https://file.aiquickdraw.com/mathbot/file/1726576979625b6fj98kj.jpg

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.

In the adjoining figure, AC is a diameter of the circle. AP = 3 cm, PB = 4 cm, and QP ⊥ AB. If the area of ΔAPQ is 18 cm², then the area of the shaded portion QPBC is:

Solve the problem of finding the area of the shaded portion QPBC in a circle where AC is the diameter and QP ⊥ AB. Key concepts include geometry, area of triangles, and Pythagoras' Theorem. Suitable for students in Grades 9-12.

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