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.
Area of a rectangular park
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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.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.
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.
>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
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