To solve this problem, let’s analyze the given data and proceed step by step.

Given:

1. ABCD is a rectangle with a breadth of 12 cm and a length that is twice its breadth. So:

   • Length = $2 \times 12 = 24$ cm.

2. $AF = 12 \, \text{cm}$, which divides the top side into equal halves (since $AB = 24 \, \text{cm}$).

3. $AE = ED$, meaning $AE$ and $ED$ each equal half the breadth, $6 \, \text{cm}$ (since $AD = 12 \, \text{cm}$).

To Find:

The area of triangle $CFE$ (shaded).

Step-by-Step Solution:

1. Vertices of Triangle $CFE$:

   • $C(24, 0)$
   • $F(12, 12)$
   • $E(6, 6)$

2. Using the formula for the area of a triangle given vertices: $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ Here, $(x_1, y_1) = (24, 0)$, $(x_2, y_2) = (12, 12)$, $(x_3, y_3) = (6, 6)$.

3. Substituting values: $\text{Area} = \frac{1}{2} \left| 24(12 - 6) + 12(6 - 0) + 6(0 - 12) \right|$ $\text{Area} = \frac{1}{2} \left| 24 \cdot 6 + 12 \cdot 6 - 6 \cdot 12 \right|$ $\text{Area} = \frac{1}{2} \left| 144 + 72 - 72 \right|$ $\text{Area} = \frac{1}{2} \times 144 = 72 \, \text{cm}^2$

Final Answer:

The area of the shaded triangle $CFE$ is 72 cm².

Let me know if you'd like more details or further clarifications.

Related Questions:

1. How would the area change if $AB$ was tripled instead of doubled?
2. Can we solve this using coordinate geometry for non-rectangular shapes?
3. How do the dimensions of the rectangle affect the location of $CFE$?
4. What if $AF \neq 12 \, \text{cm}$, how would that affect the calculation?
5. What alternative methods (e.g., geometric reasoning) can we use to find this area?

Tip:

For coordinate geometry, always double-check the arrangement of points and the orientation before applying formulas.