Hint
Use the stated units and precision. Use base area and perpendicular height.
Worked solution
(a)
Use base area and perpendicular height.
One third of base area 144 times height 8 is \(384\) \(\displaystyle \mathrm{cm}^{3}\).
Answer: \(384\)
(b)
Use the height within the triangular face, not the pyramid vertical height.
Half the base-side and face-slant-height product is \(60\) \(\displaystyle \mathrm{cm}^{2}\).
Answer: \(60\)
(c)
Sum one square and four triangular faces.
The net contains the square base and four equal triangles, totalling \(\displaystyle 384 \mathrm{cm}^{2}\).
Answer: \(384\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 2 marks
- 1 mark: Use base area and perpendicular height.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
(b) · 2 marks
- 1 mark: Use the height within the triangular face, not the pyramid vertical height.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Check the answer against the original conditions.
(c) · 2 marks
- 1 mark: Sum one square and four triangular faces.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
Total: 6 marks.