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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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06 Mar 2010, 05:21
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Question Stats:

70% (02:01) correct 30% (02:10) wrong based on 653 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Nov 2013, 01:59, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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06 Mar 2010, 05:45
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aljatar wrote:
Hi everyone, I am struggling with this one, found the answer but I am looking for a fast way to do it ?

Can anyone help ?

R.

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

[Reveal] Spoiler:
OA is A

Hi, and welcome to Gmat Club.

The question you posted can be solved as follows:
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: $$(x^2+2xy+y^2)-(x^2+y^2)=280^2-200^2$$ --> $$2xy=(280-200)(280+200)$$ --> $$2xy=80*480$$ --> $$xy=40*480=19200$$.

Hope it helps.
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14 May 2010, 03:04
I have a small confusion. The diagonal is given and it divides the rectangle into two 30-60-90 triangles. Can't we find the measure of two other sides? What am i missing here???

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14 May 2010, 09:44
great explanation thanks

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14 May 2010, 10:14
bibha wrote:
I have a small confusion. The diagonal is given and it divides the rectangle into two 30-60-90 triangles. Can't we find the measure of two other sides? What am i missing here???
Does the diagonal of a rectangle always divide it into two 30-90-60 triangles ? Think again...!!!
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26 Dec 2010, 01:10
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The fastest way:

we know 560 is an integer (no fraction) and hence probability of sides of rectangle being integer is quite high
200 is diagonal - recognizing it from pythagorean patterns it seems to be a multiple of 10 (10 - 8 - 6)
= 10 * 2 * 10
hence other sides of the pythagorean triplet will be:
8 * 2 * 10
and
6 * 2 * 10
= 160
= 120
bingo - (160 + 120 ) * 2 = 560
hence area = 160 * 120 = 19200

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21 Nov 2013, 14:06
sleekmover wrote:
The fastest way:

we know 560 is an integer (no fraction) and hence probability of sides of rectangle being integer is quite high
200 is diagonal - recognizing it from pythagorean patterns it seems to be a multiple of 10 (10 - 8 - 6)
= 10 * 2 * 10
hence other sides of the pythagorean triplet will be:
8 * 2 * 10
and
6 * 2 * 10
= 160
= 120
bingo - (160 + 120 ) * 2 = 560
hence area = 160 * 120 = 19200

Yeah there has to be a shorter way to solve this one but where did you get the 10*2*10 stuff for each side on the (10 - 8 - 6)?

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26 Mar 2014, 09:05
1
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hideyoshi wrote:
bibha wrote:
I have a small confusion. The diagonal is given and it divides the rectangle into two 30-60-90 triangles. Can't we find the measure of two other sides? What am i missing here???
Does the diagonal of a rectangle always divide it into two 30-90-60 triangles ? Think again...!!!

No I don't think so buddy, there's nothing clear when it is a rectangle. If its a square I'm pretty sure that the two right triangles that are divided by the diagonal are in fact two 45-45-90 triangles, but with the rectangle I don't think one can be sure about the angle

Hope this helps
Cheers
J

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

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11 Dec 2014, 04:47
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It is a really good exercise.

Solving this one in a standard way brings a lot of confusion.
In my opinion it is a helpful here to know, that a square has a biggest area if the sum of length of bases is the same.
For example: If sum of bases is 8, the biggest possible area is 16, which is a square. (Other options 5x3=15; 6x2=12; 8x1=8)

Knowing this, we could easily eliminate answer choices that are other 200 because we know that diagonal is 200 and therefore the maximum area is 200^2/2
Moreover, knowing the rule of square diagonal we could remove 200, because square diagonal would be base * sqrt(2) (90-45-45 formula)

And also 196 is a (14*14) so this means that diagonal is not integer and we could remove it also.

