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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.

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. Try area=$\frac12 \sqrt{a^2c^2-\left(\dfrac{a^2+c^2-b^2}{2}\right)^2}$ $\endgroup$
    – J. W. Tanner
    May 5, 2019 at 5:44
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    $\begingroup$ 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 $\endgroup$
    – Jane Doé
    May 5, 2019 at 5:46

3 Answers 3

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

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    $\begingroup$ Nice observation! $\endgroup$
    – user376343
    May 5, 2019 at 6:05
  • $\begingroup$ 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. $\endgroup$
    – Dr_Donut
    May 6, 2019 at 4:12
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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.

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  • $\begingroup$ The computation is particularly easy if you take $b=\sqrt{13}$ $\endgroup$
    – J. W. Tanner
    May 5, 2019 at 5:56
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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 $$A = \frac{1}{2}bc\sin(\alpha) = \frac{1}{2} \cdot\sqrt{50}\cdot\frac{\sqrt{98}}{10}=\frac{7}{2}$$

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