Hint
Count the bottom, two vertical sides and semicircular arc. The shared diameter is not an extra outside edge.
Worked solution
(a)
The exposed perimeter is 12+2r+πr. Thus \(\displaystyle (2+\pi )r=8+4\pi =4(2+\pi )\), giving \(\displaystyle r=4 \mathrm{cm}\).
Answer: \(r=4\text{ cm}\).
(b)
The rectangle area is \(\displaystyle 8(6)=48 \mathrm{cm}^{2}\) and the semicircle area is \(\displaystyle (\frac{1}{2})\pi (4^{2})= 8\pi \mathrm{cm}^{2}\). Total area is \(\displaystyle 48+8\pi \mathrm{cm}^{2}\).
Answer: \(48+8\pi\text{ cm}^2\).
(c)
The perimeter increase is \(\displaystyle \frac{20+ 4\pi }{10}= 2+ \frac{2\pi }{5}\mathrm{cm}\). It is shared between the two vertical sides, so the height increases by \(\displaystyle 1+ \frac{\pi }{5}\). The new height is \(\displaystyle 7+ \frac{\pi }{5}\mathrm{cm}\). The area is then \(\displaystyle 8(7+ \frac{\pi }{5})+ 8\pi = 56+ \frac{48\pi }{5}\mathrm{cm}^{2}\).
Answer: Height \(\displaystyle 7+ \frac{\pi }{5}\text{ cm}\); area \(\displaystyle 56+ \frac{48\pi }{5}\text{ cm}^{2}\).
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: Use outside perimeter 12+2r+πr.
- 1 mark: Solve to \(\displaystyle r=4 \mathrm{cm}\).
Alternative methods: An explicitly labelled boundary sum is accepted.
Common error: Counting the shared diameter in addition to the rectangle bottom.
(b) · 2 marks
- 1 mark: Add the correct rectangle and semicircle areas.
- 1 mark: Give exact area \(\displaystyle 48+8\pi \mathrm{cm}^{2}\).
Alternative methods: An equivalent factorised exact expression is accepted.
Common error: Using a full-circle area for the curved top.
(c) · 3 marks
- 1 mark: Relate the perimeter increase to twice the height increase.
- 1 mark: Obtain new height \(\displaystyle 7+ \frac{\pi }{5}\mathrm{cm}\).
- 1 mark: Obtain new area \(\displaystyle 56+ \frac{48\pi }{5}\mathrm{cm}^{2}\).
Alternative methods: Solving the new perimeter equation directly is accepted.
Common error: Increasing every length by 10% even though the radius must remain fixed.
Total: 7 marks.