Calculating the length of an arc and the area of a sector of a circle, and calculating the volume of standard solids including the prism, cylinder, pyramid, cone and sphere.

What this dot point is asking

The SQA wants you to find the length of an arc and the area of a sector of a circle, treating each as a fraction of the whole circle, and to calculate the volume of the standard solids: the prism, cylinder, pyramid, cone and sphere. These are calculator-paper staples that reward careful substitution and correct rounding.

Arcs and sectors as fractions of a circle

A full circle turns through $360^\circ$. An arc subtending an angle $\theta$ at the centre is the fraction $\dfrac{\theta}{360}$ of the whole circumference, and a sector with that angle is the same fraction of the whole area.

You can also work backwards: if you know the arc length or sector area, set up the same equation and solve for the missing angle or radius.

Volume of solids

A prism has the same cross-section all along its length, so its volume is the area of that cross-section times the length. A cylinder is a prism with a circular cross-section.

Composite solids and working backwards

Real objects are often made from two or more standard solids joined together, and the total volume is the sum of the parts. A pencil sharpened to a point is a cylinder plus a cone; an ice-cream cone topped with a scoop is a cone plus a hemisphere (half a sphere, volume $\tfrac{2}{3}\pi r^3$). Find each piece separately, then add.

Questions also run in reverse: you may be told the volume and asked for a missing length. Substitute the known volume into the formula and rearrange. For a cylinder of volume $200\pi$ cm$^3$ and radius $5$ cm, $200\pi = \pi \times 25 \times h$, so $h = \dfrac{200\pi}{25\pi} = 8$ cm.

Examples in context

These formulae size up real objects. A grain silo modelled as a cylinder topped by a cone has its capacity found by adding $\pi r^2 h$ for the cylinder and $\tfrac{1}{3}\pi r^2 h$ for the cone. A windscreen wiper sweeps a sector of a circle, so the glass it cleans is $\dfrac{\theta}{360} \times \pi r^2$. Keeping symbolic answers like $36\pi$ until the final step avoids rounding errors building up.

Try this

Q1. Find the length of an arc of radius $10$ cm and angle $90^\circ$. [2 marks]

• Cue. $\dfrac{90}{360} \times 2\pi \times 10 = 5\pi = 15.7$ cm.

Q2. Find the volume of a cone of radius $4$ cm and height $9$ cm. [2 marks]

• Cue. $\tfrac{1}{3}\pi \times 16 \times 9 = 48\pi = 151$ cm$^3$.

Q3. Find the area of a sector of radius $6$ cm and angle $120^\circ$. [2 marks]

• Cue. $\dfrac{120}{360} \times \pi \times 36 = 12\pi = 37.7$ cm$^2$.