| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Proving angle relationships |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining cosine rule, arc length, and sector area—all standard C2 techniques. Part (i) is given as 'show that', part (ii) requires simple perimeter calculation (two sides + arc), and part (iii) uses triangle area minus sector area. No novel insight required, just methodical application of formulas. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(13^2 = 10^2 + 14^2 - 2 \times 10 \times 14 \times \cos\theta\) | M1 |
| \(\cos\theta = 0.4536\) | ||
| \(\theta = 1.10\) A.G. | A1 | Obtain 1.10 radians (allow 1.1 radians); SR B1 only for verification of 1.10, unless complete method |
| (ii) | arc \(EF = 4 \times 1.10 = 4.4\) | B1 |
| perimeter \(= 4.4 + 10 + 13 + 6\) | M1 | Attempt perimeter of region - sum of arc and three sides with attempt to subtract 4 from at least one relevant side |
| \(= 33.4\) cm | A1 | Obtain 33.4 cm |
| (iii) | area \(AEF = \frac{1}{2} \times 4^2 \times 1.1\) | M1 |
| \(= 8.8\) | A1 | Obtain 8.8 |
| area \(ABC = \frac{1}{2} \times 10 \times 14 \times \sin 1.1\) | M1 | Attempt use of \((\frac{1}{2})ab\sin\theta\), sides consistent with angle used |
| \(= 62.4\) | A1 | Obtain 62.4 or better (allow 62.38 or 62.39) |
| hence total area \(= 53.6\) cm² | A1 | Obtain total area as 53.6 cm² |
(i) | $13^2 = 10^2 + 14^2 - 2 \times 10 \times 14 \times \cos\theta$ | M1 | Attempt to use correct cosine rule in $\triangle ABC$ |
| $\cos\theta = 0.4536$ | | |
| $\theta = 1.10$ A.G. | A1 | Obtain 1.10 radians (allow 1.1 radians); SR B1 only for verification of 1.10, unless complete method |
(ii) | arc $EF = 4 \times 1.10 = 4.4$ | B1 | State or imply $EF = 4.4$cm (allow $4 \times 1.10$) |
| perimeter $= 4.4 + 10 + 13 + 6$ | M1 | Attempt perimeter of region - sum of arc and three sides with attempt to subtract 4 from at least one relevant side |
| $= 33.4$ cm | A1 | Obtain 33.4 cm |
(iii) | area $AEF = \frac{1}{2} \times 4^2 \times 1.1$ | M1 | Attempt use of $(\frac{1}{2})r^2\theta$, with $r = 4$ and $\theta = 1.10$ |
| $= 8.8$ | A1 | Obtain 8.8 |
| area $ABC = \frac{1}{2} \times 10 \times 14 \times \sin 1.1$ | M1 | Attempt use of $(\frac{1}{2})ab\sin\theta$, sides consistent with angle used |
| $= 62.4$ | A1 | Obtain 62.4 or better (allow 62.38 or 62.39) |
| hence total area $= 53.6$ cm² | A1 | Obtain total area as 53.6 cm² |7\\
\includegraphics[max width=\textwidth, alt={}, center]{9362eb16-88c9-4279-97aa-907b4916b965-3_469_673_1720_737}
The diagram shows triangle $A B C$, with $A B = 10 \mathrm {~cm} , B C = 13 \mathrm {~cm}$ and $C A = 14 \mathrm {~cm} . E$ and $F$ are points on $A B$ and $A C$ respectively such that $A E = A F = 4 \mathrm {~cm}$. The sector $A E F$ of a circle with centre $A$ is removed to leave the shaded region $E B C F$.\\
(i) Show that angle $C A B$ is 1.10 radians, correct to 3 significant figures.\\
(ii) Find the perimeter of the shaded region $E B C F$.\\
(iii) Find the area of the shaded region $E B C F$.
\hfill \mbox{\textit{OCR C2 2010 Q7 [10]}}