SOLUTION: Consider a rectangle that has a diagonal of 20 feet and a perimeter of 52 feet. Find the area of this rectangle.

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Question 657795: Consider a rectangle that has a diagonal of 20 feet and a perimeter of 52 feet. Find the area of this rectangle.
Answer by kevwill(135) About Me  (Show Source):
You can put this solution on YOUR website!
let l be the length of the rectangle and w be the width of the rectangle. With the diagonal being 20 feet long, using the Pythagorean Theormem, we get
sqrt%28l%5E2+%2B+w%5E2%29+=+20
l%5E2+%2B+w%5E2+=+400
With a perimeter of 52 feet, we have
2%2Al+%2B+2%2Aw+=+52
l+%2B+w+=+26
l+=+26+-+w
Substituting, we get
%2826-w%29%5E2+%2B+w%5E2+=+400
%28676-52w%2Bw%5E2%29+%2B+w%5E2+=+400
2%2Aw%5E2+-+52w+%2B+276+=+0
We can solve for w using the quadratic equation with a=2, b=-52, and c=276:
w+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
w+=+%28-%28-52%29+%2B-+sqrt%28+%28-52%29%5E2-4%2A2%2A276%29%29%2F%282%2A2%29+

w+=+13+%2B-+sqrt%2831%29+
+w+=+13+%2B+sqrt%2831%29+ and +l+=+26-w+=+26+-+%2813%2Bsqrt%2831%29%29+=+13-sqrt%2831%29+, or
+w+=+13+-+sqrt%2831%29+ and +l+=+26-w+=+26+-+%2813-sqrt%2831%29%29+=+13%2Bsqrt%2831%29+
Both of these are basically the same solution with the l and w values being switched. So the area is:

The area is 138 square feet.