# Can one deduce whether a given quantity is possible as the area of a triangle when supplied with the length of two of its sides?

Recently I have found a question like following:

In triangle $ABC$, $AB=AC=2$. Which of the following could be the area of triangle $ABC$? Indicate all possible areas:

[A] $0.5$ [B] $1.0$ [C] $1.5$ [D] $2.0$ [E] $2.5$ [F] $3.0$

From my points of view, I can only guess one answer if I assume the triangle is a right angle triangle. In that case, the area will be $2$.

But the answer showed the result, [A][B][C][D].

So, my question this is there any axiom that the area of the right angle will be the highest area of any type of triangle with the expressing two lengths of it?

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Not an axiom, but yes: The area of a triangle is $\frac{1}{2}ab \sin \theta$, where $a$ and $b$ are two known side lengths and $\theta$ is the angle between the two given side lengths. Since $\sin \theta$ reaches its maximum value when $\theta=90°$ (in which case we have $\sin 90° = 1$), the largest possible area is indeed attained for the case of a right angle.

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and since $\sin \theta$ is continuous and $\sin 0 = 0$ ; all values between $0$ and $1$ are possible(intermediate value theorem) –  Cruncher Aug 27 '14 at 12:56

Hint: The area of $\Delta ABC$ is $\dfrac{1}{2} \cdot AB \cdot AC \cdot \sin A$. You know that $AB = AC = 2$ and that $0 < \sin A \le 1$, since $0 < A < 180^{\circ}$. So, what is the range of possible areas of $\Delta ABC$?

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The area of the parallelogram proportional to it's height. It is maximal for a rectangle. The triangle is the half parallelogram

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I really do not understadn this answer at all. Could you explain this a bit more? –  KBusc Aug 27 '14 at 15:09
@KBusc The sentence "It is maximal for a rectangle" should read "For fixed side lengths the area is maximal for a rectangle". –  Taemyr Aug 28 '14 at 7:23
@Taemyr ohh ty very much that makes sense now –  KBusc Aug 28 '14 at 11:41
I like this answer much better than my own. –  mweiss Aug 28 '14 at 13:05

The peak area will be the area when these two legs are perpendicular to each other. But smaller areas are possible, all the way down to $0$, as the two legs become closer and closer to being parallel. If you know that the area is $\frac12(\text{base})(\text{height})$ then you can use one of these legs as the base, and see that height is maximized when the other leg is perpendicular to the base.

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Given two sides have the same length = 2, the minimum area of the triangle will be 0. And maximum area will be 2 (when the triangle is a right triangle).

Thus possible area are A, B, C and D (<=2). E and F are not possible (>2)

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In computation of area of a triangle least reqiurement is all sides, otherwise area remains parametric, therefore we may just say that area is atmost 2 in the present case.

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No it is not possible to deduce just by two sides, because we don't know the angle between those two sides, which will decide how much area triangle will cover. Take two lines of $5$cm each and keep on changing the angle $(b,w)$ from $0$ to $180°$, you'll find that just two sides isn't enough.

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Please reread the full question :) –  Khandaker Mustakimur Rahman Aug 27 '14 at 6:52