# Finding area based on only the sides

I came across this simple looking question:

The perimeter of a triangle is $$42$$ cm. One side of a triangle is $$8$$ cm longer than the smallest side and the third side is $$1$$ cm less than $$3$$ times the smallest side. Find the area of the triangle.

At first it seemed easy enough. The smallest side is $$7$$ (which can be derived from:$$\frac{42-8+1}{5}$$, where subtracting $$8$$ takes care of the long side and now we have $$2$$ units of the shortest side, then adding one takes care of the third side now we have $$5$$ units of the third side and dividing by $$5$$ we can get the length of the smallest side, $$7$$) and with that knowledge the length of the first and third side is $$15$$ and $$20$$ respectively.

That's when I got stuck. The question gives us no indication of the angles in the triangle. I thought we could use the Pythagorean theorem to see if the triangle is right-angled, but it's not. Is there a rule we can use to find the area? Yes, I'm familiar with trigonometry and $$sin()$$, $$cos()$$ and $$tan()$$ so you can use those in your answer.

If you don't mind can you give me the rule to finding the area, not the answer to the problem? Because I want to solve this question myself. Many thanks to whoever answer's this!

• By the way Alex, I am SO impressed by how much you know at the age of $13$!!! I'm $16$, but comparatively your knowledge is probably quite a bit more advanced than mine! Feb 15, 2021 at 21:49
• Thanks @A-LevelStudent :-) Feb 15, 2021 at 22:33
• You're welcome, you deserve it! Good luck in your studies :) Feb 15, 2021 at 22:34

Use Heron's formula. $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ where $$s$$ is half of the perimeter.

• Wow that was fast, thanks! Feb 15, 2021 at 6:14
• I have to wait 12 minutes before I can accept the answer Feb 15, 2021 at 6:15
• Ok, glad to help. I am not in a rush.. Feb 15, 2021 at 6:17
• @AlexJaynMF would you like a proof of Heron's formula? Feb 15, 2021 at 14:29
• @A-LevelStudent Sure, I would love too! Feb 15, 2021 at 20:54

Here is a proof for Heron's formula, as requested.

We begin with the Cosine rule: $$a^2=b^2+c^2-2bc\cos A\iff\cos A=\frac{b^2+c^2-a^2}{2bc}$$ Recall the identity $$\sin^2x+\cos^2x\equiv 1\iff\sin^2x\equiv1-\cos^2x$$ This means that \begin{align} \sin^2A&=1-\cos^2A\\ &=1-\frac{(b^2+c^2-a^2)^2}{(2bc)^2}\\ &=\frac{(2bc)^2-(b^2+c^2-a^2)^2}{(2bc)^2}\\ &=\frac{(2bc)^2-(a^2-b^2-c^2)^2}{(2bc)^2}\\ &=\frac{(a^2-(b^2-2bc+c^2))((b^2+2bc+c^2)-a^2)}{(2bc)^2}\\ &=\frac{(a^2-(b-c)^2)((b+c)^2-a^2)}{(2bc)^2}\\ &=\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{(2bc)^2} \end{align} So that factors really nicely! (Just for clarification: I have used the formulae $$(x+y)^2=x^2+2xy+y^2$$ and $$x^2-y^2=(x+y)(x-y)$$ quite a few times in the manipulations above.)

Now consider the expression for the 'semi-perimeter' $$s$$: $$s=\frac{a+b+c}{2}$$ This means that $$s-b=\frac{a+b+c}{2}-\frac{2b}{2}=\frac{a-b+c}{2}$$ and so on for $$a$$ and $$c$$ as well. Thus, we have \begin{align} \sin^2A&=\frac{2s \times 2(s-a)\times 2(s-b)\times 2(s-c)}{(2bc)^2}\\ &=\frac{4^2s(s-a)(s-b)(s-c)}{(2bc)^2} \end{align}

Now, in a triangle, the $$\sin$$ of any angle will be positive. This is because any angle $$\theta$$ in a triangle must satisfy $$0<\theta<180^\circ$$, and for $$0<\theta<180^\circ$$ we know that $$\sin\theta$$ is positive. Hence when we now square root both sides we can (and must) take the positive square root. So we have \begin{align} \sin A&=\frac{4}{2bc}\sqrt{s(s-a)(s-b)(s-c)}\\ &=\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)}\\ \end{align} But wait! We know that the area of a triangle, let's call it $$S$$, is equal to $$\frac{1}{2}bc\sin A$$!! This means that \begin{align} S&=\frac{1}{2}bc\sin A\\ &=\frac{1}{2}bc\times\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{s(s-a)(s-b)(s-c)} \end{align} So we finally have our truly remarkable endpoint:

$$\mathbf{S=\sqrt{s(s-a)(s-b)(s-c)}}$$

as required!!

I hope you enjoyed that as much as I enjoyed writing it! :) If you have any questions please don't hesitate to ask.

• Nice work. Imagine how hard it must've been for Heron, without modern algebra. :) Feb 15, 2021 at 22:06
• @PM2Ring thanks :) I have no idea how Heron would've done it without modern algebra, it's mind boggling. Feb 15, 2021 at 22:08
• @PM2Ring according to Wikipedia, the original proof made use of cyclic quadrilaterals (I realize that is extremely vague though.) Feb 15, 2021 at 22:09
• Right. I've occasionally wondered about the details of how he did it. Those Alexandrians had really sharp geometric intuition. At least we have some of Archimedes' stuff, which is pretty awesome to read (in translation), eg his analysis of the volume of the sphere. Back in those days, you could be a master of all known mathematics, but modern mathematics covers so much territory that it's no longer possible for someone to be an expert of all of it. :) I'm glad we've made so much progress, but I think it would've been fun to be a good mathematician in earlier times. Feb 15, 2021 at 22:19
• @PM2Ring Agreed, there'd have been so much to discover. Although: with no use of modern algebraic notation? That'd be ANNOYING, to say the least :) Feb 15, 2021 at 22:24