3. Given is $\triangle A B C$ where $\measuredangle C$ is obtuse. Points $D \in A C, E \in A B$, and $F \in B C$ are marked on its sides such that quadrilateral $C D E F$ is a rhombus. If $A E=140, B E=84$, and $r=15 \sqrt{3}$, where $r$ is the radius of the circle inscribed in the rhombus $C D E F$, calculate the area of $\triangle A B C$.
3. Let $\alpha, \beta$ and $\gamma$ be the angles of the observed triangle, and let $M$ and $N$ be the feet of the perpendiculars from $E$ to $AC$ and $BC$, respectively. Since the height of the rhombus is equal to the diameter of the inscribed circle, it follows that $E M = E N = 30 \sqrt{3}$. Now we get $\sin \alpha = \frac{E M}{E A} = \frac{3 \sqrt{3}}{14}$ and $\sin \beta = \frac{E N}{E B} = \frac{5 \sqrt{3}}{14}$, and from there we calculate $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \frac{13}{14}$ and $\cos \beta = \sqrt{1 - \sin^2 \beta} = \frac{11}{14}$ (here we used that $\alpha$ and $\beta$ are acute, since $\gamma$ is obtuse). From there we have

$$
\sin \gamma = \sin (180^\circ - \gamma) = \sin (\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha = \frac{98 \sqrt{3}}{196} = \frac{\sqrt{3}}{2}
$$

since $\gamma$ is obtuse, it follows that $\gamma = 120^\circ$. Further, since we have $AB = 140 + 84 = 224$, from $\frac{AB}{\sin \gamma} = \frac{BC}{\sin \alpha} = \frac{AC}{\sin \beta}$ we get $BC = 96$ and $AC = 160$. From there we find $P_{\triangle ABC} = \frac{BC \cdot AC \cdot \sin \gamma}{2} = 3840 \sqrt{3}$.