Area, Surface Area and Volume: Question 7

Worked solution

The sector is a fraction of a full circle. That fraction is the sector angle over $360^\circ$: $\frac{\theta}{360} = \frac{210}{360} = \frac{7}{12} = 0.58333\ldots$

We will use radius $r = 12\text{ cm}$ and $\pi = 3.142$ throughout.

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Part (a): Length of the arc

The arc length is that fraction of the full circumference $2\pi r$: $\text{arc} = \frac{\theta}{360}\times 2\pi r = \frac{210}{360}\times 2 \times 3.142 \times 12.$

Work out the circumference first: $2 \times 3.142 \times 12 = 75.408\ \text{cm}.$

Then take the fraction: $\text{arc} = \frac{210}{360}\times 75.408 = \frac{7}{12}\times 75.408 = 43.988\ \text{cm}.$

$\boxed{\text{arc} \approx 44.0\ \text{cm}\ (3\text{ s.f.})}$

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Part (b): Area of the sector

The sector area is the same fraction of the full circle area $\pi r^2$: $A = \frac{\theta}{360}\times \pi r^2 = \frac{210}{360}\times 3.142 \times 12^2.$

Work out the full-circle area first: $3.142 \times 12^2 = 3.142 \times 144 = 452.448\ \text{cm}^2.$

Then take the fraction: $A = \frac{210}{360}\times 452.448 = \frac{7}{12}\times 452.448 = 263.928\ \text{cm}^2.$

$\boxed{A \approx 264\ \text{cm}^2\ (3\text{ s.f.})}$

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Part (c): Perimeter of the sector (length of edging)

The outside boundary of a sector is made of three parts: the two straight radii and the curved arc.

Two straight radii: $12 + 12 = 24\ \text{cm}.$

Curved arc (from part (a), keep the full value $43.988\text{ cm}$, not the rounded $44.0$): $\text{arc} = 43.988\ \text{cm}.$

Total perimeter: $P = 24 + 43.988 = 67.988\ \text{cm}.$

$\boxed{P \approx 68.0\ \text{cm}\ (3\text{ s.f.})}$

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Method note

Because both an arc length and a sector area are just the fraction $\tfrac{\theta}{360}$ of the whole circle, it is efficient to compute the full circumference ($75.408\text{ cm}$) and full area ($452.448\text{ cm}^2$) once, then multiply each by $\tfrac{7}{12}$. In part (c), remember that a sector’s perimeter is not just the arc; you must also add the two radii that form the straight edges.