SOLUTION: Find the dimensions of a rectangle that has an area 48 m^2 and a diagonal of length 10 m.

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Question 182199: Find the dimensions of a rectangle that has an area 48 m^2 and a diagonal of length 10 m.
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the length of the rectangle and y = the width of the rectangle.
The area is x*y = 48 sq.m.
The diagonal is 10 m, so x%5E2%2By%5E2+=+10%5E2 from the Pythagorean theorem.
So we have two equations:
1) x%2Ay+=+48 Rewrite this equation as: x+=+48%2Fy and substitute into equation 2).
2) x%5E2%2By%5E2+=+100
2a) %2848%2Fy%29%5E2%2By%5E2+=+100 Simplifying this, we get:
2a) 2304%2Fy%5E2+%2B+y%5E2+=+100
2a) %282304+%2B+y%5E4%29%2Fy%5E2+=+100 Multiply both sides by y%5E2
2a) 2304+%2B+y%5E4+=+100y%5E2 Rearrange this into standard "quadratic" form.
y%5E4-100y%5E2%2B2034+=+0 Solve by factoring, noting that y%5E4+=+%28y%5E2%29%5E2
%28y%5E2%29%5E2+-+100%28y%5E2%29%2B2304+=+0
%28%28y%5E2%29-64%29%28%28y%5E2%29-36%29+=+0 Applying the zero product rule, we get:
y%5E2-64+=+0 or y%5E2-36+=+0 Factoring the left sides, we get:
%28y%2B8%29%28y-8%29+=+0 or %28y%2B6%29%28y-6%29+=+0 From which we get:
y+=+-8
y+=+8
y+=+-6
y+=+6
Discard the negative solutions as the lengths of the rectangle sides can only be positive.
So we have:
y = 6 or 8 and x = 8 or 6
The dimensions of the rectangle would be:
Length = 8m and width = 6m
Please note that while we solved only for y, the x-value is obtained by substituting y = 8 into eqation 1):
x%2Ay+=+48 Substitute y+=+8
x%2A%288%29+=+48 Dividing both sides by 8.
x+=+6