Hint
Subtract the two sector areas. The two straight edges each have length 7−4.
Worked solution
(a)
\(\displaystyle A= (\frac{1}{2})(7^{2}- 4^{2})(1.5)= (\frac{1}{2})(33)(1.5)= 24.75\mathrm{cm}^{2}\).
Answer: \(24.75\text{ cm}^2\).
(b)
The arcs total \(\displaystyle (7+4)(1.5)=16.5 \mathrm{cm}\). The straight edges total \(\displaystyle 2(7-4)=6 \mathrm{cm}\). The complete perimeter is 22.5 cm.
Answer: 22.5 cm.
(c)
The area ratio is \(\displaystyle \frac{99}{24.75}= 4\), so the positive length scale is \(\displaystyle \sqrt{4}=2\). Its perimeter is \(\displaystyle 2(22.5)=45 \mathrm{cm}\). Similarity preserves the angle, which remains 1.5 rad.
Answer: Factor 2; perimeter 45 cm; angle 1.5 rad.
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) · 2 marks
- 1 mark: Subtract the inner-sector area from the outer-sector area.
- 1 mark: Obtain \(\displaystyle 24.75 \mathrm{cm}^{2}\).
Alternative methods: The equivalent factorised expression \(\displaystyle (\frac{1}{2})(R- r)(R+ r)\theta\) is accepted.
Common error: Subtracting radii before squaring rather than subtracting their squares.
(b) · 2 marks
- 1 mark: Include both arcs and two thickness-length connectors.
- 1 mark: Obtain 22.5 cm.
Alternative methods: Adding all four boundary pieces individually is accepted.
Common error: Counting only the outer arc or using two whole outer radii.
(c) · 3 marks
- 1 mark: Obtain length scale 2 from the area ratio.
- 1 mark: Give perimeter 45 cm.
- 1 mark: State unchanged angle 1.5 rad.
Alternative methods: An equivalent radius-based similarity argument is accepted.
Common error: Multiplying the perimeter or the angle by the area ratio 4.
Total: 7 marks.