# What is the area of a triangle with sides $\sqrt{5}$, $\sqrt{10}$, $\sqrt{13}$?

I found a "fun algebra problem" that asks you to find the area of a triangle whose sides are $$\sqrt{5}$$, $$\sqrt{10}$$, $$\sqrt{13}$$. After some algebra hell trying to work with Heron's formula, I plugged the question into Wolfram and it spit out 3.5.

Is there some elegant way to reach this? My algebra kungfu has so far been too weak.

• Welcome to Mathematics Stack Exchange. Try area=$\frac12 \sqrt{a^2c^2-\left(\dfrac{a^2+c^2-b^2}{2}\right)^2}$ May 5, 2019 at 5:44
• Well, there are statements of Heron's formula that involve only squares of the side lengths, which seems helpful when side lengths are square roots of integers: en.wikipedia.org/wiki/Heron%27s_formula For instance, $A=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} = \frac{1}{4}\sqrt{(5+10+13)^2-2(25+100+169)}= \frac{1}{4}\sqrt{28^2-598}=\frac{14}{4}$. But maybe there's a more clever way May 5, 2019 at 5:46

Hint: Observe that $$5 = 1^2 + 2^2, 10 = 1^2+3^2$$ and $$13 = 2^2 + 3^2,$$

• Nice observation! May 5, 2019 at 6:05
• This was a great visualization! I was able to use this with the shoelace formula to get the area of each triangle and subtract them from the larger square. May 6, 2019 at 4:12

Hint:

$$\frac12 \sqrt{a^2c^2-\left(\dfrac{a^2+c^2-b^2}{2}\right)^2}$$

• The computation is particularly easy if you take $b=\sqrt{13}$ May 5, 2019 at 5:56
For another interesting approach, consider the Law of Cosines, $$a^2 = b^2 + c^2 - 2bc\cos(\alpha)$$. If we let $$a = \sqrt{13}$$, $$b = \sqrt{5}$$, and $$c = \sqrt{10}$$, then we find that $$13 = 15 - 10\sqrt{2}\cos(\alpha)$$, and thus that $$\cos(\alpha) = \sqrt{2}/10$$. Using $$\cos(\alpha)$$, we can calculate $$\sin(\alpha)$$ through some basic trigonometric manipulation to find that $$\sin(\alpha) = \sqrt{98}/10$$. Using the area formula for triangles $$A =\frac{1}{2}bc\sin(\alpha)$$, we find that $$A = \frac{1}{2}bc\sin(\alpha) = \frac{1}{2} \cdot\sqrt{50}\cdot\frac{\sqrt{98}}{10}=\frac{7}{2}$$