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A small, rectangular park has a perimeter of 560 feet and 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 18 Oct 2018, 17:29
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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, 20:11
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.
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 30 Oct 2018, 18:05
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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Re: A small, rectangular park has a perimeter of 560 feet and a  [#permalink]

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New post 13 Sep 2019, 14:01
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, 06:51
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Re: A small, rectangular park has a perimeter of 560 feet and a   [#permalink] 23 Sep 2019, 06:51

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