Derive the formula for the area of a segment of a circle. Write the expression for the area of the minor segment $APB$ of a circle with centre $O$, radius $r$, and sector angle $\theta$.

Consider a circle with centre $O$, radius $r$, and a minor sector $OAPB$ of angle $\theta$. The chord $AB$ divides the minor sector into two regions: the triangle $\triangle OAB$ and the minor segment $APB$. Therefore: $$\text{Area of minor segment } APB = \text{Area of sector } OAPB - \text{Area of } \triangle OAB$$ Substituting the sector area formula: $$\text{Area of minor segment } APB = \frac{\theta}{360} \times \pi r^2 - \text{Area of } \triangle OAB$$ For the major segment $AQB$: $$\text{Area of major segment } AQB = \pi r^2 - \text{Area of minor segment } APB$$ The area of $\triangle OAB$ depends on the angle $\theta$ and the radius $r$. For specific angles, standard triangle area formulas apply: When $\theta = 90^\circ$, $\triangle OAB$ is right-angled, area $= \frac{1}{2}r^2$. When $\theta = 60^\circ$, $\triangle OAB$ is equilateral, area $= \frac{\sqrt{3}}{4}r^2$. When $\theta = 120^\circ$, use $\frac{1}{2}r^2\sin 120^\circ$ or split into two right triangles.