Chapter 11: Areas Related to Circles

Earlier classes introduced two friends , the circumference $C = 2\pi r$ and the area $A = \pi r^2$ , for a full circle of radius $r$. This chapter generalises both: instead of using the whole circle, use only a slice of it. A slice cut by two radii is a sector; a slice cut by a chord is a segment. Then we combine these slices with rectangles, triangles, and other circles to compute areas of intricate shaded regions , the diagrams you see in the board exam, often with a clock face or a flower pattern.

Three skills are tested here. First, fluency with the formulas for arc length, sector area, and segment area. Second, recognising the combination of shapes in a diagram , what to add and what to subtract. Third, careful arithmetic with $\pi$, where the textbook usually wants either $\pi = 22/7$ or $\pi = 3.14$ depending on the question's request.

For the board exam this is a high-yield chapter , almost every paper has at least one $3$- or $4$-mark question that asks you to compute the area of a shaded region built from circles, sectors, and polygons. The reasoning is mostly arithmetic; mistakes usually come from using the wrong formula or forgetting a unit.

Real-world applications abound: pizza slices, flower-bed designs, the swept area of a windscreen wiper, the rotating cone of a lighthouse beam, the cross-section of a pipe. The same calculations underpin gear design, surveying, and astronomy (the angular size of a planet seen from Earth).

What's inside

• Circumference and area refresher , values of $\pi$.
• Arc length and sector area , formulas in degrees and the proof by proportion.
• Segment area , minor and major.
• Combined figures , add, subtract, overlap.
• Word problems , wipers, clock hands, flower beds.

Key results / Formula card

• Circumference: $C = 2\pi r$. Area: $A = \pi r^2$.
• Arc length subtending angle $\theta$ (in degrees) at the centre: $\ell = \dfrac{\theta}{360^\circ} \cdot 2\pi r$.
• Sector area (angle $\theta$): $A_{\text{sec}} = \dfrac{\theta}{360^\circ} \cdot \pi r^2 = \dfrac{1}{2} r \ell$.
• Segment area (minor): $A_{\text{seg}} = A_{\text{sec}} - A_{\text{triangle}}$, where the triangle is the isosceles triangle formed by the two radii and the chord.
• Major segment $= \pi r^2 -$ (minor segment).
• $\pi \approx 22/7$ or $3.14$ as the problem dictates.