4400 Higher June 2008 Paper 3 Q22
22
Diagram NOT accurately drawn
The diagram shows a kite \(ABCD\).
\(AB = AD = 10\) cm.
\(CB = CD = 2\) cm.
Angle \(BCD = 90^\circ\).
Calculate the area of the kite.
(6)
| Scheme | Marks |
|---|---|
| Area of \(\Delta BCD = 2\) | B1 |
\((BD^2 =)\ 2^2 + 2^2\) or \(\left(\dfrac{BD}{2}\right)^2 + \left(\dfrac{BD}{2}\right)^2 = 2^2\) or \(\dfrac{BD/2}{2} = \cos 45^\circ\) or \(\sin 45^\circ\) or \(\dfrac{BD}{2} = 2\cos 45^\circ\) or \(2\sin 45^\circ\) | M1 |
\((BD =)\ \sqrt{8}\) or \(2\sqrt{2}\) or 2.83 or better (2.8284...) or \(\left(\dfrac{BD}{2}\right) = \sqrt{2}\) or \(\dfrac{\sqrt{8}}{2}\) or 1.41 or better (1.4142...) | A1 |
second part method 1 - uses Pythagoras to find \(AM\), where \(M\) is midpoint of \(BD\) \(AM^2 = 10^2 - \left(\dfrac{BD}{2}\right)^2\) | M1 |
| \(AM = \sqrt{98}\) or \(7\sqrt{2}\) or 9.90 or better (9.8994...) | A1 |
| 16 | A1 |
| (6) | |
| (6 marks) |
Notes
B1: for area of triangle \(BCD\)
M1: for correct start to Pythagoras or trig for finding \(BD\) or \(\left(\dfrac{BD}{2}\right)\)
A1: for lengths \(BD\) or \(\left(\dfrac{BD}{2}\right)\) correct
A1: for \(\sqrt{98}\) or \(7\sqrt{2}\) 9.90 or better
A1: for 16 or answer rounding to 16.0
Second part method 2
finds angle \(A\) either using Cosine Rule or by first finding \(\frac{A}{2}\) using trig
\(\cos A = \dfrac{10^2 + 10^2 - BD^2}{2 \times 10 \times 10}\) or \(\dfrac{192}{200}\) or 0.96 or \(\sin\dfrac{A}{2} = \dfrac{BD/2}{10}\) or \(\dfrac{\sqrt{8}}{20}\) or 0.141 or better (0.14142...) M1
\((A =)\) 16.3 or better (16.2602...) A1 for angle \(A\) correct
16 A1 for 16 or answer rounding to 16.0
Second part method 3
finds angle \(ABD\) (or angle \(ADB\)) using trig or Cosine Rule
\((\cos\angle ABD =)\ \dfrac{BD/2}{10}\) or \((\cos\angle ABD =)\ \dfrac{10^2 + BD^2 - 10^2}{2 \times 10 \times BD}\) or \(\cos\angle ABD = \dfrac{\sqrt{8}}{20}\) or 0.141 or better (0.14142...) M1
\((\angle ABD =)\ 81.9^\circ\) or better (81.8698...) A1
16 A1 for 16 or answer rounding to 16.0