9. \( A B C D \) is a square. \( M \) is the midpoint of \( A B \) and \( N \) is the midpoint of \( A D \). A point is selected at random in the square. Find the probability that the point lies in the triangle MAN.

To find the probability that a randomly selected point in square \( ABCD \) lies in triangle \( MAN \), we first need to determine the area of both the triangle \( MAN \) and the square. Let's denote the side length of the square \( ABCD \) as \( s \). The area of square \( ABCD \) is then: \[ \text{Area of square } ABCD = s^2 \] Next, we need to identify the coordinates of points \( M \) and \( N \): - If \( A \) is at \( (0, s) \), \( B \) is at \( (s, s) \), \( C \) is at \( (s, 0) \), and \( D \) is at \( (0, 0) \), - The coordinates of the midpoints are: - Point \( M \) (midpoint of \( AB \)): \[ M = \left(\frac{0+s}{2}, \frac{s+s}{2}\right) = \left(\frac{s}{2}, s\right) \] - Point \( N \) (midpoint of \( AD \)): \[ N = \left(\frac{0+0}{2}, \frac{s+0}{2}\right) = \left(0, \frac{s}{2}\right) \] Now the vertices of triangle \( MAN \) are at points \( M\), \( A\), and \( N \): - \( A(0, s) \) - \( M\left(\frac{s}{2}, s\right) \) - \( N(0, \frac{s}{2}) \) The area of triangle \( MAN \) can be calculated using the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\): \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Plugging in our coordinates: - \( A(0, s) \) - \( M\left(\frac{s}{2}, s\right) \) - \( N(0, \frac{s}{2}) \) We find the area as follows: \[ \text{Area} = \frac{1}{2} \left| 0\left(s - \frac{s}{2}\right) + \frac{s}{2}\left(\frac{s}{2} - s\right) + 0\left(s - s\right) \right| \] \[ = \frac{1}{2} \left| 0 + \frac{s}{2}\left(-\frac{s}{2}\right) + 0 \right| \] \[ = \frac{1}{2} \left| -\frac{s^2}{4} \right| = \frac{s^2}{8} \] Now, we have the area of triangle \( MAN \) as \( \frac{s^2}{8} \) and the area of square \( ABCD \) as \( s^2 \). Finally, the probability \( P \) that a randomly chosen point in the square lies within the triangle is given by the ratio of the area of the triangle to the area of the square: \[ P = \frac{\text{Area of triangle } MAN}{\text{Area of square } ABCD} = \frac{\frac{s^2}{8}}{s^2} = \frac{1}{8} \] Therefore, the probability that the point lies in triangle \( MAN \) is \( \frac{1}{8} \).