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by GmatMathPro » Sun Dec 18, 2011 7:34 am
Let the lengths of the sides be x and y. We want to find the area, which is x*y.

The perimeter is 340, so 2x+2y=340 or x+y=170.

Also, from pythagorean theorem, x^2+y^2=130^2

We're looking for the value of x*y, so we want to find a way to get that into the mix. If you remember that (x+y)^2=x^2+2xy+y^2, then you'll see that you can introduce this term by squaring both sides of the perimeter equation:

(x+y)^2=170^2 ---> x^2 + 2xy + y^2 =170^2

Now, subtract the equation x^2+y^2=130 to get 2xy=170^2-130^2

2xy=(170+130)(170-130) (factoring by difference of squares)

2xy=300(40)

2xy=12000

xy=6000

aditya.j wrote:A rectangular park has a perimeter of 340 feet and a diagonal measurement of 130 feet.What is its area, in square feet?



2500

1440

6000

7040

8080
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by Brent@GMATPrepNow » Sun Dec 18, 2011 7:41 am
aditya.j wrote:A rectangular park has a perimeter of 340 feet and a diagonal measurement of 130 feet.What is its area, in square feet?

2500
1440
6000
7040
8080
Let x and y equal the length and width of the rectangle.
1. perimeter=340 --> 2x+2y=340 --> x+y=170
2. diagonal=130 --> x^2 + y^2 = 130^2 (from Pythagoras)
Our goal is to find the value of xy (since this equals the area of the rectangle)

If we take x+y=170 and square both sides we get (x+y)^2 = 170^2
If we expand this, we get x^2 + 2xy + y^2 = 170^2

Here's the trick: Notice that we have x^2 + y^2 on the left-hand-side.

From 2, we know that x^2 + y^2 = 130^2
So, we'll take x^2 + 2xy + y^2 = 170^2 and replace x^2 + y^2 with 130^2 to get:

2xy + 130^2 = 170^2
This means that 2xy = 170^2 - 130^2
And this means that xy = (170^2 - 130^2)/2
Which means xy = 6000 = C

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by shankar.ashwin » Sun Dec 18, 2011 7:43 am
2 * (l+b) = 340

l+b = 170

Given diagonal = 130. ('l' and 'b' form a right triangle with diagonal 130 and l+b = 170)

Look for special triangles :

5-12-13 is a special triangle so 50-120-130 is also a right triangle with 50+120 = 170

therefore l=120 and b=50.

Area = 120*50 = 6000
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by Brent@GMATPrepNow » Sun Dec 18, 2011 7:48 am
shankar.ashwin wrote:2 * (l+b) = 340

l+b = 170

Given diagonal = 130. ('l' and 'b' form a right triangle with diagonal 130 and l+b = 170)

Look for special triangles :

5-12-13 is a special triangle so 50-120-130 is also a right triangle with 50+120 = 170

therefore l=120 and b=50.

Area = 120*50 = 6000
Awesome solution, shankar.ashwin!!

I didn't notice the 50-120-130 triangle.

Cheers,
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by tomada » Sun Dec 18, 2011 4:00 pm
We can assume that the other two sides are 50 and 120 when we're told that the diagonal is 130? I see that the dimensions satisfy the perimeter that we're given but, in general, could a triangle with a hypotenuse of 13 have legs of lengths other than 5 and 12?
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by GmatMathPro » Sun Dec 18, 2011 4:06 pm
tomada wrote:We can assume that the other two sides are 50 and 120 when we're told that the diagonal is 130? I see that the dimensions satisfy the perimeter that we're given but, in general, could a triangle with a hypotenuse of 13 have legs of lengths other than 5 and 12?
Yes, there are infinitely many right triangles with a hypotenuse of 13 that have legs of lengths other than 5 and 12, but in this case the constraint of the perimeter forces it to be a 5-12-13 triangle.
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by tomada » Sun Dec 18, 2011 4:13 pm
Thanks, Pete.
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by tomada » Sun Dec 18, 2011 4:19 pm
Thanks, Pete.
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