Can area of rectangle be greater than the square of its diagonal? 
Q: A wall, rectangular in shape, has a perimeter of 72 m. If the length of its diagonal is 18 m, what is the area of the wall ?

The answer given to me is area of 486 m2. This is the explanation given to me



Is it possible to have a rectangle of diagonal 18 m and area greater than the area of a square of side 18 m ?

In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2\ge 2ab$ if orthogonal sides' lengths are $a,\,b$. Why? Because the difference is $(a-b)^2\ge 0$.
