SOLUTION: if the perimeter of a rectangle is 20 feet and the diagonal is 2 sqrt13 feet, then what are the length and width?

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Question 173514: if the perimeter of a rectangle is 20 feet and the diagonal is 2 sqrt13 feet, then what are the length and width?
Found 2 solutions by solver91311, Mathtut:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!
I just did this one a couple of hours ago. See the solution to problem 174453.
Answer by Mathtut(3670)   (Show Source): You can put this solution on YOUR website!
P=2(L+W)...where L and W are length and width respectively
:
we are given the diagonal the hypothenuse of a right triangle where L and W are the legs
:
...eq 1
................eq 2
:
lets re write eq 1 for L. 2L=20-2W---->L=(20-2W)/2-->L=10-W
:
now lets plug in this L value into eq 2 and solve for W
:

:
multiplying out
:
combining like terms on left side
:
dividing by 2
:

:
W can be either 4 or 6 but when
:
W=4 then L=6 and when W=6 the L=4....
:
:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=4 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 6, 4. Here's your graph:
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