SOLUTION: The perimeter of a rectangle is 22, and its diagonal is \sqrt{61}. Find its dimensions and area, accurate to two decimal places.

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Question 947304: The perimeter of a rectangle is 22, and its diagonal is \sqrt{61}. Find its dimensions and area, accurate to two decimal places.
Answer by macston(5194)   (Show Source): You can put this solution on YOUR website!
Let L=length; Let W=Width; Perimeter=P=2(L+W); Diagonal=D=hypotenuse
L and W are two legs of a right triangle with the diagonal as hypotenuse, so:
P=2(L+W)
22=2(L+W) Divide each side by 2
11=L+W Solve for W, subtract L from each side
11-L=W


Substitute for W


Subtract 61 from each side

Solved by pluggable solver: SOLVE quadratic equation with variable
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, 5. Here's your graph:

ANSWER:Length is 6 or 5
If L=6 units, then W=11-L=11-6=5 units
If L=5 units, then W=11-L=11-5=6 units
ANSWER: The dimensions are 6 units x 5 units (or 5 units x 6 units)
The area is the same either way: Area =L x W=6 units x 5 units= 30 square units
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