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Manager
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A rectangular garden has a perimeter of 560 feet. If the [#permalink]
 12 Jul 2006, 06:15
A rectangular garden has a perimeter of 560 feet. If the garden has a diagonal fence of 200 feet, what is the area of the garden, in square feet?
A) 19,200
B) 19'600
C) 20'000
D) 20'400
E) It cannot be calculated with the information given.
Please explain.
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Director
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You know that a+b=280, c=200 , a^2+2ab+b^2=280^2, use pitagoras to see that a^2+b^2=40000 2ab=38 400 and area is 19 200
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VP
Joined: 02 Jun 2006
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A) 19,200
Given 2(l+w) = 560 => (l+w) = 280
Diagonal = 200
i.e. 200 x 200 = l^2 + w^2
i.e 200x200 = (l+w)^2 - 2lw
i.e. 280x280 - 200x200 = 2lw
i.e 480 x 80 /2 = lw
i.e. 19,200 = lw
Answer : A
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Manager
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It is A indeed.
Thank you guys.
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GMAT Club Legend
Joined: 07 Jul 2004
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Let one side of the rectangle be x ft
The other side will be (560-2x)/2 = 280-x ft
So x^2 + (280-x)^2 = 200^2
x^2 + 280^2 - 560x + x^2 = 200^2
2x^2 - 560x + 38400 = 0
x^2 - 280x + 19200 = 0
x = 140 +/- 20 --> x can be 120 or 160
If x = 120, the other side = 160 --> Area = 19200
If x = 160, the other side = 120 --> Area = 19200
So both gives us an area of 19200
Ans A
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SVP
Joined: 30 Mar 2006
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A 19200
Perimeter = 560
2(l+b) = 560
l+b = 280
(l+b)^2 = 280^2
l^2 + b^2 + 2lb = 280^2
Also diagonal = 200
l^2 + b^2 = 200^2
Substituting
280^2 - 2lb = 200^2
280^2 - 200^2 = 2lb
Lb = 19200
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