Higher June 2025 Paper 5 Q22
22 The diagram shows a quadrilateral ABCD.
DB is a diagonal of the quadrilateral.
Not to scale
AD = 5 cm, AB = 7 cm and BC = 14 cm.
Angle DAB = \(x\)° and angle DBC = 45°.
\(\cos x^\circ = 0.8\) and \(\sin x^\circ = 0.6\).
Work out the area of triangle BDC.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 21 with correct working | 6 | Correct working requires evidence of at least M2A1B1 | |
| M2 for \(\sqrt{7^2 + 5^2 - 2 \times 7 \times 5 \times 0.8}\) oe | Allow M2 for square root soi later after correct substitution seen For M2, accept e.g. \(\sqrt{49 + 25 - 70 \times 0.8}\) | ||
| or M1 for \(7^2 + 5^2 - 2 \times 7 \times 5 \times 0.8\) oe A1 for \(\sqrt{18}\) or \(3\sqrt{2}\) | For M1 accept 49 + 25 – 70 × 0.8 After M1 and \(\sqrt{18}\) allow M2A1 | ||
| AND B1 for sin 45 = \(\frac{1}{\sqrt{2}}\) or \(\frac{\sqrt{2}}{2}\) | For B1 sin 45 must be used within the method e.g. not just seen as sin 45 = \(\frac{\sqrt{2}}{2}\) | ||
| M1 for \(\frac{1}{2} \times 14 \times \textit{their}\text{BD}\) × sin45 oe | Their BD must be clearly indicated in working or on diagram . Allow BD as the answer to their work with triangle ABD | ||
| If 0 or 1 scored instead award SC2 for answer 21 with no or insufficient working If 0 scored instead award SC1 for BD =\(\sqrt{18}\) oe | See Appendix 3 | ||
Appendix 3: Question 22
Page 7 point 12 in general guidance of mark scheme - allow other equivalent correct methods e.g.
Using right-angled trig and perpendicular from D to AB (to K) to find DK and BK = 7 – AK.
Then Pythagoras to find BD
M2 for BD = \(\sqrt{(7 - 5 \times 0.8)^2 + (5 \times 0.6)^2}\) oe
or M1 for BD\(^2\) = \((7 - 5 \times 0.8)^2 + (5 \times 0.6)^2\) oe