In the figure given below, AC is a diameter of the circle.

Given,

We know that,

Angle in a semi-circle is a right angle.

∴ ∠ABC = 90°.

In ∆APQ and ∆ABC,

⇒ ∠APQ = ∠ABC [Both equal to 90°]

⇒ ∠BAC = ∠PAQ [Common angles]

∴ ∆APQ ∼ ∆ABC (By A.A. axiom)

We know that,

The ratio of area of similar triangles is equal to the ratio of the square of the corresponding sides.

$$\begin{aligned}\Rightarrow \frac{\text{Area of ∆APQ}}{\text{Area of ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{18}{\text{Area of ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{AB^2}{AP^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC}= \frac{(4 + 3)^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{7^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{49}{9} \times 18 = 98 \text{ cm}^2.\end{aligned}$$

From figure,

Area QPBC = Area of ∆ABC − Area of ∆APQ

= 98 - 18 = 80 cm².

Hence, option 3 is the correct option.