3) Two squares with side $10 \mathrm{~cm}$ are given, one of which has a vertex at the center of the other. The area of the part common to both squares measures
(A) $20 \mathrm{~cm}^{2}$
(B) $25 \mathrm{~cm}^{2}$
(C) $40 \mathrm{~cm}^{2}$
(D) $50 \mathrm{~cm}^{2}$
(E) depends on the position.
3) The answer is $\mathbf{( B )}$.

If lines parallel to the sides of the square are drawn from $P$, it can be seen that the two shaded right triangles are equal, and thus the area of the common part is that of the square with side $5 \mathrm{~cm}$.

SECOND SOLUTION

The answer is $\mathbf{( B )}$, as can be easily deduced from the figure on the side, since the quadrilaterals into which the first square is divided are equal.

![](https://cdn.mathpix.com/cropped/2024_04_17_dcf3bd4e9dbd26cb9fe5g-1.jpg?height=719&width=319&top_left_y=1148&top_left_x=1528)