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Area of a rectangular park

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Area of a rectangular park

by crackgmat007 » Fri May 08, 2009 9:52 am
Q. 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

From above, I got l+b=280 and l^2+b^2=200^2. How do I solve to get 19,200? Pls help.
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Re: Area of a rectangular park

by dtweah » Fri May 08, 2009 11:21 am
crackgmat007 wrote:Q. 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

From above, I got l+b=280 and l^2+b^2=200^2. How do I solve to get 19,200? Pls help.
You don't need to solve this complex equation. You will have to do quadratic equation to get result which is time consuming plus you could make error. When ever the diagonal of a rectangle is an integer, you can generate the L and W from what you know about 3 4 5 right triangles. Same applies to the hypotenuse of any right triangle.

3 4 5 Double these values to infinity and you generate all possible right triangles with integer values
6 8 10
12 16 20 Stop
multiple by 10
120 160 200

120 x 160 = 19200.
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by vitaly » Fri May 08, 2009 12:58 pm
>From above, I got l+b=280 and l^2+b^2=200^2. How do I solve to get 19,200? Pls help.

This is easy. Since (l+b)^2=l^2+2lb+b^2

=> 280^2=l^2+2lb+b^2=200^2+2lb
=> 2lb = 280^2-200^2
=> lb = (280-200)(280+200)/2 = 80*480/2 = 80*240 = 19200
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