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# A rectangular park has a perimeter of 560 feet and a diagonal measurem

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Math Expert
Joined: 02 Sep 2009
Posts: 52231
A rectangular park has a perimeter of 560 feet and a diagonal measurem  [#permalink]

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16 Oct 2018, 00:19
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Difficulty:

45% (medium)

Question Stats:

65% (01:34) correct 35% (02:18) wrong based on 77 sessions

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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

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

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18 Oct 2018, 17:29
4
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

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##### General Discussion
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Joined: 23 May 2018
Posts: 332
Location: Pakistan
GMAT 1: 770 Q48 V50
GPA: 3.4
Re: A rectangular park has a perimeter of 560 feet and a diagonal measurem  [#permalink]

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16 Oct 2018, 01:20
3
Assuming that the 560 feet perimeter of the park was a square, we get that one side would be 140 feet. For any parallelogram (of equal perimeter), square has the greatest area.

With 140feet for one side, we get the area 19600 sq ft.

The park, however, is a rectangle and will thus have an area less than a square.

We have only one option for this.

A. 19200
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Re: A rectangular park has a perimeter of 560 feet and a diagonal measurem  [#permalink]

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17 Oct 2018, 01:15
MsInvBanker wrote:
Assuming that the 560 feet perimeter of the park was a square, we get that one side would be 140 feet. For any parallelogram (of equal perimeter), square has the greatest area.

With 140feet for one side, we get the area 19600 sq ft.

The park, however, is a rectangle and will thus have an area less than a square.

We have only one option for this.

A. 19200

You assume that a square cannot be called 'rectangle'. Is it legit in GMAT?
Intern
Joined: 01 Jul 2018
Posts: 7
A rectangular park has a perimeter of 560 feet and a diagonal measurem  [#permalink]

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27 Oct 2018, 06:36
Let Length be L and breadth be B

Diagonal = 200
L^2+ B^2 =40000

Perimeter= 560
2(L+B)=560

(L+B)^2= L^2+B^2 +2LB
solving the only unknown LB (i.e area)
LB=19200

(All area in Sq. feet and length,breath and diagonal are in feet)
Intern
Joined: 17 Jun 2017
Posts: 34
Re: A rectangular park has a perimeter of 560 feet and a diagonal measurem  [#permalink]

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27 Oct 2018, 21:21
MsInvBanker wrote:
Assuming that the 560 feet perimeter of the park was a square, we get that one side would be 140 feet. For any parallelogram (of equal perimeter), square has the greatest area.

With 140feet for one side, we get the area 19600 sq ft.

The park, however, is a rectangle and will thus have an area less than a square.

We have only one option for this.

A. 19200

You assumed this ?? How did you assume it to be a square in the first place ??
Manager
Joined: 20 Jun 2018
Posts: 73
Location: India
Re: A rectangular park has a perimeter of 560 feet and a diagonal measurem  [#permalink]

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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

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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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

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 rectangular park has a perimeter of 560 feet and a diagonal measurem &nbs [#permalink] 30 Oct 2018, 18:05
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