SOLUTION: Find the dimensions of a rectangle if the perimeter is 72m and area is 288m^2

Question 968372: Find the dimensions of a rectangle if the perimeter is 72m and area is 288m^2
Answer by amarjeeth123(569) (Show Source): You can put this solution on YOUR website!
Let the length be x.
Let the breadth be y.
Perimeter=2(length+breadth)=2(x+y)
2(x+y)=72
x+y=36
y=36-x
Area=length*breadth
x(36-x)=288
36x-x^2=288
x^2-36x+288=0

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=144 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 24, 12. Here's your graph:

x=24 m
y=12 m
The dimensions of the rectangle are 24m and 12m.
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