A small, rectangular park has a perimeter of 560 feet and a : PS Archive
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# A small, rectangular park has a perimeter of 560 feet and a

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A small, rectangular park has a perimeter of 560 feet and a [#permalink]

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24 Apr 2006, 12:15
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A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A. 19,200
B. 19,600
C. 20,000
D. 20,400
E. 20,800
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24 Apr 2006, 12:24
M8 wrote:
A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A. 19,200
B. 19,600
C. 20,000
D. 20,400
E. 20,800

we want to find xy = ?

2x + 2y = 560
x+y = 280

x^2 + y^2 = (200)^2

(x+y)^2 - x^2 - y^2 = 2xy

so (280)^2 - (200)^2 = 2xy

= 38400 / 2 = 19200

so A
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24 Apr 2006, 12:28
2 * (l +b) = 560 so l +b =280

also l^2 +b^2 = 200 ^2

(l +b) ^2 = l^2 +b^2 + 2* l *b

so l*b = ( (l +b) ^2 - (l^2 +b^2 )) /2

((280) ^2 - (200)^2) /2

= 19,200
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25 Apr 2006, 10:45
Yes the OA is 'A' - 19200,
Both of you guys are correct, nice explanations, but why do you use this expression (l +b) ^2 to receive a quadratic equation?
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25 Apr 2006, 11:04
conocieur wrote:
(x+y)^2 - x^2 - y^2 = 2xy

ipc302 wrote:
(l +b) ^2 = l^2 +b^2 + 2* l *b

looks like you both are doing the same thing here. Where are you getting that from? just by multiplying out (l+w)^2?
Re: PS: Area.   [#permalink] 25 Apr 2006, 11:04
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