8,9 In triangle $D E F$ angle $D E F$ is equal to $60^{\circ}$. Find the area of triangle $D E F$, if it is known that $D F=3, E F=$ $\frac{6}{\sqrt{3}}$
Applying the cosine theorem, find $D E$ from triangle $D E F$.

## Solution

Let $D E=x$. Applying the cosine theorem to triangle $D E F$, we get the equation

$$
9=x^{2}+12-2 \cdot x \cdot \frac{6}{\sqrt{3}} \cdot \frac{1}{2}, \text { or } x^{2}-2 x \sqrt{3}+3=0
$$

from which we find that $x=\sqrt{3}$. Therefore,

$$
S_{\triangle \mathrm{DEF}}=\frac{1}{2} \cdot E D \cdot E F \cdot \sin \angle D E F=\frac{1}{2} \cdot \sqrt{3} \cdot \frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-37.jpg?height=378&width=600&top_left_y=460&top_left_x=726)

## Answer

$\frac{3 \sqrt{3}}{2}$.

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