# Possible areas of an isosceles triangle

I've got the following GRE question. I would like to know how to do this using only the knowledge that is required for the GRE test.

In triangle ABC , AB=AC=2. Which of the following could be the area of triangle ABC?

a) 0.5 b) 1.0 c) 1.5 d) 2.0 e) 2.5 f) 3.0

(It seems that one could do this using calculus, by finding the maximum of the area, but I would like to know if there is any simpler method). Thanks!

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Visualize one of the two legs of length $2$ lying horizontal, and the other one hinged to it, and also lying flat. So we start with $0$ angle at the hinge between these two sides. Now gradually increase the angle the moving side makes with the horizontal side.

The height of the triangle (and therefore the area) increases until the angle at the hinge reaches $90$ degrees, and then decreases.

One does not really need calculus to see that. The free end of the moving side is tracing out a semicircle. The maximum height is reached when the free leg is pointing straight up. The maximum area is therefore when height is also $2$, giving area $2$. Any height in the interval $(0,\frac{\pi}{2}$ is achievable. Heights other than $2$ are in fact achievable twice, with an acute angle at the hinge and also with an obtuse angle.

Remark: If one is in a formula mood, the area of a triangle with two sides of length $a$ and $b$. with angle $\theta$ between them, is $\frac{1}{2}ab\sin\theta$. But $\sin\theta$ reaches a maximum of $1$ when $\theta=\frac{\pi}{2}$ ($90^\circ$).

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Let $ABC$ be a right-angled triangle at $A$. Its area is $\frac12 \cdot \textrm{base}\cdot \textrm{height},$ i.e. $\frac12\cdot 2\cdot 2=2$. It's the maximum area that $ABC$ can have, so anything less than or equal to that is valid.

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