4MA0 Higher June 2014 Paper 4R Q20
20 The diagram shows a solid cone.
Diagram NOT accurately drawn
The diameter of the base of the cone is \(10a\) cm.
The height of the cone is \(12a\) cm.
The total surface area of the cone is \(360\pi\) cm2
The volume of the cone is \(k\pi\) cm3, where \(k\) is an integer.
Find the value of \(k\).
(6)
| Scheme | Marks |
|---|---|
| (Slant Height =) \(\sqrt{(5a)^2 + (12a)^2}\) (= 13\(a\)) | M1 |
(total surface area =) \(\pi \times (5a)^2 + \pi \times 5a \times \text{“}13a\text{”}\) oe or \(\pi \times (5a)^2 + \pi \times 5a \times \sqrt{(5a)^2 + (12a)^2}\) (=\(90\pi a^2\)) | M1 |
| eg. \(90\pi a^2 = 360\pi\) oe or \(\pi \times (5a)^2 + \pi \times 5a \times \text{“}13a\text{”} = 360\pi\) oe | M1 |
| A1 | |
\(V = \dfrac{1}{3} \times \pi \times (5 \times \text{“}2\text{”})^2 \times 12 \times \text{“}2\text{”}\) (\(=100\pi a^3\)) or \(V = \dfrac{1}{3} \times \pi \times 10^2 \times 24\) oe or \(k = \dfrac{1}{3} \times (5 \times \text{“}2\text{”})^2 \times 12 \times \text{“}2\text{”}\) | M1 |
| 800 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: correct use of Pythagoras – condone missing brackets
M1: dep on first M1 – must have either \(25a^2\) or \((5a)^2\)
M1: dep on first M1
for equation formed (need not be simplified) – must have either \(25a^2\) or \((5a)^2\)
A1: \(a = 2\)
M1: dep on first M1
NB. For the award of this mark, brackets must be present or the value for \(r^2\) evaluated correctly for the candidate’s value of \(a\)
A1: cao