Question

Look at picture. Does anybody know the answer I’m lost?

Answer 1

Let $x,y$ be the dimensions of the rectangle. We know the equations for both area and perimeter:

$A=xy=36$

$P=2(x+y)=36 \iff x+y=18$

So, we have  the following system:

$\begin{cases}xy=36\\x+y=18\end{cases}$

From the second equation, we can deduce

$y=18-x$

Plug this in the first equation to get

$xy=x(18-x)=-x^2+18=36$

Refactor as

$x^2-18x+36=0$

And solve with the usual quadratic formula to get

$x=9\pm3\sqrt{5}$

Both solutions are feasible, because they're both positive.

If we chose the positive solution, we have

$x=9+3\sqrt{5} \implies y=18-x=18-9-3\sqrt{5}=9-3\sqrt{5}$

If we choose the negative solution, we have

$x=9-3\sqrt{5} \implies y=18-x=18-9+3\sqrt{5}=9+3\sqrt{5}$

So, we're just swapping the role of $x$ and $y$. The two dimensions of the rectangle are $9+3\sqrt{5}$ and $9-3\sqrt{5}$