<s>[INST] 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 ?
 [/INST] No, use the Pythagorean Theorem.
$$c^2 = a^2+b^2$$
$c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).
Recall for any real number, its square must be non-negative.
$$(a-b)^2 \geq 0 \implies a^2-2ab+b^2 \geq 0 \implies \color{blue}{a^2+b^2 \geq 2ab}$$
The area of the rectangle is $ab$, but $c^2 \geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.
Now, to find the area itself.
For the diagonal:
$$c^2 = a^2+b^2$$
$$\implies 18^2 = a^2+b^2$$
$$\color{blue}{324 = a^2+b^2} \tag{1}$$
For the perimeter:
$$2(a+b) = 72$$
$$a+b = 36$$
Now, define one variable in terms of the other.
$$\color{purple}{a = 36-b} \tag{2}$$
Combine $(1)$ and $(2)$.
$$324 = a^2+b^2 \implies 324 = (36-b)^2+b^2$$
$$324 = 36^2-2(36)b+b^2+b^2 \implies 324 = 1296-72b+2b^2 \implies 2b^2-72b+972 = 0$$
But $$\Delta = b^2-4ac$$
$$\Delta = 72^2-4(2)(972) = -2592$$
$$\implies \Delta < 0$$
Thus, there is no solution. (No such rectangle exists.)
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