In an obtuse triangle, the largest side is 4, and the smallest side is 2. Can the area of the triangle be greater than $2 \sqrt{3}$?
Express the largest side of the triangle using the Law of Cosines.

## Solution

Let in triangle $ABC$ angle $B$ be obtuse, $AB=2, AC=4$. Denote $BC=x$. By the Law of Cosines,

$$
AC^{2}=AB^{2}+BC^{2}-2 AB \cdot BC \cos \angle B, \text{ or } 16=4+x^{2}-4 x \cos \angle B,
$$

from which $x^{2}-12=4 x \cos \angle B<0$, so $x^{2}<12, x<2 \sqrt{3}$.

If $S$ is the area of the triangle, then

$$
S=\frac{1}{2} AB \cdot BC \sin \angle B=\frac{1}{2} \cdot 2 \cdot x \sin \angle B=x \sin \angle B<2 \sqrt{3} \cdot 1=2 \sqrt{3}
$$

## Answer

No.