| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector area calculation |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard formulas for arc length and sector area with radians. Part (a) requires simple arithmetic (30 - 18 = 12), parts (b) and (c) are direct formula applications (s = rθ, A = ½r²θ), and part (d) uses the standard triangle area formula ½r²sin(θ). All steps are routine with no problem-solving or insight required, making it easier than average but not trivial since it requires correct formula recall and multi-step execution. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Length of arc \(= 12\) (cm) | B1 | Mark parts (a) and (b) together |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses arc length \(12 = 9 \times \theta\) | M1 | Allow degree calculations for M mark |
| \(\theta = \dfrac{4}{3}\) or awrt \(1.33\) | A1 | Accept \(\theta = \frac{12}{9}, 1\frac{1}{3}\); allow decimals awrt \(1.33\); accept \(0.424\pi\) or \(0.425\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of sector \(= \frac{1}{2}r^2\theta = \frac{1}{2} \times 9^2 \times \theta\) | M1 | Uses/states \(A = \frac{1}{2} \times 9^2 \times \theta\) with their value of \(\theta\) in radians from (b); allow correct formula in degrees \(A = \frac{\theta}{360} \times \pi \times 9^2\) |
| Area \(= \text{awrt } 54 \text{ (cm}^2\text{)}\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of triangle \(OAB = \frac{1}{2} \times 9 \times 9 \times \sin\theta = \text{awrt } 39.4 \text{ cm}^2\) | M1, A1 | M1: correct method using \(\frac{1}{2} \times 9 \times 9 \times \sin\theta\) with their \(\theta\); alternative isosceles methods acceptable if full method shown; segment formula \(\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\) scores M0 unless subtracted from sector area; A1: awrt \(39.4 \text{ cm}^2\) |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Length of arc $= 12$ (cm) | B1 | Mark parts (a) and (b) together |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses arc length $12 = 9 \times \theta$ | M1 | Allow degree calculations for M mark |
| $\theta = \dfrac{4}{3}$ or awrt $1.33$ | A1 | Accept $\theta = \frac{12}{9}, 1\frac{1}{3}$; allow decimals awrt $1.33$; accept $0.424\pi$ or $0.425\pi$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of sector $= \frac{1}{2}r^2\theta = \frac{1}{2} \times 9^2 \times \theta$ | M1 | Uses/states $A = \frac{1}{2} \times 9^2 \times \theta$ with their value of $\theta$ in radians from (b); allow correct formula in degrees $A = \frac{\theta}{360} \times \pi \times 9^2$ |
| Area $= \text{awrt } 54 \text{ (cm}^2\text{)}$ | A1 | — |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of triangle $OAB = \frac{1}{2} \times 9 \times 9 \times \sin\theta = \text{awrt } 39.4 \text{ cm}^2$ | M1, A1 | M1: correct method using $\frac{1}{2} \times 9 \times 9 \times \sin\theta$ with their $\theta$; alternative isosceles methods acceptable if full method shown; segment formula $\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta$ scores M0 unless subtracted from sector area; A1: awrt $39.4 \text{ cm}^2$ |7.
\begin{tikzpicture}
% Define coordinates and parameters
\coordinate (O) at (0,0);
\def\r{2.5} % Drawing radius
% The perimeter is 30 cm, radius is 9 cm.
% Perimeter = 2r + r*theta => 30 = 18 + 9*theta => 9*theta = 12 => theta = 4/3 radians.
% 4/3 radians is approximately 76.39 degrees.
\def\angle{76.39}
% Draw the circle
\draw (O) circle (\r);
% Draw the radii OA and OB to form the minor sector
\draw[thick] (O) -- (\r,0) node[right] {$A$};
\draw[thick] (O) -- (\angle:\r) node[above right] {$B$};
% Mark and label the center O
\fill (O) circle (1.5pt);
\node[below left] at (O) {$O$};
% Label the radius as 9 cm
\path (O) -- (\r,0) node[midway, below] {9 cm};
% Draw and label the angle theta
\draw (0.7,0) arc (0:\angle:0.7);
\node at (\angle/2:1.0) {$\theta$};
\end{tikzpicture}
Figure 1 shows a circle with centre $O$ and radius 9 cm . The points $A$ and $B$ lie on the circumference of this circle. The minor sector $O A B$ has perimeter 30 cm and the angle between the radii $O A$ and $O B$ of this sector is $\theta$ radians.
Find
\begin{enumerate}[label=(\alph*)]
\item the length of the arc $A B$,
\item the value of $\theta$,
\item the area of the minor sector $O A B$,
\item the area of triangle $O A B$, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2015 Q7 [7]}}