Hint
Use the stated units and precision. Identify the semicircle right angle and use AB as the hypotenuse.
Worked solution
(a)
AB is a diameter and C lies on the circle, so ∠ACB=90°. Then \(BC^2=1225-441\) gives \(\displaystyle BC=28 \mathrm{cm}\).
Answer: 28 cm.
(b)
Use half the product of the two perpendicular triangle sides.
Use the perpendicular sides \(\displaystyle AC=21\) and \(\displaystyle BC=28\): half their product is \(\displaystyle 294 \mathrm{cm}^{2}\).
Answer: \(294\)
(c)
Halve the stated diameter rather than treating it as a radius.
Half the diameter 35 is \(\displaystyle \frac{35}{2}\) cm.
Answer: \(\displaystyle \frac{35}{2}\)
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) · 3 marks
- 1 mark: State the angle-in-a-semicircle theorem and locate the right angle at C.
- 1 mark: Apply Pythagoras with the diameter as hypotenuse and take the positive length.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
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 half the product of the two perpendicular triangle sides.
- 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: Halve the stated diameter rather than treating it as a radius.
- 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: 7 marks.