| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Simultaneous equations with arc/area |
| Difficulty | Standard +0.3 This is a straightforward application of sector formulas (area = ½r²θ, perimeter = 2r + rθ) leading to a simple quadratic equation. While it requires multiple steps and algebraic manipulation, the techniques are standard and the path is clear once the formulas are recalled. Slightly above average due to the multi-part structure and need to solve a quadratic, but no novel insight required. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = \frac{72.576}{r^2}\) or \(72.576\, r^{-2}\) isw | B1 [1] | eg \(\frac{9072}{125r^2}\) or \(\frac{9072r^{-2}}{125}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r\theta + 2r\) or \(r(\theta+2) = 24.48\) seen and \(\theta = \frac{24.48-2r}{r}\) or equivalent constructive step to give \(\frac{24.48}{r}-2\) AG | B1 [1] | Or \(2\pi r\frac{\theta}{2\pi}+2r=24.48\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(their\ \frac{2\times36.288}{r^2} = \frac{24.48}{r}-2\) oe | M1 | NB 72.576 |
| \(r^2 - 12.24r + 36.288\ [=0]\) | M1 | Quadratic obtained in form \(f(r)\ [=0]\) |
| \([r=]\ 5.04\) or \(7.2\) oe | A1 [3] | Allow B3 for 5.04 and 7.2 unsupported; or allow SC3 for obtaining and solving an equation for \(\theta\) and then finding both values of \(r\) |
## Question 7:
### Part (a):
$\theta = \frac{72.576}{r^2}$ or $72.576\, r^{-2}$ isw | **B1 [1]** | eg $\frac{9072}{125r^2}$ or $\frac{9072r^{-2}}{125}$
### Part (b):
$r\theta + 2r$ or $r(\theta+2) = 24.48$ seen **and** $\theta = \frac{24.48-2r}{r}$ or equivalent constructive step to give $\frac{24.48}{r}-2$ **AG** | **B1 [1]** | Or $2\pi r\frac{\theta}{2\pi}+2r=24.48$
### Part (c):
$their\ \frac{2\times36.288}{r^2} = \frac{24.48}{r}-2$ oe | **M1** | NB 72.576
$r^2 - 12.24r + 36.288\ [=0]$ | **M1** | Quadratic obtained in form $f(r)\ [=0]$
$[r=]\ 5.04$ or $7.2$ oe | **A1 [3]** | Allow **B3** for 5.04 and 7.2 unsupported; or allow **SC3** for obtaining and solving an equation for $\theta$ and then finding both values of $r$
---7 The area of a sector of a circle is $36.288 \mathrm {~cm} ^ { 2 }$. The angle of the sector is $\theta$ radians and the radius of the circle is $r \mathrm {~cm}$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\theta$ in terms of $r$.
The perimeter of the sector is 24.48 cm .
\item Show that $\theta = \frac { 24.48 } { r } - 2$.
\item Find the possible values of $r$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2019 Q7 [5]}}