| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector perimeter calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of standard arc length and sector area formulas with simple algebraic manipulation. Part (a) requires setting up perimeter = 2r + rθ = 56 and solving for r, while part (b) is direct substitution into ½r²θ. Both parts are routine textbook exercises requiring only recall and basic algebra, making this easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| 2(a) | Arc = \(r\theta\); \(1.5r + r + r\) = (56) ⟹ 3.5r = 56 ⟹ r = 16 | M1, M1, A1 |
| 2(b) | Area of sector = \(\frac{1}{2}r^2\theta\) = \(\frac{1}{2} \times 16^2(1.5) = 192\) cm² | M1, A1 |
2(a) | Arc = $r\theta$; $1.5r + r + r$ = (56) ⟹ 3.5r = 56 ⟹ r = 16 | M1, M1, A1 | For $r\theta$ or 16θ or 16×1.5 OE multiplicatio; For realising that perimeter is sum of two radii and arc; AG Completion (condone verification) |
2(b) | Area of sector = $\frac{1}{2}r^2\theta$ = $\frac{1}{2} \times 16^2(1.5) = 192$ cm² | M1, A1 | $\frac{1}{2}r^2\theta$ OE seen; Condone absent cm² |2 The diagram shows a sector $O A B$ of a circle with centre $O$ and radius $r \mathrm {~cm}$.\\
\includegraphics[max width=\textwidth, alt={}, center]{37627fc4-a90b-4f3b-9b10-0a9e200f8485-2_486_381_1686_739}
The angle $A O B$ is 1.5 radians. The perimeter of the sector is 56 cm .
\begin{enumerate}[label=(\alph*)]
\item Show that $r = 16$.
\item Find the area of the sector.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q2 [5]}}