# 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 given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18\sqrt2\ \textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason. – Théophile Nov 15 '18 at 19:09
• The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so??? – alephzero Nov 15 '18 at 20:16
• In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486! – amalloy Nov 15 '18 at 22:37
• The problem with the question is that if you solve for the side lengths you get complex numbers. – 1123581321 Nov 15 '18 at 22:53
• @1123581321 Specifically, $l$ and $b$ are $18\pm9i\sqrt2$. – Teepeemm Nov 16 '18 at 2:10

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 some fixed diagonal $$d$$. The large square has area $$d^2$$.

• +1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same. – mckenzm Nov 15 '18 at 23:26
• Here is a more dynamic, animated version of the same picture. – Xander Henderson Nov 16 '18 at 4:44

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.

• Great explanation, but that “it's” is jarring... – DaG Nov 17 '18 at 9:34
• Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks! – AlexanderJ93 Nov 17 '18 at 9:35
• +1 Thank you for the one-line proof! – DaG Nov 17 '18 at 10:01

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.

• To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $\ell$ and height $b$. – Ilmari Karonen Nov 15 '18 at 21:39
• @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions. – Mark Bennet Nov 16 '18 at 8:40

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:

• $$I^2 + B^2 = 25^2 = 625$$
• $$2I + 2B = 70$$
• $$I + B = 35$$
• $$I^2 + 2IB + B^2 = 1,225$$
• $$2IB = 600$$
• $$IB = 300$$ ,

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.

• A wall is a 2-dimensional object. – Jossie Calderon Feb 14 at 18:46