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It is quite possible this question has no answer -- that is, the area cannot be determined from the information given. It's a question I've created myself as I study for the GRE. No trigonometry is meant to be used, only basic logic about geometry.

(Please bear with me -- this GRE material is quite different than the formal math I am used to in the areas of analysis, abstract algebra, probability, etc.! -- I want to make sure my logic is sound and that I haven't made any leaps in logic or imposed any extra assumptions that weren't given!)

Please consider the figure below, not necessarily drawn to scale:

enter image description here

My question: can I determine the area of the triangle with the information given?


My idea: Draw a line from $B$ down to a new point $D$ on line $AC$ so that that it creates a right angle with $AC$:

enter image description here

Now, observe triangle $BCD$ above. By construction it is a right triangle, whose hypotenuse is length $5$. Therefore, it must be a "$3$-$4$-$5$" right triangle. (Is this step correct or have a made a jump??)

Therefore, I can determine $DC=3$ and $BD=4$.

Finally, I plug in these lengths to compute the area as $A=\frac{1}{2}\cdot 14 \cdot 4 =28$.

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  • $\begingroup$ Angle C is free to vary. That would change the length of BD while preserving the given lengths. $\endgroup$ – Kaj Hansen Aug 3 '14 at 0:06
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    $\begingroup$ You have made a "jump" assuming that a hypotenuse of 5 in a right triangle implies that the other sides have lengths 3 and 4. Try 1 and the square root of 24 for example. Sorry, it was a nice try. $\endgroup$ – Paul Sundheim Aug 3 '14 at 0:07
  • $\begingroup$ Unsolvable with the information given. $\endgroup$ – hexaflexagonal Aug 3 '14 at 0:09
  • $\begingroup$ Even if you could assume that $\triangle BCD$ is a $3$-$4$-$5$ triangle (which you cannot assume!), you could not be certain that the up-right side has length $4$, since the figure is not drawn to scale. $\endgroup$ – Blue Aug 3 '14 at 0:18
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    $\begingroup$ Thank you for your comments! These are all really enlightening points. You even answered another question I had -- if all right triangles with hypotenuse of $5$ were $3$-$4$-$5$ triangles. Study guides don't go into these subtleties -- they just tell you what to memorize, even if it can't always be applied. $\endgroup$ – Mathemanic Aug 3 '14 at 0:22
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I decided to upgrade my comment into an answer.

Angle $C$ is free to vary. That would change the length of $BD$ while preserving the given lengths, which is were you went astray (the length of $BD$ is not necessarily $4$).

Now, in general, if we know two side lengths of a triangle (call them $x$ and $y$) and the angle measure between them (call it $\theta$), then the area of the triangle is given by $A = \frac{1}{2}xy\sin(\theta)$.

Applying this to our scenario, since the angle between the two given sides can vary, then so too can the area of the triangle. This lack of uniqueness is evident in the figure shown below: Every triangle shown has the prescribed side lengths, but all have different areas. In short, we cannot determine the area with the given information.

enter image description here

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    $\begingroup$ Mind if I add a Mathematica figure to your answer, showing a variety of different allowed triangles? $\endgroup$ – Semiclassical Aug 3 '14 at 0:16
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    $\begingroup$ No problem @Semiclassical. $\endgroup$ – Kaj Hansen Aug 3 '14 at 0:28
  • $\begingroup$ Thank you! So I would need to impose another piece of information to solve this. I would need to know all three sides (that way, angle $C$ can't vary so that $BD$ can't vary either, right?). Or alternatively, if I knew the angle between the two sides, I could use the formula you gave. $\endgroup$ – Mathemanic Aug 3 '14 at 0:29
  • $\begingroup$ That's not quite right: If one chooses angle $\angle CAB$, then there are in general two possible triangles and so two possible areas (if the triangle exists---it won't, if the angle is larger than the biggest one shown on the figure added.) @Mathemanic $\endgroup$ – Semiclassical Aug 3 '14 at 0:37
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    $\begingroup$ @Semiclassical is it possible for you to post the code for you Mathematica figure somewhere? I've recently started learning it and would like to understand how such diagrams are made $\endgroup$ – Hushus46 Jul 21 '17 at 16:18
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$a^2+b^2=5^2$ has many solutions, $b=\sqrt{25-a^2}$ for any $a\le 5$.

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