10.181. In a square with side $a$, the midpoints of two adjacent sides are connected to each other and to the opposite vertex of the square. Determine the area of the inner triangle.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-640.jpg?height=418&width=418&top_left_y=95&top_left_x=235)

Fig. 10.150

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-640.jpg?height=384&width=356&top_left_y=96&top_left_x=796)

Fig. 10.151
## Solution.

Let $K, L$ be the midpoints of $A D$ and $C D$ (Fig. 10.150). Then $S_{1}=\frac{1}{2} a \cdot \frac{a}{2}=\frac{a^{2}}{4}$; $S_{2}=\frac{1}{2} \frac{a}{2} \cdot \frac{a}{2}=\frac{a^{2}}{8}$; $S_{3}=\frac{1}{2} a \cdot \frac{a}{2}=\frac{a^{2}}{4}$. The area of the square $S_{0}=a^{2}$. Therefore, $S=S_{0}-S_{1}-S_{2}-S_{3}=a^{2}-\frac{a^{2}}{4}-\frac{a^{2}}{4}-\frac{a^{2}}{8}=\frac{3 a^{2}}{8}$.

Answer: $\frac{3 a^{2}}{8}$.