## Task B-4.5.

The line whose equation is $3 x+4 y-24=0$ intersects the x-axis at point $A$, and the y-axis at point $B$. A point $S$ is chosen on the segment $\overline{A B}$. Equilateral triangles $S C A$ and $S D B$ are constructed on opposite sides of the segment $\overline{A B}$. Calculate the area of the quadrilateral $A D B C$.
## Second Solution.

We arrive at the intersection with the coordinate axes: $A(8,0)$ and $B(0,6)$,

1 point and conclude that $|A B|=10$.

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Since triangles $S C A$ and $S D B$ are equilateral, the line $C D$ intersects the line $A B$ at an angle of $60^{\circ}$, or $120^{\circ}$, so points $C, S$ and $D$ lie on the same line.

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The lengths $\overline{A B}$ and $\overline{C D}$ are the diagonals of the quadrilateral $A B C D$ with length $a+b$.

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Then the area of the quadrilateral $\mathrm{ABCD}$ can be calculated using the formula

$$
P=d_{1} \cdot d_{2} \cdot \sin \varangle\left(d_{1}, d 2\right)=\frac{1}{2}(a+b)^{2} \sin 60^{\circ}=\frac{1}{2} \cdot 100 \frac{\sqrt{3}}{2}=25 \sqrt{3}
$$

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