| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 4 |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector perimeter calculation |
| Difficulty | Moderate -0.8 This is a straightforward sector geometry question requiring only basic recall of perimeter and area formulas. Part (i) involves simple algebraic manipulation of the perimeter formula (2r + rθ = 18), and part (ii) substitutes the result into the standard area formula. Both parts are routine exercises with no problem-solving insight required, 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 |
|---|---|---|
| (i) State \(2r + r\theta = 18\) | B1 | |
| Obtain correctly \(\theta = \frac{18 - 2r}{r}\) | A1 | Accept work in degrees. Formula must be correct |
| (ii) Substitute \(S = \frac{1}{2}r^2 \left(\frac{18-2r}{r}\right)\) | M1 | Accept work in degrees. Award for substituting for \(\theta\) in correct expression for \(S\). |
| Obtain \(9r - r^2\) | A1 | [4] |
(i) State $2r + r\theta = 18$ | B1 |
Obtain correctly $\theta = \frac{18 - 2r}{r}$ | A1 | Accept work in degrees. Formula must be correct
(ii) Substitute $S = \frac{1}{2}r^2 \left(\frac{18-2r}{r}\right)$ | M1 | Accept work in degrees. Award for substituting for $\theta$ in correct expression for $S$.
Obtain $9r - r^2$ | A1 | [4] |\includegraphics{figure_2}
The diagram shows a sector $OAB$ of a circle with centre $O$ and radius $r$ cm in which angle $AOB$ is $\theta$ radians. The sector has a perimeter of 18 cm.
\begin{enumerate}[label=(\roman*)]
\item Show that $\theta = \frac{18 - 2r}{r}$. [2]
\item Find the area of the sector in terms of $r$, simplifying your answer. [2]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2011 Q2 [4]}}