Hence, one answer choice is left:)

A :D

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

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11 Dec 2014, 19:53
4
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Say length of the park = x, width = (280-x)

$$200^2 = x^2 + (280-x)^2$$

$$2x^2 - 560x + (280^2 - 200^2) = 0$$

$$2x^2 - 560x + (280 + 200)(280 - 200) = 0$$

$$2x^2 - 560x + 480 * 80 = 0$$

$$x^2 - 280x + 480*40 = 0$$

Dimensions = 160 & 120

Area = 19200

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

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02 Feb 2016, 18:56
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Bunuel wrote:
aljatar wrote:
Hi everyone, I am struggling with this one, found the answer but I am looking for a fast way to do it ?

Can anyone help ?

R.

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

[Reveal] Spoiler:
OA is A

Hi, and welcome to Gmat Club.

The question you posted can be solved as follows:
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: $$(x^2+2xy+y^2)-(x^2+y^2)=280^2-200^2$$ --> $$2xy=(280-200)(280+200)$$ --> $$2xy=80*480$$ --> $$xy=40*480=19200$$.

Hope it helps.

great solution, thanks. just wondering, why did you decide to square $$x+y=280$$? did you do it so that you could subtract $$x^2+y^2=200^2$$ from it?

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

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02 Feb 2016, 22:09
nycgirl212 wrote:
Bunuel wrote:
aljatar wrote:
Hi everyone, I am struggling with this one, found the answer but I am looking for a fast way to do it ?

Can anyone help ?

R.

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

[Reveal] Spoiler:
OA is A

Hi, and welcome to Gmat Club.

The question you posted can be solved as follows:
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: $$(x^2+2xy+y^2)-(x^2+y^2)=280^2-200^2$$ --> $$2xy=(280-200)(280+200)$$ --> $$2xy=80*480$$ --> $$xy=40*480=19200$$.

Hope it helps.

great solution, thanks. just wondering, why did you decide to square $$x+y=280$$? did you do it so that you could subtract $$x^2+y^2=200^2$$ from it?

Yes. I wanted to get xy in the end.
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Re: A small, rectangular park has a perimeter of 560 feet and a [#permalink]

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03 Feb 2016, 20:50
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you can avoid a lot of work in this problem by recognizing that, with the info provided, the diagonal forms a triangle inside the rectangle with sides that have a 3:4:5 ratio.

diagonal = 200
2x + 2y = 560, or x + y = 280
a^2 + b^2 = c^2 for each the sides of the triangle

using the ratio 3:4:5 for sides, and knowing c = 200, you can deduce the following

a=120
b=160

160x120=19,200

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

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26 Mar 2017, 12:54
mbaboop wrote:
you can avoid a lot of work in this problem by recognizing that, with the info provided, the diagonal forms a triangle inside the rectangle with sides that have a 3:4:5 ratio.

diagonal = 200
2x + 2y = 560, or x + y = 280
a^2 + b^2 = c^2 for each the sides of the triangle

using the ratio 3:4:5 for sides, and knowing c = 200, you can deduce the following

a=120
b=160

160x120=19,200

as the user above, how can you deduce that the ratio of the the sides is 3:4:5 ? (rather than just assuming and hoping to be right)
is it always the case? will a diagonal cut the rectangle in these proportions?

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

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08 Aug 2017, 21:57
Bunuel wrote:
aljatar wrote:
Hi everyone, I am struggling with this one, found the answer but I am looking for a fast way to do it ?

Can anyone help ?

R.

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

[Reveal] Spoiler:
OA is A

Hi, and welcome to Gmat Club.

The question you posted can be solved as follows:
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: $$(x^2+2xy+y^2)-(x^2+y^2)=280^2-200^2$$ --> $$2xy=(280-200)(280+200)$$ --> $$2xy=80*480$$ --> $$xy=40*480=19200$$.

Hope it helps.

I just need to say that I love the elegance of this answer. It's beautiful... Thanks, Bunuel.

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Re: A small, rectangular park has a perimeter of 560 feet and a   [#permalink] 08 Aug 2017, 21:57
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