A corner square of an $8 \times 8$ chessboard has been cut out. We want to cover the remaining area with congruent triangles in a single layer. What is the minimum number of pieces needed?
Let's choose the side of a small square on the board as the unit. Thus, the area of the remaining piece is 63 units. Consider the two triangles containing the sides running into the inner vertex of the cut-out field. The inner vertex is at least a vertex of one of them, otherwise the mentioned sides would intersect there. In the triangle where the inner vertex is also a vertex, the side on the perimeter of the cut-out piece is at most one unit, and the corresponding height is at most 7 units. Therefore, in a covering system, the area of the triangles is at most $7 / 2$ units, so the number of triangles is at least 63 : $\frac{7}{2}=18$. This many triangles are sufficient, as the covering system shown in the figure demonstrates.

![](https://cdn.mathpix.com/cropped/2024_05_02_b11be51aa0bf783d1ccfg-1.jpg?height=604&width=596&top_left_y=457&top_left_x=754)