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# Determining Area

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Determining Area [#permalink]  17 Aug 2010, 18:21
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FIGURE: Imagine a rectangle within a rectangle. Every side of the interior rectangle is 2 inches from the congruous side of the larger rectangle.

A rectangular picture is surrounded by a border, as shown in the figure above. Without the border the length of the picture is twice the width. If the area of the border is 196 square inches, what is the length, in inches, of the picture, excluding the border?

A. 10
B. 15
C. 30
D. 40
E. 60
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Re: Determining Area [#permalink]  17 Aug 2010, 18:54
kwhitejr wrote:
FIGURE: Imagine a rectangle within a rectangle. Every side of the interior rectangle is 2 inches from the congruous side of the larger rectangle.

A rectangular picture is surrounded by a border, as shown in the figure above. Without the border the length of the picture is twice the width. If the area of the border is 196 square inches, what is the length, in inches, of the picture, excluding the border?

A. 10
B. 15
C. 30
D. 40
E. 60

Length of the outer rectangle - L, Breadth of the outer rectangle - B.
Length of the inner rectangle - l, Breadth of the inner rectangle - b.

Now given that - L*B = 196 sq. unit and l = 2b.

Also l = L - 4 and b = B - 4

Hence area of the outer rectangle (border area) = 196 - Area of the inner rectangle

(L * B) = 196 - (l*b)

((l+4) * (b+4)) = 196 - (l*b)

Simplifying => (lb + 4l + 4b + 16) = (196 - lb)

(4l + 4b) = 180
(4(2b) + 4b) = 180
12b = 180

b = 15, l = 30.

Hence the length of the inner rectangle or picture is 30 units. Answer C.
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Re: Determining Area [#permalink]  17 Aug 2010, 19:15
Ans : C

Assume the length of inner rectangle as L and the width as w then L = 2W
so the length of outer rectange will be L+4 and width as W+4

The area of border is
area of outer rectangle - area of inner rectangle
[(L+4)*(W+4)]- [ L * W] =196

4L +4W +16 = 196
substituting L = 2W we get
12W = 180
W= 15
L = 2W = 30
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Re: Determining Area   [#permalink] 17 Aug 2010, 19:15
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