5. Given a circle $s$ with center $O$ and diameter $A B=4 \text{~cm}$.

a) Construct three tangents to this circle, two of which are at points $A$ and $B$, and the third such that the segment $C D$ between the first two tangents is $5 \text{~cm}$ long.

b) Determine $\Varangle C O D$.

c) Determine the area of the figure bounded by the constructed tangents and the given circle.
Solution. a) Let $t_{1}$ and $t_{2}$ be the tangents to the given circle constructed at points $A$ and $B$ (diagram on the right). Around an arbitrary point $P$ on the tangent $t_{1}$, we draw a circle with radius $r=5 \mathrm{~cm}$ and find the point $Q$ at the intersection with $t_{2}$. From point $O$, we draw a perpendicular $O R$ to the line $P Q$ and find the point $S$ at the intersection of this perpendicular with the given circle. Through point $S$, we draw a line parallel to the line $P Q$, and this is the desired third tangent to the circle.

![](https://cdn.mathpix.com/cropped/2024_06_05_b58a8127523cd0cd5361g-2.jpg?height=690&width=504&top_left_y=1049&top_left_x=993)

b) We have:

$$
\measuredangle C O D=\measuredangle C O S+\measuredangle S O D=\frac{1}{2}(\measuredangle A O S+\measuredangle S O B)=90^{\circ}
$$

c) The desired area is obtained when the area of trapezoid $A B D C$ is reduced by the area of a semicircle with radius $r=2 \mathrm{~cm}$. We have

$$
P_{1}=\frac{A C+B D}{2} \cdot A B=\frac{C S+D S}{2} \cdot A B=\frac{C D}{2} \cdot A B=10 \mathrm{~cm}^{2} \text { and } P_{2}=\frac{4 \pi}{2}=2 \pi \mathrm{cm}
$$

so $P=P_{1}-P_{2}=(10-2 \pi) \mathrm{cm}^{2}$.

## Eighth Division