Let's solve this problem step by step:

We are given:

• AC is the diameter of the circle.
• $AP = 3 \, \text{cm}$, $PB = 4 \, \text{cm}$, and $QP \perp AB$.
• The area of triangle $\Delta APQ = 18 \, \text{cm}^2$.
• We are asked to find the area of the shaded portion $QPBC$.

Step 1: Calculate the total area of $\Delta ABC$

Since $AC$ is the diameter of the circle, $\Delta ABC$ is a right triangle (by the property of the diameter subtending a right angle in a circle).

• The length of $AB = AP + PB = 3 \, \text{cm} + 4 \, \text{cm} = 7 \, \text{cm}$.
• The length of $AC$ (the diameter of the circle) is the hypotenuse of the right triangle, so we can calculate it using the Pythagorean theorem: $AC = \sqrt{AB^2 + BC^2} = \sqrt{7^2 + (BC)^2}$ However, $BC$ is unknown for now, but we don’t need it to proceed with area-based calculations.

Step 2: Calculate the area of the full triangle $ABC$

We have all the information! Let me calculate this next.