3. Given trapezoid $ABCD$ with bases $\overline{AB}=a$ and $\overline{CD}=b$. Find the length of the segment that satisfies the following conditions:

- parallel to $AB$ and $CD$
- divides the trapezoid into two parts with equal areas,
- its endpoints lie on the non-parallel sides of the trapezoid.
Solution. First Method. Let the line $p\|A B\| C D$ intersect the sides $A D$ and $B C$ at points $K$ and $L$ respectively and bisect the trapezoid $A B C D$ into two equal-area parts. Let $P$ be the area of the trapezoid $A B C D$, and $h$ be its height. Let $P_{1}$ be the area of the trapezoid $A B L K$ with height $h_{1}$, and $P_{2}$ be the area of the trapezoid $K L C D$ with height $h_{2}$. Let $\overline{K L}=x$ (see diagram). Then, we have $h=h_{1}+h_{2}$ and $P_{1}=P_{2}=\frac{1}{2} P$. From $P=\frac{a+b}{2} \cdot h, \quad P_{1}=\frac{a+x}{2} \cdot h_{1} \quad$ and $\quad P_{2}=\frac{x+b}{2} \cdot h_{2}$, we get $h=\frac{2 P}{a+b}, h_{1}=\frac{2 P_{1}}{a+x}=\frac{P}{a+x}$ and $h_{2}=\frac{2 P_{2}}{x+b}=\frac{P}{x+b}$. Substituting into $h=h_{1}+h_{2}$, we get $\frac{2 P}{a+b}=\frac{P}{a+x}+\frac{P}{x+b}$, i.e., $\frac{2}{a+b}=\frac{1}{a+x}+\frac{1}{x+b}$,

which simplifies to

$$
2(a+x)(x+b)=(a+b)(x+b)+(a+b)(a+x)
$$

Rearranging the last equation, we get $2 x^{2}=a^{2}+b^{2}$, from which

$$
x=\sqrt{\frac{1}{2}\left(a^{2}+b^{2}\right)}=\sqrt{\left(\frac{a}{\sqrt{2}}\right)^{2}+\left(\frac{b}{\sqrt{2}}\right)^{2}}
$$

Solution B. Let the line $p\|A B\| C D$ intersect the sides $A D$ and $B C$ at points $K$ and $L$ respectively and bisect the trapezoid $A B C D$ into two equal-area parts. Let $\overline{K L}=x$. Extend the sides $A D$ and $B C$ to their intersection $M$ (see diagram). Then, the triangles $\triangle A B M, \triangle K L M$ and $\triangle D C M$ are similar, so

$$
P_{A B M}: P_{K L M}: P_{D C M}=a^{2}: x^{2}: b^{2}
$$

i.e., $P_{A B M}=k a^{2}, P_{K L M}=k x^{2}$ and $P_{D C M}=k b^{2}$. From the condition $P_{A B L K}=P_{K L C D}$, we get

$$
P_{A B M}-P_{K L M}=P_{K L M}-P_{D C M}, \text{ i.e., } k a^{2}-k x^{2}=k x^{2}-k b^{2}
$$

from which we get $x^{2}=\frac{1}{2}\left(a^{2}+b^{2}\right)$, so $x=\sqrt{\left(\frac{a}{\sqrt{2}}\right)^{2}+\left(\frac{b}{\sqrt{2}}\right)^{2}}$.