Surface Areas and Volumes

This chapter is a formula-heavy mensuration chapter. Memorise the table once and the problems become arithmetic. Common 4-mark question: a composite shape (e.g. cylinder + hemisphere). The trick is knowing exactly which surfaces are exposed.

1. The complete formula sheet

For all formulas, $\pi \approx \frac{22}{7}$ unless stated.

Cuboid (length $l$, breadth $b$, height $h$)

$\text{Lateral surface area (LSA)}=2h(l+b)$ $\text{Total surface area (TSA)}=2(lb+bh+hl)$ $\text{Volume}=lbh$ $\text{Diagonal}=\sqrt{l^2+b^2+h^2}$

Cube (side $a$)

$\text{LSA}=4a^2,\text{TSA}=6a^2,\text{Volume}=a^3,\text{Diagonal}=a\sqrt{3}$

Right circular cylinder (radius $r$, height $h$)

$\text{Curved surface area (CSA)}=2\pi rh$ $\text{TSA}=2\pi r(r+h)$ $\text{Volume}=\pi r^{2}h$

Right circular cone (radius $r$, height $h$, slant $\ell$)

$\ell =\sqrt{r^2+h^2}$ $\text{CSA}=\pi r\ell$ $\text{TSA}=\pi r(r+\ell )$ $\text{Volume}=\frac{1}{3}\pi r^{2}h$

Sphere (radius $r$)

$\text{Surface area}=4\pi r^2$ $\text{Volume}=\frac{4}{3}\pi r^3$

Hemisphere (radius $r$)

$\text{CSA}=2\pi r^2$ $\text{TSA}=3\pi r^2\text{(adds the flat circular face)}$ $\text{Volume}=\frac{2}{3}\pi r^3$

2. Worked example — cylinder

A cylindrical pillar has radius 0.7 m and height 4 m. Find the curved surface area and volume.

CSA $=2\pi rh=2\times \frac{22}{7}\times 0.7\times 4=17.6$ m².

Volume $=\pi r^{2}h=\frac{22}{7}\times 0.49\times 4=6.16$ m³.

3. Worked example — composite shape

A toy is in the shape of a cone mounted on a hemisphere. Both have the same radius 3 cm, and the cone's height is 4 cm. Find the total surface area.

The exposed surfaces are:

• The cone's curved surface (not the base — it's joined to the hemisphere).
• The hemisphere's curved surface (not the flat face — it's joined to the cone).

Cone slant $\ell =\sqrt{3^2+4^2}=5$.

${\text{CSA}}_{\text{cone}}=\pi r\ell =\pi \times 3\times 5=15\pi$. ${\text{CSA}}_{\text{hemisphere}}=2\pi r^2=2\pi \times 9=18\pi$.

Total $=33\pi \approx 103.7$ cm².

4. The "conservation of volume" trick

When you melt one solid and recast it as another, volume is conserved (mass = volume × density and density is unchanged).

Example. A solid cylindrical iron rod of radius 2 cm and height 16 cm is melted and recast into a sphere. Find the sphere's radius.

$\pi r_{\text{cyl}}^{2}h=\frac{4}{3}\pi r_{\text{sph}}^3$ $\pi \cdot 4\cdot 16=\frac{4}{3}\pi r^3$ $r^3=\frac{3\cdot 64}{4}=48\Rightarrow r=\sqrt[3]{48}\approx 3.63\text{ cm}.$

5. Tips for marks

• Always state the formula first before substituting. Examiners give 1 mark for the formula alone.
• Mind the units. Convert everything to the same unit before computing. Volume answers go in cm³ or m³.
• For composite shapes, draw a sketch and mark which surfaces are exposed. The figure earns 1 mark.

What's next

Statistics — the final chapter — switches gears entirely into data: mean, median, mode, bar graphs, histograms and frequency polygons.