Problem 6. Plot the figure on the coordinate plane defined by the system of inequalities and find its area $\left\{\begin{array}{c}|x+5|+\sqrt{3}|y-1| \leq 3 \\ y \leq \sqrt{4-4 x-x^{2}}+1 \\ |2 y-1| \leq 5\end{array}\right.$.
# Solution:

The first inequality defines the area inside a rhombus centered at the point $(-5 ; 1)$ with diagonals of 6 and $2 \sqrt{3}$.

The second inequality defines the area below the upper semicircle centered at the point $(-2 ; 1)$ with a radius of $2 \sqrt{2}$.

The third inequality: the strip $-2 \leq y \leq 3$.

The intersection is a figure consisting of two regions - a sector of the circle $C A D$ and the triangle $C A B$. (See puc.)

Let's find the area of the sector $S_{1}$.

The tangent of the angle $C A D$ is the ratio of the diagonals of the rhombus $\operatorname{tg} \angle C A D=\frac{\sqrt{3}}{3}$, hence $\angle C A D=30^{\circ}$, then $S_{1}=\frac{1}{12} S_{\text {circle}}$, where $S_{\text {circle}}=8 \pi$. We get $S_{1}=\frac{2}{3} \pi$.

Let's find the area of the triangle $S_{2}$.

$S_{2}=\frac{1}{2} A C \cdot B C$

$A C=R=2 \sqrt{2}, B C=A C \cdot \operatorname{tg} \angle D A C=2 \sqrt{2} \cdot \frac{\sqrt{3}}{3}=\frac{2}{3} \sqrt{6}$

$S_{2}=\frac{1}{2} \cdot 2 \sqrt{2} \cdot \frac{2}{3} \sqrt{6}=\frac{4}{3} \sqrt{3}$

The desired area $S=S_{1}+S_{2}=\frac{2}{3} \pi+\frac{4}{3} \sqrt{3}$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7a16de7ae2a30fbf517g-5.jpg?height=1140&width=1607&top_left_y=1229&top_left_x=196)

Answer: $\frac{2}{3} \pi+\frac{4}{3} \sqrt{3}$

Grading criteria for 10th grade assignments

| Assignment | 1 | 2 | 3 | 4 | 5 | 6 | Total |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Points | 15 | 15 | 15 | 15 | 20 | 20 | 100 |