two rectangles are drawn such that they share a center and that the shaded rectangle is 2 inches from the edge of the larger rectangle as shown above. If the perimeter of the shaded rectangle above is 27 inches, then the area of the larger rectangle is how many square inches more than the area of the shaded one?


A. 16
B. 53
C. 54
D. 70
E. 108

Attachment:
image040.jpg


Let the length and breadth of the shaded rectangle be l and b respectively.
length and breadth of the larger rectangle = l+4 and b+4 respectively.
Perimeter of shaded rectangle =2(l+b) = 27

Difference of areas = (l+4)(b+4)-lb
=lb+4(l+b)+16-lb = 2x27+16 = 70 [ putting value of 2(l+b) = 27]
Answer D.
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Two rectangles are drawn such that they share a center and that the sh  [#permalink]

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New post 14 Sep 2018, 11:28
Length of shaded rectangle = L
Breadth of shaded rectangle = B
Given:
Perimeter of shaded rectangle = 2(L + B) = 27
Length of outer rectangle = L + 4
Breadth of outer rectangle = B + 4

We need to find
= Area of outer rectangle - Area of shaded region
= \((L + 4) * (B + 4) - LB\)
= \(LB + 4L + 4B + 16 - LB\)
= \(4(L + B) + 16\)
= \(2 * 2(L + B) + 16\)
= \(2 * 27 + 16\)
= 70

Answer: D
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Two rectangles are drawn such that they share a center and that the sh   [#permalink] 14 Sep 2018, 11:28
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