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# A small, rectangular park has a perimeter of 560 feet and a diagonal m

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Manager
Joined: 26 Mar 2008
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A small, rectangular park has a perimeter of 560 feet and a diagonal m [#permalink]

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27 Sep 2008, 13:05
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45% (medium)

Question Stats:

75% (01:41) correct 25% (02:47) wrong based on 110 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-small-rectangular-park-has-a-perimeter-of-560-feet-and-a-91118.html
[Reveal] Spoiler: OA

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VP
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Re: A small, rectangular park has a perimeter of 560 feet and a diagonal m [#permalink]

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27 Sep 2008, 13:45
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arorag 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

2 (l+w) = 560 means l+w = 280

Sqrt ( l^2 + w^2) = 200 means (l2^+w^2) = 40000

square both sides of (l+w) =280

l^2 + w^2 +2lw = 78400

means 2lw= 38400 lw=19,200

Hence A

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Senior Manager
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Re: A small, rectangular park has a perimeter of 560 feet and a diagonal m [#permalink]

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27 Sep 2008, 13:45
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A:

I thought the 200 as the hypotenuse was a little 'fishy', so I started thinking of it as a 3-4-5 triangle: 200 as the hypotenuse, 120 and 160 as the legs and it worked: 120*160 = 19,200.

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Manager
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Re: A small, rectangular park has a perimeter of 560 feet and a diagonal m [#permalink]

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27 Sep 2008, 14:08
same here:

a+b = 280

(a+b)(a+b) =78'400

a^2+2ab+b^2 = 78'400

a^2+b^2 = 200^2 = 40'000

2ab = 38'400

ab = 19'200

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

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13 Nov 2014, 13:03
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Re: A small, rectangular park has a perimeter of 560 feet and a diagonal m [#permalink]

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13 Nov 2014, 13:09
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arorag 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

Given:
(1) $$2x+2y=560$$ (perimeter) --> $$x+y=280$$
(2) $$x^2+y^2=200^2$$ (diagonal, as per Pythagoras).

Question: $$xy=?$$

Square (1) --> $$x^2+2xy+y^2=280^2$$. Now subtract (2) fro this: $$(x^2+2xy+y^2)-(x^2+y^2)=280^2-200^2$$ --> $$2xy=(280-200)(280+200)$$ --> $$2xy=80*480$$ --> $$xy=40*480=19200$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-small-rectangular-park-has-a-perimeter-of-560-feet-and-a-91118.html
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Kudos [?]: 132821 [4], given: 12378

Re: A small, rectangular park has a perimeter of 560 feet and a diagonal m   [#permalink] 13 Nov 2014, 13:09
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