Question: A sector has radius $6$ cm and angle $70°$ at the centre. Find the arc length and the area of the sector.

Step 1. A sector is just a fraction of the whole circle, and the fraction is $\dfrac{70°}{360°}$.

Arc length = fraction × circumference; sector area = fraction × circle area.

Step 2. Arc length:

$$\text{Arc} = \dfrac{70}{360} \times 2\pi(6) = 7.3303\ldots \approx 7.33 \text{ cm} \quad (3 \text{ s.f.})$$

Step 3. Sector area:

$$\text{Area} = \dfrac{70}{360} \times \pi(6)^2 = 21.991\ldots \approx 22.0 \text{ cm}^2 \quad (3 \text{ s.f.})$$

Step 4. If the perimeter of the sector were asked, add the two radii to the arc:

$$\text{Perimeter} = 7.3303\ldots + 6 + 6 \approx 19.3 \text{ cm}$$

Check. $70°$ is about $\frac{1}{5}$ of the circle. Circumference $2\pi(6) \approx 37.7$, and $\frac{1}{5}$ of that is $\approx 7.5$ — close to $7.33$. ✓

⚠ Watch out: Arc length uses the circumference $2\pi r$; sector area uses the circle area $\pi r^2$ — mixing them up is the most common slip. And a sector's perimeter is arc $+\ 2r$: the two straight radii count, not the arc alone.

Question: The volume of a sphere is $523\frac{17}{21}$ cm$^3$. Taking $\pi = \frac{22}{7}$, find the radius and the surface area of the sphere.

Step 1. Form an equation from the volume formula:

$$\dfrac{4}{3}\pi r^3 = 523\dfrac{17}{21}$$

Step 2. Substitute $\pi = \frac{22}{7}$ and solve for $r^3$:

$$\dfrac{4}{3}\left(\dfrac{22}{7}\right) r^3 = \dfrac{11000}{21}$$ $$\dfrac{88}{21}\, r^3 = \dfrac{11000}{21} \;\Rightarrow\; r^3 = \dfrac{11000}{88} = 125$$

Step 3. Cube-root — not square-root:

$$r = \sqrt[3]{125} = 5 \text{ cm (exact)}$$

Step 4. Now the surface area, with the radius found:

$$\text{Surface area} = 4\pi r^2 = 4\left(\dfrac{22}{7}\right)(5)^2 = \dfrac{2200}{7} = 314\dfrac{2}{7} \text{ cm}^2$$

Check. Volume back: $\frac{4}{3} \cdot \frac{22}{7} \cdot 125 = \frac{11000}{21} = 523\frac{17}{21}$. ✓

⚠ Watch out: Volume gives $r^3$, so undo it with a cube root; surface area gives $r^2$, undone with a square root. And if a question gives the diameter, halve it first — substituting $d$ straight into $4\pi r^2$ quadruples the answer.

Question: A cone has base radius $5$ cm and vertical height $14$ cm. Find its total surface area.

Step 1. Total surface area of a cone = curved surface + base circle:

$$\text{TSA} = \pi r l + \pi r^2$$

The formula needs the slant height $l$ — which is not given, so find it first.

Step 2. The radius, the vertical height and the slant height form a right-angled triangle, with $l$ as the hypotenuse:

$$l = \sqrt{14^2 + 5^2} = \sqrt{221} = 14.866\ldots \text{ cm}$$

Step 3. Curved surface area:

$$\pi r l = \pi(5)(14.866\ldots) = 233.51\ldots \text{ cm}^2$$

Step 4. Add the base circle:

$$\pi r^2 = \pi(5)^2 = 78.539\ldots \text{ cm}^2$$ $$\text{TSA} = 233.51\ldots + 78.539\ldots = 312.05\ldots \approx 312 \text{ cm}^2 \quad (3 \text{ s.f.})$$

⚠ Watch out: The slant height is $l = \sqrt{r^2 + h^2}$, never $r + h$ — and keep the two heights in their own formulas: volume $\frac{1}{3}\pi r^2 h$ uses the vertical height, curved surface $\pi r l$ uses the slant height. Read the question for "curved" vs "total": $\pi rl$ alone is only the curved part.

