10.318. The bases of a trapezoid are equal to $a$ and $b$. Determine the length of the segment parallel to the bases and dividing the trapezoid into two equal-area parts.
Solution.

Let in trapezoid $ABCD$ (Fig. 10.107) $BC\|AD, EF\|BC$, the areas $S_{1}$ and $S_{2}$ of trapezoids $AEFD$ and $EBCF$ are equal, $BC=a, AD=b$. Draw $BL$ and $EN$ - the heights of trapezoids $EBCF$ and $AEFD$, respectively. Let $EF=x$, $BL=h_{1}, EN=h_{2}$. Draw $BK\|CF, EM\|CD$. Then $KF=a$, $EK=x-a, MD=x, AM=b-x$. $\triangle AEM \sim \Delta EBK$. Then $\frac{EK}{BL}=\frac{AM}{EN}$, $\frac{x-a}{h_{1}}=\frac{b-x}{h_{2}}$ (a). $S_{1}=\frac{BC+EF}{2} \cdot BL=\frac{a+x}{2} \cdot h_{1} \cdot S_{2}=\frac{AD+EF}{2} \cdot EN=\frac{b+x}{2} \cdot h_{2} \cdot$ Then $\frac{a+x}{2} h_{1}=\frac{b+x}{2} h_{2}$ (b). Multiplying equalities (a) and (b), we get: $\frac{x^{2}-a^{2}}{2}=\frac{b^{2}-x^{2}}{2}$. Hence $2 x^{2}=a^{2}+b^{2} ; x=\sqrt{\frac{a^{2}+b^{2}}{2}}$. Answer: $\sqrt{\frac{a^{2}+b^{2}}{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0791.jpg?height=434&width=503&top_left_y=78&top_left_x=96)

Fig. 10.108

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0791.jpg?height=425&width=665&top_left_y=60&top_left_x=588)

Fig. 10.109