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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

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).

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???

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...!!!
_________________

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

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)?

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

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.

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 ?

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

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).

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

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).

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

Show Tags

03 Feb 2016, 20:50

2

This post received KUDOS

3

This post was BOOKMARKED

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

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?

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 ?

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

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).