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 ?
The area of the square built on the diagonal must be at least twice the area of the rectangle:

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A wall has a thickness. 

Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36. 

Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.

The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.

We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.

This t can be used to calculate the area of the wall as (72 - 4t) * t.

Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.