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:
$\hskip 4 cm$ 
Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has fixed diagonal $d$. The large square has area $d^2$. I have created a more dynamic, animated version using Desmos.
A simple explanation without proof or pictures:
The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.
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 $(a-b)^2\ge 0$.
You can prove that no such rectangle exists as follows:
Let $l \ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2l\ge l+b= 36$ so $l \ge 18$
Then the diagonal of a right-angled triangle is the longest side so $d\gt l\ge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.
The answer given, though arithmetically correct does not represent a real wall.
I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.
Another PWW (noted by AlexanderJ93 and others):
$\hspace{5cm}$
No. As others have said.
What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.
If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.
If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:
Total perimeter: 70 Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)
This should now give the solution of:
which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)
Hope that helps!
-Van
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.)
No. Using Pythagoras and a simple inequality we get $$d^2=a^2+b^2\geq 2ab\geq ab$$ If $a,b$ are the sides and $d$ the diagonal
Adjusted the PWW given by farruhota (and ripping off their very image) to improve by a factor of $2$ (cf. also Théophile's answer):
$A=lw$
$P=2(l+w)$
$d=\sqrt{l^2+w^2}$
Can $A>d^2$?
Can $lw>l^2+w^2$?
$-lw>l^2-2wl+w^2=(l-w)^2$
Width and length are necessarily positive. The square of their difference also must be positive.
So we have a negative number that must be greater than a positive number. A contradiction.
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.
