| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Compound shape area |
| Difficulty | Standard +0.3 This is a straightforward C2 question testing area and arc length formulas for circles/sectors. Part (a) requires equating two areas (rectangle + semicircle = sector) and algebraic manipulation, but the steps are routine. Parts (b)-(d) involve direct application of perimeter formulas with substitution. The question is slightly easier than average as it's highly structured with clear guidance through each step, requiring only standard formula recall and basic algebra. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Area of \(X = 2d^2 + \frac{1}{2}\pi d^2\), Area of \(Y = \frac{1}{2}(4d^2)\theta\) | B1, M1 A1 | |
| Equate and divide by \(d^2: 2 + \frac{1}{2}\pi = 2\theta\), \(\theta = 1 + \frac{1}{4}\pi\) | M1 A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(12 + 3\pi\) | B1 B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(4d + r\theta = 12 + 6(1 + \frac{1}{4}\pi) = 18 + \frac{3}{2}\pi\) | M1, A1, A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Difference \(= 13\) mm (or 12.9 mm) or 12.88 mm | M1 A1 | (2 marks) |
## (a)
Area of $X = 2d^2 + \frac{1}{2}\pi d^2$, Area of $Y = \frac{1}{2}(4d^2)\theta$ | B1, M1 A1 |
Equate and divide by $d^2: 2 + \frac{1}{2}\pi = 2\theta$, $\theta = 1 + \frac{1}{4}\pi$ | M1 A1 | (5 marks)
## (b)
$12 + 3\pi$ | B1 B1 | (2 marks)
## (c)
$4d + r\theta = 12 + 6(1 + \frac{1}{4}\pi) = 18 + \frac{3}{2}\pi$ | M1, A1, A1 | (3 marks)
## (d)
$X: 12 + 3\pi = 21.425$ cm, $Y: 18 + \frac{3}{2}\pi = 22.712$ cm
Difference $= 13$ mm (or 12.9 mm) or 12.88 mm | M1 A1 | (2 marks)
**Total: (12 marks)**
---\includegraphics{figure_3}
Fig. 3 shows the cross-sections of two drawer handles. Shape $X$ is a rectangle $ABCD$ joined to a semicircle with $BC$ as diameter. The length $AB = d$ cm and $BC = 2d$ cm. Shape $Y$ is a sector $OPQ$ of a circle with centre $O$ and radius $2d$ cm. Angle $POQ$ is $\theta$ radians. Given that the areas of the shapes $X$ and $Y$ are equal,
\begin{enumerate}[label=(\alph*)]
\item prove that $\theta = 1 + \frac{1}{4}\pi$. [5]
\end{enumerate}
Using this value of $\theta$, and given that $d = 3$, find in terms of $\pi$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item the perimeter of shape $X$, [2]
\item the perimeter of shape $Y$. [3]
\item Hence find the difference, in mm, between the perimeters of shapes $X$ and $Y$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q7 [12]}}