| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Verify shape type from coordinates |
| Difficulty | Standard +0.8 This is a geometric proof question requiring students to establish angle relationships in a circle theorem context and then prove an algebraic relationship (likely involving similar triangles or Pythagoras). While it requires geometric insight and formal proof rather than routine calculation, the similar triangles approach and geometric mean result are standard A-level techniques. The multi-part nature and proof requirement place it above average difficulty but not at the extreme end. |
| Spec | 1.02r Proportional relationships: and their graphs1.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Angle BDC \(= 90 - \theta\) (angles of triangle) | M1 | Including reason; Reasons can be given in either order |
| Angle CDA \(= \theta\) (Angle ADB \(= 90°\) as it is the angle in a semicircle) | E1 | Answer given so mark is for reason |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Triangle ACD: \(\tan\theta = \frac{h}{b}\); Triangle BCD: \(\tan\theta = \frac{a}{h}\) | M1 | At least one correct expression for \(\tan\theta\); Alternative method: triangle ACD is similar to triangle DBC |
| \(\frac{a}{h} = \frac{h}{b} \Rightarrow h^2 = ab \Rightarrow h = \sqrt{ab}\) | E1 | Setting expressions equal and correct completion to given answer; AG |
## Question 14(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Angle BDC $= 90 - \theta$ (angles of triangle) | M1 | Including reason; Reasons can be given in either order |
| Angle CDA $= \theta$ (Angle ADB $= 90°$ as it is the angle in a semicircle) | E1 | Answer given so mark is for reason |
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## Question 14(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Triangle ACD: $\tan\theta = \frac{h}{b}$; Triangle BCD: $\tan\theta = \frac{a}{h}$ | M1 | At least one correct expression for $\tan\theta$; Alternative method: triangle ACD is similar to triangle DBC |
| $\frac{a}{h} = \frac{h}{b} \Rightarrow h^2 = ab \Rightarrow h = \sqrt{ab}$ | E1 | Setting expressions equal and correct completion to given answer; AG |
---14 (i) In Fig. C1.3, angle CBD $= \theta$. Show that angle CDA is also $\theta$, as given in line 23 .\\
(ii) Prove that $h = \sqrt { a b }$, as given in line 24 .
\hfill \mbox{\textit{OCR MEI Paper 3 2018 Q14 [4]}}