## Task B-4.2.

Calculate the area of a triangle with the lengths of two sides being $13 \mathrm{~cm}$ and $14 \mathrm{~cm}$, and the length of the angle bisector between these two sides being $\frac{28 \sqrt{13}}{9} \mathrm{~cm}$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_1020492bf6e1d2c949fdg-16.jpg?height=454&width=506&top_left_y=358&top_left_x=772)

Let \(a, b\) be the sides of the given triangle and \(s\) the angle bisector between them. Then the area of the triangle is

$$
P=\frac{1}{2} a b \cdot \sin \gamma \text { or } P=\frac{1}{2} a s \cdot \sin \frac{\gamma}{2}+\frac{1}{2} b s \cdot \sin \frac{\gamma}{2}=\frac{1}{2} s \cdot(a+b) \cdot \sin \frac{\gamma}{2}
$$

It follows that:

$$
\begin{aligned}
& \frac{1}{2} a b \cdot \sin \gamma=\frac{1}{2} s \cdot(a+b) \cdot \sin \frac{\gamma}{2} \\
& \left(2 \cdot \sin \frac{\gamma}{2} \cdot \cos \frac{\gamma}{2}\right) \cdot 13 \cdot 14=\frac{28 \sqrt{13}}{9} \cdot 27 \cdot \sin \frac{\gamma}{2}
\end{aligned}
$$

Then we have

$$
\begin{aligned}
& \cos \frac{\gamma}{2}=\frac{3 \sqrt{13}}{13}, \sin \frac{\gamma}{2}=\sqrt{1-\cos ^{2} \frac{\gamma}{2}}=\frac{2 \sqrt{13}}{13} \\
& P=\frac{1}{2} \cdot 13 \cdot 14 \cdot 2 \cdot \frac{3 \sqrt{13}}{13} \cdot \frac{2 \sqrt{13}}{13}=84 \mathrm{~cm}^{2}
\end{aligned}
$$