3.1. Znayka cut out a semicircle from paper. Neznaika marked a point $D$ on the diameter $A B$ of this semicircle and cut off two semicircles from Znayka's semicircle with diameters $A D$ and $D B$. Find the area of the remaining figure if the length of the part of the chord lying inside it, passing through point $D$ perpendicular to $A B$, is 6. If necessary, round your answer to two decimal places.
Answer: 28.27 (exact value: $9 \pi$).

Solution. The area $S$ of the resulting figure is

$$
S=\frac{\pi}{2}\left(\left(\frac{A B}{2}\right)^{2}-\left(\frac{A D}{2}\right)^{2}-\left(\frac{D B}{2}\right)^{2}\right)=\frac{\pi}{8}\left(A B^{2}-A D^{2}-B D^{2}\right)
$$

Since $A B=A D+D B$, we get

$$
S=\frac{\pi}{4} A D \cdot B D
$$

Let $C$ be a point on the arc of the semicircle such that $D C \perp A B$. Since $D C^{2}=A D \cdot B D$, then $S=\pi\left(\frac{D C}{2}\right)^{2}$. Given that $D C=6$, it follows that $S=9 \pi \approx 28.27$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a656db1b08002c401269g-2.jpg?height=422&width=717&top_left_y=1468&top_left_x=1206)