We are given a triangle $XYZ$ with a line segment $PQ$ parallel to the base $YZ$. We are given that $|XP| = 2$ cm, $|PY| = 3$ cm, $|PQ| = 6$ cm, and the area of triangle $XPQ$ is 24 cm$^2$. We are asked to find the area of the trapezoid $PQZY$.

1. Problem Description

We are given a triangle $XYZ$ with a line segment $PQ$ parallel to the base $YZ$. We are given that $|XP| = 2$ cm, $|PY| = 3$ cm, $|PQ| = 6$ cm, and the area of triangle $XPQ$ is 24 cm$^2$. We are asked to find the area of the trapezoid $PQZY$.

2. Solution Steps

Since $PQ \parallel YZ$, triangles $XPQ$ and $XYZ$ are similar. The ratio of their corresponding sides is given by

$\frac{XP}{XY} = \frac{2}{2+3} = \frac{2}{5}$.

The ratio of their areas is the square of the ratio of corresponding sides:

$\frac{\text{Area}(XPQ)}{\text{Area}(XYZ)} = \left(\frac{XP}{XY}\right)^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25}$.

We are given that the area of triangle $XPQ$ is 24 cm$^2$. Therefore, we have

$\frac{24}{\text{Area}(XYZ)} = \frac{4}{25}$.

$\text{Area}(XYZ) = \frac{24 \cdot 25}{4} = 6 \cdot 25 = 150$ cm$^2$.

The area of the trapezoid $PQZY$ is the difference between the area of triangle $XYZ$ and the area of triangle $XPQ$.

Area$(PQZY) = \text{Area}(XYZ) - \text{Area}(XPQ) = 150 - 24 = 126$ cm$^2$.

3. Final Answer

The area of the trapezium $PQZY$ is 126 cm$^2$.