| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | January |
| Marks | 11 |
| 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 application of cosine rule to find an angle, then using standard arc length and sector area formulas. The compound shape area requires subtracting a triangle from a sector, which is a routine technique. All steps are standard C2 content with no novel insight required, making it slightly easier than average. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05b Sine and cosine rules: including ambiguous case1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(15^2 = 10^2 + 10^2 - 2 \times 10 \times 10\cos\angle BOC\) | M1 | Uses cosine rule; must be correct or other correct trig e.g. \(2\times\theta\) where \(\sin\theta = \frac{7.5}{10}\) |
| \(\cos\angle BOC = \frac{10^2+10^2-15^2}{2\times10\times10}\) or \(\frac{-25}{200}\) or \(-0.125\) | A1 | Makes cos subject correctly, or uses \(2\times\sin^{-1}\left(\frac{7.5}{10}\right)\) |
| \(\angle BOC = 1.696\) (N.B. 97.2 degrees is A0) | A1 | Accept awrt 1.696; answer in degrees is A0; if 1.70 (3sf) then A0 but remaining marks available |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(s = 22\theta\) with \(\theta\) from part (a), not \(-(2\pi-\theta)\) | M1 | Uses \(s=r\theta\) with \(\theta\) in radians |
| \(r\theta = 22\times1.696 = 37.3(15)\) | A1 | Accept awrt 37.3 |
| Perimeter \(= r\theta + 15 + x + x = 39 + \text{their arc length}\) [76.3 (m)] | M1 A1ft | Adds arc length to 15 and two further equal lengths; accept awrt 76.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of sector \(= \frac{1}{2}(22)^2\theta\), not \(-(2\pi-\theta)\) | B1 | Formula used with \(\theta\) in radians; allow miscopy of angle |
| Area of triangle \(= \frac{1}{2}(10)^2\sin\theta\) | B1 | Correct formula; may use half base times height |
| Area of paved area \(= \frac{1}{2}(22)^2\theta - \frac{1}{2}(10)^2\sin\theta = 410.432 - 49.6\) or \(410.432 - \frac{75\sqrt{7}}{4} = 360.8\), awrt \(361\ \text{m}^2\) | M1 A1 | Subtracts correct triangle from sector; awrt 361 |
## Question 12:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $15^2 = 10^2 + 10^2 - 2 \times 10 \times 10\cos\angle BOC$ | M1 | Uses cosine rule; must be correct or other correct trig e.g. $2\times\theta$ where $\sin\theta = \frac{7.5}{10}$ |
| $\cos\angle BOC = \frac{10^2+10^2-15^2}{2\times10\times10}$ or $\frac{-25}{200}$ or $-0.125$ | A1 | Makes cos subject correctly, or uses $2\times\sin^{-1}\left(\frac{7.5}{10}\right)$ |
| $\angle BOC = 1.696$ (N.B. 97.2 degrees is A0) | A1 | Accept awrt 1.696; answer in degrees is A0; if 1.70 (3sf) then A0 but remaining marks available |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $s = 22\theta$ with $\theta$ from part (a), not $-(2\pi-\theta)$ | M1 | Uses $s=r\theta$ with $\theta$ in radians |
| $r\theta = 22\times1.696 = 37.3(15)$ | A1 | Accept awrt 37.3 |
| Perimeter $= r\theta + 15 + x + x = 39 + \text{their arc length}$ [76.3 (m)] | M1 A1ft | Adds arc length to 15 and two further equal lengths; accept awrt 76.3 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of sector $= \frac{1}{2}(22)^2\theta$, not $-(2\pi-\theta)$ | B1 | Formula used with $\theta$ in radians; allow miscopy of angle |
| Area of triangle $= \frac{1}{2}(10)^2\sin\theta$ | B1 | Correct formula; may use half base times height |
| Area of paved area $= \frac{1}{2}(22)^2\theta - \frac{1}{2}(10)^2\sin\theta = 410.432 - 49.6$ or $410.432 - \frac{75\sqrt{7}}{4} = 360.8$, awrt $361\ \text{m}^2$ | M1 A1 | Subtracts correct triangle from sector; awrt 361 |
---12.
\begin{tikzpicture}[>=stealth, scale=0.85]
\def\R{5.5}
\coordinate (O) at (0, 0);
\coordinate (Top) at (0, \R);
\coordinate (B) at ({\R*1/2.2*cos(240)},{\R*1/2.2*sin(240)});
\coordinate (C) at ({\R*1/2.2*cos(300)},{\R*1/2.2*sin(300)});
\coordinate (A) at ({\R*cos(240)},{\R*sin(240)});
\coordinate (D) at ({\R*cos(300)},{\R*sin(300)});
% --- Shaded region: B -- C -- D -- arc(D to A through bottom) -- A -- cycle ---
\fill[gray!60]
(B) -- (C) -- (D) arc(300:240:\R) -- cycle;
% --- Circle ---
\draw (O) circle (\R);
% --- Lines from O to B and C ---
\draw[dashed] (O) -- (B);
\draw[dashed] (O) -- (C);
% --- Horizontal line B--C ---
\draw (B) -- (C);
% --- Vertical sides A--B and D--C ---
\draw (A) -- (B);
\draw (D) -- (C);
% --- Dashed vertical line from Top to O ---
\draw[dashed] (Top) -- (O);
% --- Labels ---
\node[right, xshift=2pt] at ($(O)!0.5!(Top)$) {$22\,\mathrm{m}$};
\node[above, font=\small] at ($(O)!0.5!(B)$) {$10\,\mathrm{m}$};
\node[above, font=\small] at ($(O)!0.5!(C)$) {$10\,\mathrm{m}$};
\node[below, font=\small] at ($(B)!0.5!(C)$) {$15\,\mathrm{m}$};
\node[above right] at (O) {$O$};
\node[above left] at (B) {$B$};
\node[above right] at (C) {$C$};
\node[below left] at (A) {$A$};
\node[below right] at (D) {$D$};
\end{tikzpicture}
Diagram NOT drawn to scale
Figure 1 shows the plan for a pond and platform. The platform is shown shaded in the figure and is labelled $A B C D$.
The pond and platform together form a circle of radius 22 m with centre $O$.\\
$O A$ and $O D$ are radii of the circle. Point $B$ lies on $O A$ such that the length of $O B$ is 10 m and point $C$ lies on $O D$ such that the length of $O C$ is 10 m . The length of $B C$ is 15 m .
The platform is bounded by the arc $A D$ of the circle, and the straight lines $A B , B C$ and $C D$.
Find
\begin{enumerate}[label=(\alph*)]
\item the size of the angle $B O C$, giving your answer in radians to 3 decimal places,
\item the perimeter of the platform to 3 significant figures,
\item the area of the platform to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q12 [11]}}