Welcome to the World of Areas and Volumes!

Hello everyone! In S1 you mastered straight-edged shapes and boxes. This year we go round in circles — literally! This chapter covers circles, arcs and sectors, then moves into 3D with prisms and cylinders. (Pyramids, cones and spheres are the stars of S3 — don't worry about them yet.)

Part 1: The Circle — Circumference and Area

Everything in this chapter is powered by two famous formulas. For a circle with radius \(r\):

  • Circumference (the distance around): \(C = 2\pi r\) (or \(\pi d\), where \(d = 2r\) is the diameter)
  • Area: \(A = \pi r^2\)

Watch out: questions love to give you the diameter when the formula needs the radius. Always halve the diameter first!

Example: A circular pizza has diameter 30 cm, so \(r = 15\) cm. Circumference \(= 2\pi \times 15 \approx 94.2\) cm; Area \(= \pi \times 15^2 \approx 707\ \text{cm}^2\).

Key Takeaway for Part 1

\(C = 2\pi r\) and \(A = \pi r^2\). Radius, not diameter!


Part 2: Slices of the Pie — Arcs and Sectors

A sector is a pizza slice of a circle, and an arc is its curved crust. The whole idea of this part is one word: fraction. A sector with angle \(\theta\) at the centre is exactly \(\frac{\theta}{360^\circ}\) of the whole circle.

Arc Length

\(\text{Arc length} = \frac{\theta}{360^\circ} \times 2\pi r\)

Area of a Sector

\(\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2\)

Step-by-Step Example

A sector has radius 12 cm and angle \(60^\circ\). Find its arc length and area.

  1. The fraction of the circle: \(\frac{60^\circ}{360^\circ} = \frac{1}{6}\)
  2. Arc length: \(\frac{1}{6} \times 2\pi \times 12 = 4\pi \approx 12.6\) cm
  3. Sector area: \(\frac{1}{6} \times \pi \times 12^2 = 24\pi \approx 75.4\ \text{cm}^2\)

Perimeter trap: the perimeter of a sector is the arc plus the two radii: here \(4\pi + 12 + 12 \approx 36.6\) cm. Forgetting the two straight edges is the classic mistake!

Key Takeaway for Part 2

Arc and sector formulas are just "fraction of the circle": multiply \(2\pi r\) or \(\pi r^2\) by \(\frac{\theta}{360^\circ}\).


Part 3: Prisms — The "Stackable" Solids

A prism is a solid with the same cross-section all the way through — like a bar of chocolate or a Toblerone box. If you know the area of that cross-section (the base), volume is easy:

\(\text{Volume of prism} = \text{base area} \times \text{height}\)

Example: A triangular prism has a cross-section of area \(20\ \text{cm}^2\) and length 15 cm. \(V = 20 \times 15 = 300\ \text{cm}^3\).

Surface Area of a Prism

Total surface area = the areas of all the faces added together: the two identical ends plus every rectangular side. Imagine unfolding the solid into its net and adding up each face — the net never lies!

Key Takeaway for Part 3

Prism volume = base area × height. Surface area = add up every face of the net.


Part 4: Cylinders — Circular Prisms

A cylinder is really a prism whose base is a circle, so the same logic applies. For radius \(r\) and height \(h\):

  • Volume: \(V = \pi r^2 h\) (base area \(\pi r^2\), times height)
  • Curved surface area: \(2\pi r h\) (unroll the label of a tin — it's a rectangle of width \(2\pi r\) and height \(h\)!)
  • Total surface area (closed cylinder): \(2\pi r h + 2\pi r^2\) (curved surface + top and bottom circles)

Step-by-Step Example

A drink can has radius 3 cm and height 12 cm.

  1. Volume: \(V = \pi \times 3^2 \times 12 = 108\pi \approx 339\ \text{cm}^3\) — about 339 mL. Sounds right for a can!
  2. Curved surface: \(2\pi \times 3 \times 12 = 72\pi \approx 226\ \text{cm}^2\)
  3. Total surface: \(72\pi + 2\pi \times 3^2 = 72\pi + 18\pi = 90\pi \approx 283\ \text{cm}^2\)

Read the question carefully: an open cylinder (like a cup) has only ONE circle to add, and a pipe has none!

Key Takeaway for Part 4

Cylinder = circular prism: \(V = \pi r^2 h\), curved surface \(= 2\pi r h\).


Part 5: Composite Figures and Common Mistakes

Real exam questions mix these pieces: a running track is a rectangle plus two semicircles; a machine part is a cylinder with a cylindrical hole drilled out. The strategy is the same as S1: split or subtract, then add up areas or volumes.

Example: A washer is a disc of radius 5 cm with a hole of radius 2 cm. Area \(= \pi \times 5^2 - \pi \times 2^2 = 25\pi - 4\pi = 21\pi \approx 66.0\ \text{cm}^2\).

Common Mistakes to Avoid

  • Using the diameter where the formula needs the radius.
  • Forgetting the two radii in the perimeter of a sector.
  • Mixing up \(2\pi r\) (circumference) and \(\pi r^2\) (area) — length has one \(r\), area has \(r^2\).
  • Adding the top and bottom circles to an open cylinder's surface area.
  • Leaving answers in wrong units — area in \(\text{cm}^2\), volume in \(\text{cm}^3\), always.
Chapter Summary
  • Circle: \(C = 2\pi r\), \(A = \pi r^2\)
  • Sector of angle \(\theta\): arc \(= \frac{\theta}{360^\circ} \times 2\pi r\), area \(= \frac{\theta}{360^\circ} \times \pi r^2\)
  • Prism: \(V = \text{base area} \times \text{height}\); surface area from the net
  • Cylinder: \(V = \pi r^2 h\); curved surface \(2\pi r h\); closed total \(2\pi r h + 2\pi r^2\)

Next year in S3 you'll meet the pointy and round solids — pyramids, cones and spheres — plus similar figures. The circle skills you built here are exactly what they need. Great work!