OCR MEI Paper 3 Specimen — Question 13 3 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
SessionSpecimen
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeDistance between two points
DifficultyModerate -0.8 This is a straightforward coordinate geometry verification problem requiring students to calculate distances between vertices of a regular hexagon. While it involves multiple distance calculations and the use of Pythagoras' theorem or distance formula, the approach is direct with no conceptual difficulty or novel insight required—students simply need to find side lengths and sum them, making it easier than average but not trivial due to the computational work involved.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta
13 Show that the larger regular hexagon in Fig. C1 has perimeter \(4 \sqrt { 3 }\).
Question 13:
AnswerMarks Guidance
AnswerMarks Guidance
Show diagram; Angle \(A'N = \tan 30°\) OR \(\tan 30° = \frac{1}{A'N}\)M1 soi
\(A'N = \tan 30° = \frac{1}{\sqrt{3}}\)A1
Alternative: using equilateral triangle \(OA'B'\) of side length \(2a\): \((2a)^2 = a^2 + 1 \Rightarrow a^2 = \frac{1}{3}\)M1
\(a = A'N = \frac{1}{\sqrt{3}}\)M1
Evidence of \(6 \times A'B'\) or \(12 \times A'N = 4\sqrt{3}\)E1 AG 
# Question 13:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Show diagram; Angle $A'N = \tan 30°$ OR $\tan 30° = \frac{1}{A'N}$ | M1 | soi |
| $A'N = \tan 30° = \frac{1}{\sqrt{3}}$ | A1 | |
| **Alternative:** using equilateral triangle $OA'B'$ of side length $2a$: $(2a)^2 = a^2 + 1 \Rightarrow a^2 = \frac{1}{3}$ | M1 | |
| $a = A'N = \frac{1}{\sqrt{3}}$ | M1 | |
| Evidence of $6 \times A'B'$ or $12 \times A'N = 4\sqrt{3}$ | E1 | AG |

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13 Show that the larger regular hexagon in Fig. C1 has perimeter $4 \sqrt { 3 }$.

\hfill \mbox{\textit{OCR MEI Paper 3  Q13 [3]}}
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