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

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Target Test Prep Representative
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Re: A rectangular park has a perimeter of 560 feet and a diagonal measurem  [#permalink]

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New post 18 Oct 2018, 18:29
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2
Bunuel wrote:
A rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

(A) 19200
(B) 19600
(C) 20000
(D) 20400
(E) 20800


We can let L = the length of the park and W = the width of the park. Our goal is to find the area of the park, i.e., the value of LW.

We can create the equations:

Perimeter = 560

2L + 2W = 560

L + W = 280

and

L^2 + W^2 = 200^2

Squaring the first equation, we have:

(L + W)^2 = 280^2

L^2 + W^2 + 2LW = 280^2

Substituting, we have:

200^2 + 2LW = 280^2

2LW = 280^2 - 200^2

2LW = (280 - 200)(280 + 200)

2LW = 80 x 480

LW = 40 x 480 = 19,200

Answer: A
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Re: A rectangular park has a perimeter of 560 feet and a diagonal measurem  [#permalink]

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New post 28 Oct 2018, 21:11
ScottTargetTestPrep wrote:
Bunuel wrote:
A rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

(A) 19200
(B) 19600
(C) 20000
(D) 20400
(E) 20800


We can let L = the length of the park and W = the width of the park. Our goal is to find the area of the park, i.e., the value of LW.

We can create the equations:

Perimeter = 560

2L + 2W = 560

L + W = 280

and

L^2 + W^2 = 200^2

Squaring the first equation, we have:

(L + W)^2 = 280^2

L^2 + W^2 + 2LW = 280^2

Substituting, we have:

200^2 + 2LW = 280^2

2LW = 280^2 - 200^2

2LW = (280 - 200)(280 + 200)

2LW = 80 x 480

LW = 40 x 480 = 19,200

Answer: A





Shouldn't L square plus W square be equal to 200 rather than being equal to 200 square? Please assist.
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Re: A rectangular park has a perimeter of 560 feet and a diagonal measurem  [#permalink]

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New post 30 Oct 2018, 19:05
1
Shbm wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
A rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

(A) 19200
(B) 19600
(C) 20000
(D) 20400
(E) 20800


We can let L = the length of the park and W = the width of the park. Our goal is to find the area of the park, i.e., the value of LW.

We can create the equations:

Perimeter = 560

2L + 2W = 560

L + W = 280

and

L^2 + W^2 = 200^2

Squaring the first equation, we have:

(L + W)^2 = 280^2

L^2 + W^2 + 2LW = 280^2

Substituting, we have:

200^2 + 2LW = 280^2

2LW = 280^2 - 200^2

2LW = (280 - 200)(280 + 200)

2LW = 80 x 480

LW = 40 x 480 = 19,200

Answer: A





Shouldn't L square plus W square be equal to 200 rather than being equal to 200 square? Please assist.



200^2 is correct since it’s based on the Pythagorean theorem: a^2 + b^2 = c^2. Here, L^2 + W^2 = D^2 where D is the diagonal. Notice that on a rectangle, the short and long side together with the diagonal form a right triangle; that’s why the Pythagorean theorem is applicable.
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Scott Woodbury-Stewart

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Scott@TargetTestPrep.com
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122 Reviews

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self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

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GMAT 5: 650 Q48 V31
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Re: A small, rectangular park has a perimeter of 560 feet and a  [#permalink]

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New post 13 Sep 2019, 15:01
2l+2w = 560 (as we are told)
l+w = 280

(l+w)^2 = 280^2
l^2+ 2lw +w^2 = 280^2
we know the diagonal
d = root(l^2+w^2)
200 =root(l^2+w^2)
200^2 = l^2 +w^2

sub in

2lw + 200^2 = 280^2
2lw =280^2-200^2
2lw = (280-200)(280+200)
lw = ((80)(480))/2
lw = 19,200
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Re: A small, rectangular park has a perimeter of 560 feet and a  [#permalink]

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New post 23 Sep 2019, 07:51
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Re: A small, rectangular park has a perimeter of 560 feet and a   [#permalink] 23 Sep 2019, 07:51

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