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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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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
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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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Re: Rectangular park [#permalink]
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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 ? Thanks in advance 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 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: \((x^2+2xy+y^2)(x^2+y^2)=280^2200^2\) > \(2xy=(280200)(280+200)\) > \(2xy=80*480\) > \(xy=40*480=19200\). Answer: A. Hope it helps.
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Re: Rectangular park [#permalink]
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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 306090 triangles. Can't we find the measure of two other sides? What am i missing here???



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Re: Rectangular park [#permalink]
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14 May 2010, 09:44
great explanation thanks



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Re: Rectangular park [#permalink]
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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 306090 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 309060 triangles ? Think again...!!!
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Re: Rectangular park [#permalink]
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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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Re: Rectangular park [#permalink]
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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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Re: Rectangular park [#permalink]
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hideyoshi wrote: bibha wrote: I have a small confusion. The diagonal is given and it divides the rectangle into two 306090 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 309060 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 454590 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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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) (904545 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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Say length of the park = x, width = (280x) \(200^2 = x^2 + (280x)^2\) \(2x^2  560x + (280^2  200^2) = 0\) \(2x^2  560x + (280 + 200)(280  200) = 0\) \(2x^2  560x + 480 * 80 = 0\) \(x^2  280x + 480*40 = 0\) Dimensions = 160 & 120 Area = 19200 Answer = A
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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 ? Thanks in advance 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 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: \((x^2+2xy+y^2)(x^2+y^2)=280^2200^2\) > \(2xy=(280200)(280+200)\) > \(2xy=80*480\) > \(xy=40*480=19200\). Answer: A. 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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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 ? Thanks in advance 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 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: \((x^2+2xy+y^2)(x^2+y^2)=280^2200^2\) > \(2xy=(280200)(280+200)\) > \(2xy=80*480\) > \(xy=40*480=19200\). Answer: A. 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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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
A is the answer.



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