Task B-2.2. In a circle with radius $r=10 \mathrm{~cm}$, two parallel chords of lengths $16 \mathrm{~cm}$ and $12 \mathrm{~cm}$ are drawn. If the center of the circle is inside the trapezoid whose bases are these chords, calculate the perimeter of the trapezoid.
Solution. Sketch

Consider the isosceles triangle $A B S$. The height to the base is equal to

$$
v_{1}=\sqrt{10^{2}-8^{2}}=6 \text{ cm}
$$

From triangle $S C D$ we get

$$
v_{2}=\sqrt{10^{2}-6^{2}}=8 \text{ cm}
$$

The length of the leg of the trapezoid is then

$$
b=\sqrt{\left(v_{1}+v_{2}\right)^{2}+\left(\frac{a-c}{2}\right)^{2}}=\sqrt{14^{2}+2^{2}}=\sqrt{200}=10 \sqrt{2}
$$

The perimeter of the trapezoid is $o=a+2 b+c=4(7+5 \sqrt{2})$.

![](https://cdn.mathpix.com/cropped/2024_05_30_cf2c4989ba380bc8684dg-07.jpg?height=772&width=753&top_left_y=278&top_left_x=659)