Foundation June 2023 Paper 2R Q24
24 The diagram shows a rectangular sheet of metal \(ABCD\)
Diagram NOT accurately drawn
\(BD = 50\) cm and angle \(BDC = 32^\circ\)
Nasser joins side \(AD\) to side \(BC\) to form a cylinder.
\(BC\) is the height of the cylinder.
\(DC\) is the circumference of the cross section of the cylinder.
Work out the volume, in cm3, of the cylinder.
Give your answer correct to 3 significant figures.
(6)
| Scheme | Marks |
|---|---|
\(\sin 32 = \dfrac{(BC)}{50}\) or \(\cos 32 = \dfrac{(CD)}{50}\) or \(\dfrac{(BC)}{\sin 32} = \dfrac{50}{\sin 90}\) oe or \(\dfrac{(CD)}{\sin(90 - 32)} = \dfrac{50}{\sin 90}\) oe | M1 |
\((BC =)\;50\sin 32\;(= 26.4(959...))\) or \((BC =)\sqrt{50^2 - (50\cos 32)^2}\;(= 26.4(959...))\) or \((BC =)\sqrt{50^2 - \text{“}42.4...\text{”}^2}\;(= 26.4(998...))\) or \((BC =)\;\dfrac{50}{\sin 90} \times \sin 32\) oe | M1 |
\((CD =)\;50\cos 32\;(= 42.4(024)...)\) or \((CD =)\sqrt{50^2 - (50\sin 32)^2}\;(= 42.4(024...))\) or \((CD =)\sqrt{50^2 - \text{“}26.4...\text{”}^2}\;(= 42.4(622...))\) or \((CD =)\;\dfrac{50}{\sin 90} \times \sin(90 - 32)\) | M1 |
| \((r =)\) “42.4(024…)” ÷ \(2\pi\) (= 6.74(855…)) | M1 |
| \((V =)\;\pi \times\) “6.74(855…)”2 × “26.4(959…)” | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 3790 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for finding \(BC\) or \(AD\)
Can be written on the diagram
M1: for finding \(CD\) or \(BA\)
Can be written on the diagram
M1: for finding the radius of the cylinder
M1: dep on previous M mark for the use of \(\pi r^2h\)
A1: allow answers in the range 3737 – 3794
Accept answers in standard form