Question

The area of a rectangle is 102 square feet and the perimeter is 46 feet. Find the dimensions of the rectangle

Answer 1

The are is given by \(A=x*y\) and perimeter is \(P=2x+2y\)

So we have \(102=xy\implies x=\frac{102}{y}\)

And we also have \(2x+2y=46\)

So we plug \(x=\frac{102}{y}\) for \(x\) and solve for \(y\).

\(2\frac{102}{y}+2y=46\implies \frac{204}{y}+2y=46\implies 204+2y^2=46y\)

So we have \(2y^2-46y+204=0\).

Can you solve this equation for \(y\)?