Question: In trapezium $ABCD$, the parallel sides are $AD = 9$ cm and $BC = 23$ cm, and the area is $192$ cm$^2$. $AE$ is the perpendicular height between the parallel sides. Find $AE$.

[diagram in original sheet: trapezium with AD on top, BC at the bottom, height AE drawn perpendicular to BC]

Step 1. Write down the trapezium area formula:

$$\text{Area} = \dfrac{1}{2}(\text{sum of parallel sides}) \times h$$

Step 2. Substitute what is known — the area and the two parallel sides — leaving the height as the unknown:

$$\dfrac{1}{2}(9 + 23) \times AE = 192$$

Step 3. Simplify the bracket first, then solve the equation:

$$\dfrac{1}{2}(32) \times AE = 192$$ $$16 \times AE = 192$$ $$AE = \dfrac{192}{16} = 12 \text{ cm}$$

Check. Substitute back: $\frac{1}{2}(9+23)(12) = 16 \times 12 = 192$ cm$^2$. ✓

⚠ Watch out: $h$ must be the perpendicular gap between the two parallel sides — a slanted side of the trapezium is never the height. And the "sum of parallel sides" means exactly the two parallel ones ($9$ and $23$), not any other pair.

Question: A solid is made by joining a cone (base radius $3$ cm, height $4$ cm, slant height $5$ cm) to a cylinder, which is joined to a hemisphere, all of radius $3$ cm. The total height of the solid is $14$ cm. Find, in terms of $\pi$, the volume and the total surface area.

[diagram in original sheet: cone on top of a cylinder on top of a hemisphere, total height marked 14 cm]

Step 1. Find the hidden dimension — the cylinder's height:

$$14 - 4 - 3 = 7 \text{ cm} \quad (\text{cone height } 4, \text{ hemisphere radius } 3)$$

Step 2. Volume = cone + cylinder + hemisphere:

$$V = \dfrac{1}{3}\pi(3)^2(4) + \pi(3)^2(7) + \dfrac{2}{3}\pi(3)^3$$ $$V = 12\pi + 63\pi + 18\pi = 93\pi \text{ cm}^3$$

Step 3. Surface area — only the exposed faces. The joins hide the flat circles, so it is three curved surfaces:

$$\text{SA} = \pi(3)(5) + 2\pi(3)(7) + 2\pi(3)^2$$ $$\text{SA} = 15\pi + 42\pi + 18\pi = 75\pi \text{ cm}^2$$

⚠ Watch out: For a composite solid, never add the full surface areas of each piece — the glued faces (the cone's base, the cylinder's two ends, the hemisphere's flat face) are inside the solid. Count curved surfaces only, and remember the hemisphere's curved surface is $2\pi r^2$, half of a sphere's $4\pi r^2$.

Question: A hemisphere of radius $7$ cm is placed flat-side down on top of a rectangular block $20$ cm by $15$ cm by $10$ cm. Taking $\pi = \frac{22}{7}$, find the area of the uncovered top of the block, the volume of the solid, and its mass in kg if the density is $20$ g/cm$^3$.

Step 1. The hemisphere's flat face covers a circle of radius $7$ on the top:

$$\text{Uncovered top} = 20(15) - \dfrac{22}{7}(7)^2 = 300 - 154 = 146 \text{ cm}^2$$

Step 2. Volume = block + hemisphere (half a sphere):

$$V = (20)(15)(10) + \dfrac{2}{3}\left(\dfrac{22}{7}\right)(7)^3 = 3000 + 718\dfrac{2}{3} = 3718\dfrac{2}{3} \text{ cm}^3$$

Step 3. Density $20$ g/cm$^3$ means every $1$ cm$^3$ has mass $20$ g:

$$\text{Mass} = 20 \times 3718\dfrac{2}{3} = 74373.3\ldots \text{ g}$$

Step 4. Convert to kg ($1$ kg $= 1000$ g) and round:

$$74373.3\ldots \text{ g} = 74.373\ldots \text{ kg} \approx 74.4 \text{ kg} \quad (3 \text{ s.f.})$$

⚠ Watch out: Hemisphere volume is $\frac{2}{3}\pi r^3$ — half of $\frac{4}{3}\pi r^3$, not half of $4\pi r^2$. And mass = density × volume comes out in grams here; convert to kg at the end, not before.