June 2019 Paper 2 Q12
12 The diagram shows a solid hemisphere.
The diameter is \(12a\) cm
The volume is \(486\pi\) cm3
Work out the value of \(a\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\dfrac{1}{2} \times \dfrac{4}{3} \times \pi \times (6a)^3\) or \(\dfrac{2}{3} \times \pi \times 216a^3\) or \(144\pi a^3\) | M1 | oe eg \(\dfrac{1}{2} \times \dfrac{4}{3} \times \pi \times \left(\dfrac{12a}{2}\right)^3\) or \(\dfrac{2}{3} \times \pi \times (6a)^3\) |
| \(a^3 = \dfrac{486\pi}{144\pi}\) or \(a^3 = \dfrac{27}{8}\) or \(a^3 = 486 \div \left(\dfrac{2}{3} \times 6^3\right)\) or \(a^3 = 3.375\) or \(\sqrt[3]{3.375}\) | A1 | oe equation of form \(a^3 =\) or calculation allow \((6a)^3 = 729\) or \(6a = 9\) |
| \(\dfrac{3}{2}\) or \(1\dfrac{1}{2}\) or 1.5 | A1 | SC1 answer 0.75 oe or answer 1.19… or answer 4.95… |
| Alternative method 2 | ||
| \(r^3 = \dfrac{486\pi}{\frac{2}{3}\pi}\) or \(r^3 = 729\) or \(\sqrt[3]{729}\) or 9 | M1 | oe equation of form \(r^3 =\) or calculation |
| \(6a = \sqrt[3]{\dfrac{486\pi}{\frac{2}{3}\pi}}\) or \(6a = 9\) or \(9 \div 6\) | A1 | oe equation or calculation allow \((6a)^3 = 729\) |
| \(\dfrac{3}{2}\) or \(1\dfrac{1}{2}\) or 1.5 | A1 | SC1 answer 0.75 oe or answer 1.19… or answer 4.95… |
Additional guidance
| Allow recovery of missing brackets | |
| Allow use of \(\pi = [3.14,\ 3.142]\) |