Hint
Combine \(\displaystyle A= \frac{r^{2}\theta }{2}\) and \(\displaystyle s=r\theta\) to obtain \(\displaystyle A= \frac{rs}{2}\).
Worked solution
(a)
Since \(\displaystyle A= \frac{rs}{2}, 54= \frac{r(12)}{2}= 6r\). Hence \(\displaystyle r=9 \mathrm{cm}\).
Answer: \(9\text{ cm}\).
(b)
\(\displaystyle \theta = \frac{s}{r}= \frac{12}{9}= \frac{4}{3}\) rad, which is between 0 and 2π.
Answer: \(\displaystyle \frac{4}{3}\text{ rad}\).
(c)
The perimeter is \(\displaystyle 12+2(9)=30 \mathrm{cm}\). The full disk area is \(\displaystyle 81\pi \mathrm{cm}^{2}\), so the fraction is \(\displaystyle \frac{54}{81\pi }= \frac{2}{3\pi }\). This also equals \(\displaystyle \frac{\theta }{2\pi }\).
Answer: Perimeter 30 cm; fraction \(\displaystyle \frac{2}{3\pi }\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Relate area and arc length using \(\displaystyle A= \frac{rs}{2}\) or equivalent elimination.
- 1 mark: Set up and solve \(\displaystyle 54=6r\).
- 1 mark: Give radius 9 cm.
Alternative methods: A simultaneous solution of the two standard sector equations is accepted.
Common error: Using the arc length as a perimeter or a diameter.
(b) · 2 marks
- 1 mark: Use \(\displaystyle \theta = \frac{s}{r}\) with the recovered radius.
- 1 mark: Give exact angle \(\displaystyle \frac{4}{3}\) rad.
Alternative methods: Substitution into the sector-area formula is accepted.
Common error: Using \(\displaystyle \theta = \frac{r}{s}\).
(c) · 2 marks
- 1 mark: Give complete perimeter 30 cm.
- 1 mark: Give the dimensionless disk fraction \(\displaystyle \frac{2}{3\pi }\).
Alternative methods: Using the angle fraction \(\displaystyle \frac{\theta }{2\pi }\) is accepted.
Common error: Dividing the arc by radius and treating that radian angle as the fraction of a full disk.
Total: 7 marks.