This problem involves calculating the probability that a random point chosen within a square falls in a shaded triangular region.

Given Information:

1. A square $ABCD$ is divided into three parts.
2. $E$ is the midpoint of $AB$.
3. The shaded region is the triangle $ADE$.
4. The task is to find the probability that a random point within the square lies in the shaded triangular area.

Solution Steps:

1. Calculate the total area of the square: $\text{Area of square} = \text{Side}^2$

   Let the side length of the square be $s$.

   $\text{Area of square} = s^2$

2. Calculate the area of the triangle $ADE$: The base of triangle $ADE$ is $AB = s$, and the height of $ADE$ is the perpendicular distance from $D$ to $AB$, which is $s$.

   Since $E$ is the midpoint of $AB$, the area of triangle $ADE$ is: $\text{Area of } \triangle ADE = \frac{1}{2} \cdot \text{Base} \cdot \text{Height} = \frac{1}{2} \cdot s \cdot s = \frac{s^2}{2}$

3. Calculate the probability: The probability is the ratio of the shaded area (triangle $ADE$) to the total area of the square: $\text{Probability} = \frac{\text{Area of shaded region}}{\text{Total area of square}} = \frac{\frac{s^2}{2}}{s^2} = \frac{1}{2}$

4. Convert to percentage: \text{Probability (in %)} = \frac{1}{2} \cdot 100 = 50\%

Final Answer:

The probability that a randomly chosen point lies in the shaded region is 50%.

Would you like a further breakdown of this solution or any related questions?

Related Questions:

1. How would the probability change if $E$ were not the midpoint of $AB$?
2. What is the relationship between the area of the triangle $ADE$ and the area of triangle $BCE$?
3. How can this problem be solved geometrically without direct area calculation?
4. How does the symmetry of the square affect the probability distribution?
5. What if the point was restricted to the upper triangle $BCE$, what would the new probability be?

Tip:

When dealing with probabilities in geometric shapes, always simplify calculations by leveraging symmetry and known geometric formulas.