June 2019 Paper 1 Q11
11 A cone has base radius \(r\) cm, perpendicular height \(h\) cm and slant height \(l\) cm
The curved surface area is \(60\pi\) cm2
\(l = 3r\)
Work out the value of \(h\).
Give your answer in the form \(a\sqrt{10}\) where \(a\) is an integer greater than 1
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\pi \times r \times 3r = 60\pi\) | M1 | oe |
| \(r^2 = 20\) or \(r = \sqrt{20}\) or \(r = 2\sqrt{5}\) | A1 | oe |
| \((l =)\ 3\sqrt{20}\) or \((l =)\ 6\sqrt{5}\) or \((l =)\ \sqrt{180}\) or \(l^2 = 180\) | A1 | oe |
| \((h^2 =)\ (3\sqrt{20})^2 - (\sqrt{20})^2\) or \((h^2 =)\ (6\sqrt{5})^2 - (2\sqrt{5})^2\) or \((h^2 =)\ (\sqrt{180})^2 - (\sqrt{20})^2\) or \((h^2 =)\ 160\) | M1 | oe using their \(l\) and \(r\) (this is independent so \(l\) and \(r\) can be anything) condone missing brackets |
| \((h =)\ 4\sqrt{10}\) | A1 | |
| Alternative method 2 | ||
| \(\pi \times \dfrac{l}{3} \times l = 60\pi\) | M1 | oe |
| \(l^2 = 180\) or \(l = \sqrt{180}\) or \(l = 3\sqrt{20}\) or \(l = 6\sqrt{5}\) | A1 | oe |
| \(r^2 = 20\) or \((r =)\ \sqrt{20}\) or \((r =)\ 2\sqrt{5}\) | A1 | oe |
| \((h^2 =)\ (3\sqrt{20})^2 - (\sqrt{20})^2\) or \((h^2 =)\ (6\sqrt{5})^2 - (2\sqrt{5})^2\) or \((h^2 =)\ (\sqrt{180})^2 - (\sqrt{20})^2\) or \((h^2 =)\ 160\) | M1 | oe using their \(l\) and \(r\) (this is independent so \(l\) and \(r\) can be anything) condone missing brackets |
| \((h =)\ 4\sqrt{10}\) | A1 | |
| Alternative method 3 | ||
| \(\pi \times r \times 3r = 60\pi\) or \(\pi \times \dfrac{l}{3} \times l = 60\pi\) | M1 | oe |
| \(r^2 = 20\) or \(r = \sqrt{20}\) or \(r = 2\sqrt{5}\) or \(l = 3\sqrt{20}\) or \(l = 6\sqrt{5}\) or \(l = \sqrt{180}\) or \(l^2 = 180\) | A1 | oe |
| \(r^2 + h^2 = (3r)^2\) or \((h^2 =)\ 9r^2 - r^2\) or \(\left(\dfrac{l}{3}\right)^2 + h^2 = l^2\) or \((h^2 =)\ l^2 - \dfrac{l^2}{9}\) | M1 | oe to form an equation with only 2 variables using their \(l\) or \(r\) (this is independent so \(l\) and \(r\) can be anything) |
| \((h =)\ r\sqrt{8}\) or \((h^2 =)\ 160\) | A1 | oe |
| \((h =)\ 4\sqrt{10}\) | A1 | |
Additional guidance
| Second M mark is independent of first M mark | |
| Answer with no working will not gain any marks | |
| Minimum working for full marks would be a correct expression in the second M mark for alt method 1 and alt method 2. In this the candidate would show \(l\) and \(r\) so the first M mark would be implied. On alt method 3 they would need to show correct evidence in the first A mark and second M mark as a minimum expectation | M1, A1, A1, M1, A1 |