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This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The answer was $2\sqrt{6}$ if it helps.


Edited to add that it was a multiple choice question, with possible answers:

a. $2\sqrt{6}$
b. $5$
c. $3\sqrt{6}$
d. $5\sqrt{6}$

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  • $\begingroup$ Are you familiar with the Law of Cosines? $\endgroup$
    – 2012ssohn
    Apr 28, 2014 at 23:37
  • $\begingroup$ @2012ssohn Of course i am, but how can i apply that formula to it? I even tried it at the time, but couldnt get anything useful about area from it $\endgroup$
    – Asimov
    Apr 28, 2014 at 23:38
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    $\begingroup$ Did you actually try to use Heron's formula? The formula simplifies pretty nicely. $\endgroup$
    – Nate
    Apr 28, 2014 at 23:41
  • $\begingroup$ When i tried it i got an ugly glob of roots and addition that just wouldn't simplify for me. Maybe i just need to practice simplification $\endgroup$
    – Asimov
    Apr 28, 2014 at 23:44
  • $\begingroup$ @JohnJPershing: Try multiplying-out factors of Heron's formula before substituting the specific side-lengths. Note that, if you think of the factors as something like $$\frac{1}{16}\;\cdot\;((b+c)+a)\;((b+c)-a)\;\cdot\;(a-(b-c))\;(a+(b-c))$$ then you can save a little mental energy recognizing that a couple of differences-of-squares come into play during the process. $\endgroup$
    – Blue
    Apr 28, 2014 at 23:54

5 Answers 5

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From the law of cosines ($C^2 = A^2 + B^2 - 2AB\cos \theta$), we get that $(\sqrt{33})^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cos \theta$.

Simplifying this, we get $33 = 29 - 20 \cos \theta$, which means that $\displaystyle \cos \theta = -\frac{1}{5}$

Because $\cos^2 \theta + \sin^2 \theta = 1$, we get that $\displaystyle \sin^2 \theta = \frac{24}{25}$. This means that $\displaystyle \sin \theta = \frac{2\sqrt{6}}{5}$ (note that, because $0 \le \theta \le \pi$, $\sin \theta \ge 0$).

The area of the triangle is $\displaystyle \frac{1}{2} A B \sin \theta = \frac{1}{2} \cdot 2 \cdot 5 \cdot \frac{2\sqrt{6}}{5} = 2\sqrt{6}$.

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  • $\begingroup$ Thank you, this is clear, concise, and could be done quickly and easily in the situation. This is probably what the judges/writers of it wanted readers to do. $\endgroup$
    – Asimov
    Apr 28, 2014 at 23:48
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Use the $\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula.

$$\begin{align} & \frac{1}{4}\sqrt{4\cdot4\cdot25-(4+25-33)^2} \\ = & \frac{1}{4}\sqrt{4^2\cdot25-4^2} \\ = & \sqrt{25-1} \\ = & 2\sqrt{6} \end{align}$$

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    $\begingroup$ This is the way I would have done it, nice. $\endgroup$
    – Sawarnik
    Apr 29, 2014 at 6:40
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You could've used Heron's formula straight away, actually.

$$\begin{align} T & = \tfrac{1}{4} \sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\ & = \tfrac{1}{4} \sqrt{(2+5-\sqrt{33})(2-5+\sqrt{33})(-2+5+\sqrt{33})(2+5+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7-\sqrt{33})(-3+\sqrt{33})(3+\sqrt{33})(7+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7+\sqrt{33})(7-\sqrt{33})(\sqrt{33}+3)(\sqrt{33}-3)} \\ & = \tfrac{1}{4} \sqrt{(49-33)(33-9)} \\ & = \tfrac{1}{4} \sqrt{16 \cdot 24} \\ & = \tfrac{1}{4} \sqrt{64 \cdot 6} \\ & = \tfrac{8}{4} \sqrt{6} \\ & = 2 \sqrt{6} \end{align}$$

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Since this was multiple-choice, I think it's worth noting that you could have guessed the right answer without doing (much) arithmetic: the diagonal of a right triangle with sides $2$ and $5$ has length $\sqrt{2^2+5^2} = \sqrt{29}$; since $\sqrt{33}$ is fairly close to this, the answer should be close to the area of a $2-5-\sqrt{29}$ triangle, which is of course $\frac12(2)(5)=5$. If you imagine how to 'stretch out' the $\sqrt{29}$ diagonal to $\sqrt{33}$, it's clear that the right angle will have to become obtuse, and this in turn means that the area of the $2-5-\sqrt{33}$ triangle will have to be less than $5$; of the provided answers, only $2\sqrt{6}\approx4.472$ is less than $5$ (and of course very close to it).

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  • $\begingroup$ I like this response, and used a similar estimation method, using the areas of slightly bigger and smaller triangles that were easier to calculate $\endgroup$
    – Asimov
    Apr 30, 2014 at 0:05
  • $\begingroup$ I hate it when there are two good answers, and both are right and good, and you wish you could accept both as correct $\endgroup$
    – Asimov
    Apr 30, 2014 at 1:41
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    $\begingroup$ @JohnJPershing For what it's worth, I think 2012ssohn's answer is much better than mine; it explains how to actually derive the correct value, rather than merely how to answer the multiple-choice question. Both are valuable, but that one's likely to be more broadly applicable; I just wanted to offer an alternate approach for the test. $\endgroup$ Apr 30, 2014 at 3:24
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Let be $ABC$ the triangle. Consider the altitude $AH$ over the greatest side, the one whose length is $\sqrt{33}$. Now call $x=BH$, so $\sqrt{33}-x=CH$. Apply Pythagoras' theorem to get $$\left\{ \begin{array}{rcl} x^2+h^2&=&4\\ \left(\sqrt{33}-x\right)^2+h^2&=&25 \end{array} \right.$$

Solve for $h$ and you are (almost) done.

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