Finding area based on only the sides

Question

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!

Answer 1

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

Answer 2

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

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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!!

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