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.
$\endgroup$ 23 Answers
$\begingroup$Hint:Observe that $5 = 1^2 + 2^2, 10 = 1^2+3^2$ and $13 = 2^2 + 3^2,$
Hint:
This formula equivalent to Heron's
$$\frac12 \sqrt{a^2c^2-\left(\dfrac{a^2+c^2-b^2}{2}\right)^2}$$
is useful in your situation.
$\endgroup$ 1 $\begingroup$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 $$A = \frac{1}{2}bc\sin(\alpha) = \frac{1}{2} \cdot\sqrt{50}\cdot\frac{\sqrt{98}}{10}=\frac{7}{2}$$
$\endgroup$
