Given the problem, we need to find the area of the shaded region inside the square, which has a side length of 1 unit.

Explanation:

1. Square Properties:

   • The square $ABCD$ has a side length of 1 unit. Therefore, the area of the square is $1 \times 1 = 1$ square unit.

2. Line Intersection:

   • The lines $AC$ and $BD$ intersect at the center of the square, which is point $O$. Since the square has a side length of 1 unit, the coordinates of point $O$ are $\left( \frac{1}{2}, \frac{1}{2} \right)$.

3. Triangle Analysis:

   • Point $M$ is the midpoint of side $AB$, so $M$ has coordinates $\left( \frac{1}{2}, 0 \right)$.
   • The shaded region is a triangle with vertices $A$, $O$, and $M$.

4. Area of the Shaded Triangle $\Delta AOM$:

   • The formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is: $\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$
   • Substituting the coordinates of points $A(0, 0)$, $O\left(\frac{1}{2}, \frac{1}{2}\right)$, and $M\left(\frac{1}{2}, 0\right)$: $\text{Area} = \frac{1}{2} \left| 0 \left(\frac{1}{2} - 0\right) + \frac{1}{2}(0 - 0) + \frac{1}{2}(0 - \frac{1}{2}) \right|$ $= \frac{1}{2} \left| 0 + 0 - \frac{1}{8} \right| = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}$

Hence, the area of the shaded region is $\frac{1}{16}$ square units.

Would you like more details or have any questions?

Related Questions:

1. How do you calculate the area of a triangle given its vertices?
2. What is the formula for the area of a square?
3. How do you find the midpoint of a line segment in a coordinate plane?
4. What is the relationship between diagonals in a square?
5. How does changing the side length of a square affect the area of shapes inside it?

Tip:

When calculating the area of a triangle using vertices, always use the absolute value to ensure you get a positive result.