1.10. Find the area of an isosceles triangle if its base is equal to $a$, and the length of the height drawn to the base is equal to the length of the segment connecting the midpoints of the base and the lateral side.
1.10. By the condition, $D K-$ is the midline of $\triangle A B C$ (Fig. 1.9). Since $B D=D K=\frac{1}{2} B C$, then $\angle C=30^{\circ}$ and $B C=2 B D$. In $\triangle B C D$ we have $C D^{2}=B C^{2}-B D^{2}$, or $\frac{a^{2}}{4}=4 B D^{2}-B D^{2}$, from which $B D=\frac{a}{2 \sqrt{3}}$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-418.jpg?height=244&width=560&top_left_y=1566&top_left_x=320)

Fig. 1.9

$$
S=\frac{1}{2} A C \cdot B D=C D \cdot B D=\frac{a^{2} \sqrt{3}}{12}
$$

Answer: $\frac{a^{2} \sqrt{3}}{12}$.