| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Circular arc problems |
| Difficulty | Moderate -0.3 This is a straightforward application of standard arc length (s = rθ), sector area, and basic trigonometry formulas with clearly given values (r = 2.1m, θ = 1.8 rad, length = 5.5m). While it requires multiple steps across three parts, each step uses direct formula substitution with minimal problem-solving or geometric insight needed, making it slightly easier than a typical A-level question. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) arc \(AC = 2.1 \times 1.8\) | M1 | \(\frac{103}{360} \times 2\pi \times 2.1\) |
| \(= 3.78\) c.a.o. | A1 | 3.78 must be seen but may be embedded in area formula |
| area \(=\) their \(3.78 \times 5.5\) \(= 20.79\) or \(20.8\) i.s.w. | M1 | dependent on first M1 |
| A1 | ||
| (ii) \(BD = 2.1 \cos(\pi - 1.8)\) or \(2.1\cos 1.3(4159,….)\) or \(2.1\sin 0.2(292…)\) r.o.t to 1 d.p. or more | M2 | M1 for \(\cos(\pi - 1.8) = \frac{BD}{2.1}\) o.e. |
| \(= 0.48\) | A1 | allow any answer which rounds to 0.48 |
| (iii) sector area \(= 3.969\) | M2 | M1 for \(\frac{1}{2} \times 2.1^2 \times 1.8\) |
| triangle area \(= 0.487\) to \(0.491\) | M2 | M1 for \(\frac{1}{2} \times 2.1 \times\) their \(0.48 \times \sin(\pi - 1.8)\) or \(\frac{1}{2} \times\) their \(0.48 \times 2.045..\) r.o.t to 3 s.f. or better |
| \(24.5\) | A1 | allow any answer which rounds to 24.5 |
**(i)** arc $AC = 2.1 \times 1.8$ | M1 | $\frac{103}{360} \times 2\pi \times 2.1$ | 103° or better
$= 3.78$ c.a.o. | A1 | 3.78 must be seen but may be embedded in area formula
area $=$ their $3.78 \times 5.5$ $= 20.79$ or $20.8$ i.s.w. | M1 | dependent on first M1
| A1 |
**(ii)** $BD = 2.1 \cos(\pi - 1.8)$ or $2.1\cos 1.3(4159,….)$ or $2.1\sin 0.2(292…)$ r.o.t to 1 d.p. or more | M2 | M1 for $\cos(\pi - 1.8) = \frac{BD}{2.1}$ o.e. | M2 for $BD = 2.1 \cos 76.8675…°$ or $2.1\sin 13.1324…$rounded to 2 or more sf or M2 for $CD = 2.045…$ r.o.t to 3 s.f. or better and $BD = \sqrt(2.1^2 - 2.045^2)$
$= 0.48$ | A1 | allow any answer which rounds to 0.48
**(iii)** sector area $= 3.969$ | M2 | M1 for $\frac{1}{2} \times 2.1^2 \times 1.8$ | or equivalent with degrees for first two Ms; N.B. $5.5 \times 3.969 = 21.8295$ so allow M2 for 21.8295
triangle area $= 0.487$ to $0.491$ | M2 | M1 for $\frac{1}{2} \times 2.1 \times$ their $0.48 \times \sin(\pi - 1.8)$ or $\frac{1}{2} \times$ their $0.48 \times 2.045..$ r.o.t to 3 s.f. or better | may be sin $1.8$ instead of $\sin(\pi - 1.8)$; N.B. $5.5 \times$ area $= 2.6785$ to $2.7005$ so allow M2 for a value in this range
$24.5$ | A1 | allow any answer which rounds to 24.513 Fig. 13.1 shows a greenhouse which is built against a wall.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{97ed9d1d-b9e5-47d6-a451-b14757c0e19d-4_606_828_347_358}
\captionsetup{labelformat=empty}
\caption{Fig. 13.1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{97ed9d1d-b9e5-47d6-a451-b14757c0e19d-4_401_350_529_1430}
\captionsetup{labelformat=empty}
\caption{Fig. 13.2}
\end{center}
\end{figure}
The greenhouse is a prism of length 5.5 m . The curve AC is an arc of a circle with centre B and radius 2.1 m , as shown in Fig. 13.2. The sector angle ABC is 1.8 radians and ABD is a straight line. The curved surface of the greenhouse is covered in polythene.\\
(i) Find the length of the arc AC and hence find the area of polythene required for the curved surface of the greenhouse.\\
(ii) Calculate the length BD .\\
(iii) Calculate the volume of the greenhouse.
\hfill \mbox{\textit{OCR MEI C2 2011 Q13 [12]}}