The area of a sector is 100. What is the radius if the perimeter is minimized?
Solution. The perimeter of the circular segment $OAB$ is:

$$
K=2 r+r \alpha
$$

The area of the circular segment is 100 units, so:

$$
T=\frac{r^{2} \alpha}{2}=100
$$

From (1), $\alpha=\frac{K-2 r}{r}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_984333977a780f169f12g-1.jpg?height=347&width=241&top_left_y=602&top_left_x=931)

Substituting the value of $\alpha$ into (2):

$$
100=\frac{r^{2} \cdot \frac{K-2 r}{r}}{2}, \quad \text { which gives } \quad r K=2 r^{2}+200, \quad K=2 r+\frac{200}{r}
$$

We can solve the problem using the inequality between the arithmetic mean and the geometric mean. The arithmetic mean is minimal when it equals the geometric mean, and in this case, the two values are equal. Therefore,

$$
K=2 r+\frac{200}{r} \geq 2 \sqrt{2 r \cdot \frac{200}{r}}=2 \cdot \sqrt{400}=40
$$

The perimeter $K$ will be minimal when $K=40$. This occurs when $2 r=\frac{200}{r}$, which gives $r=10$ (clearly $r>0$).

Of course, we can also obtain the value of $r$ from the equation $K=40$, i.e., $2 r+\frac{200}{r}=40$. After rearranging, we get $(r-10)^{2}=0$, so $r=10$.

Based on the above, the radius of the circular segment with an area of 100 units and minimal perimeter is: $r=10$ units.