4400 Higher November 2009 Paper 4 Q21
21
Diagram NOT accurately drawn
\(A\), \(B\) and \(L\) are points on a circle, centre \(O\).
\(AB\) is a chord of the circle.
\(M\) is the midpoint of \(AB\).
\(LOM\) is a straight line.
\(AB = 24\) cm.
\(LM = 18\) cm.
Calculate the diameter of the circle.
(4)
| Scheme | Marks |
|---|---|
| \(12 \times 12 = 18(d - 18)\) or \(12 \times 12 = 18x\) | M1 |
| \(144 = 18d - 324\) or \(x = 8\) | M1 |
| \(18d = 468\) or \((d =)\ 8 + 18\) | M1 |
| 26 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: or for \(r^2 = 12^2 + (18 - r)^2\)
M1: or for \(r^2 = 144 + 324 - 18r - 18r + r^2\)
M1: or for \(36r = 468\)
A1: dep on all method marks
Alternative 1: \(AM = 12\), \(OM = 5\), \(OA = 13\) and \(13 + 5 = 18\) or \(18 - 5 = 13\): M3 for use of Pythagorean triple 5-12-13 or use of \(OM = 5\) Pythagoras to obtain \(\sqrt{5^2 + 12^2} = 13\) for \(OA\) and justification; 26 A1 dep on M3
Alternative 2: M1 for a complete, correct method for finding \(\angle AOM\) or \(\angle BOM\) or \(\angle OAB\) or \(\angle OBA\), eg \(\tan\angle ALM = \frac{12}{18}\), \(\angle ALM = 33.69°\), \(\angle AOM = 2 \times 33.69° = 67.38°\)
or \(AL = \sqrt{12^2 + 18^2} = \sqrt{468} = 21.63\) cm, \(\cos\angle ALB = \frac{468 + 468 - 576}{2 \times 468} = 0.3846\), \(\angle ALB = 67.38°\), \(\angle ALM = 33.69°\), \(\angle AOM = 67.38°\)
or \(\tan\angle ALM = \frac{12}{18}\), \(\angle ALM = 33.69°\), \(\angle OAM = 90° - 2 \times 33.69° = 22.62°\)
M1 for a correct numerical expression for the length of \(OA\) or \(OM\), eg \(\frac{12}{\sin 67.38°}\) or \(\frac{24\sin 22.62°}{\sin 134.76°}\) (= 13) or \(\frac{12}{\tan 67.38°}\) or \(12\tan 22.62°\) (= 5)
M1 for the length of \(OA\) or \(OM\) used to find the diameter, eg 2 × “13” or 2 × (18 − “5”), dep on both previous M1s
26 A1 dep on all method marks. Accept answer rounding to 26.0