Higher November 2020 Unit 2 Q12
12. Calculate the perpendicular height of a cone with a volume of 5533 cm3 and a base area of 825 cm2. [3]
From the formula list: Volume of cone \(= \dfrac{1}{3}\pi r^2 h\)
| Scheme | Marks | Comments |
|---|---|---|
| (Height =) \(\dfrac{3 \times 5533}{825}\) OR \(\dfrac{5533}{\frac{1}{3} \times 825}\) | M2 | M1 for \(5533 = 1/3 \times \text{height} \times 825\) or equivalent. |
| = 20·1(2 cm) | A1 | Allow an answer of 20(cm) from correct working. |
Alternative method (finding the radius first):
| Scheme | Marks | Comments |
|---|---|---|
| Use \(A = \pi r^2\) to evaluate \(r\) or \(r^2\). | Allow use of \(\pi\) = 3·14, 3·142 or 3·14(59…). When using the \(\pi\) button on the calculator, \(r\) = 16·2(05…) OR \(r^2\) = 262·6(…). There will be no FT for any radius other than \(r\) = 16cm, from working seen. | |
| (Height =) \(\dfrac{3 \times 5533}{\pi \times 16.2(05\ldots)^2}\) OR \(\dfrac{5533}{\frac{1}{3} \times \pi \times 16.2(05\ldots)^2}\) OR \(\dfrac{3 \times 5533}{\pi \times 262.6(\ldots)}\) OR \(\dfrac{5533}{\frac{1}{3} \times \pi \times 262.6(\ldots)}\) | M2 | M1 for \(5533 = 1/3 \times \text{height} \times \pi \times 16.2(05\ldots)^2\) or equivalent. Allow M1 for use of \(r\) = 16 (cm) |
| = 20·1(2… cm) | A1 | Allow an answer of 20(cm) from correct working. Accept an answer in the range 20·10 to 20·143(cm) FT base radius = 16 cm: Allow an answer in the range 20·6(cm) to 20·65(cm) OR 21(cm) from correct working. |