SOLUTION: Geomatry. The length of a rectangle is 1 cm longer than its width. If the diagnonal of the rectangle is 4cm what are the dimensions (the length and teh width) of the rectangle?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Geomatry. The length of a rectangle is 1 cm longer than its width. If the diagnonal of the rectangle is 4cm what are the dimensions (the length and teh width) of the rectangle?      Log On


   



Question 100153: Geomatry. The length of a rectangle is 1 cm longer than its width. If the diagnonal of the rectangle is 4cm what are the dimensions (the length and teh width) of the rectangle?
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
The statement that the "length is 1 cm longer than the width" means that we can define the length (l) = w + 1. Width (w) is simply = to w. So, the rectangle has sides of w, w+1, w, and w+1. The diagonal is known to be 4 cm (given).
Now we simply apply the Pythagorean formula. Let's call the diagonal d. That means
d%5E2+=+w%5E2+%2B+%28w%2B1%29%5E2
%28w%2B1%29%5E2+=+w%5E2+%2B+2w+%2B+1
Therefore, d%5E2+=+w%5E2+%2B+w%5E2+%2B+2w+%2B+1+=+2w%5E2+%2B+2w+%2B+1
Of course, d%5E2 is the square of the diagonal, and we were told the diagonal was 4 cm, so d%5E2+=+16.
Substituting what we know:
16+=+2w%5E2+%2B+2w+%2B+1
Subtracting 16 from both sides:
0+=+2w%5E2+%2B+2w+-+15
This does not factor easily, so you can use the quadratic equation to find w.
Quadratic formula: x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
Substituing, we have w+=+%28-2+%2B-+sqrt%28+2%5E2-4%2A2%2A%28-15%29+%29%29%2F%282%2A2%29+