| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Circle or Circular Arc Area |
| Difficulty | Challenging +1.2 This question requires understanding of equilateral triangle geometry, circle sector areas, and careful geometric decomposition. While it involves multiple steps (finding angles, calculating sector areas, subtracting triangle area), the approach is methodical and uses standard formulas. The symmetry simplifies the problem, but students must visualize the construction and recognize that the shaded region equals triangle area minus three circular sectors. More challenging than routine integration but less demanding than proof-based or multi-technique problems. |
| Spec | 1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of triangle \(= \dfrac{1}{2}\times(2r)^2\sin\left(\dfrac{\pi}{3}\right)\) or \(\dfrac{1}{2}\times r^2 \sin\left(\dfrac{\pi}{3}\right)\) | M1 | Correct method for area of either triangle. Ignore reference to which triangle |
| Area of sector \(= \dfrac{1}{2}\times r^2 \times \dfrac{\pi}{3}\) | M1 | Use of sector formula \(\dfrac{1}{2}r^2\theta\) with \(\theta = \dfrac{\pi}{3}\), which may be embedded within a segment |
| Area \(R\) = Sector + 2 Segments \(= \dfrac{1}{2}r^2\times\dfrac{\pi}{3} + 2\left(\dfrac{1}{2}r^2\times\dfrac{\pi}{3} - \dfrac{1}{2}r^2\times\dfrac{\sqrt{3}}{2}\right)\) or equivalent correct methods (Triangle + 3 Segments, 3 Sectors \(-\) 2 Triangles, Big Triangle \(-\) 3 White bits) | M1A1 | M1: fully correct method (may be implied by final answer awrt \(0.705r^2\)). A1: correct exact expression — \(\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}\) must be seen |
| \(= \dfrac{1}{2}\pi r^2 - \dfrac{\sqrt{3}}{2}r^2 = r^2\left(\dfrac{1}{2}\pi - \dfrac{\sqrt{3}}{2}\right)\) | A1 | Cso. Allow \(\dfrac{r^2}{2}(\pi - \sqrt{3})\) or any exact equivalent with \(r^2\) taken out as common factor |
## Question 15:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of triangle $= \dfrac{1}{2}\times(2r)^2\sin\left(\dfrac{\pi}{3}\right)$ or $\dfrac{1}{2}\times r^2 \sin\left(\dfrac{\pi}{3}\right)$ | M1 | Correct method for area of either triangle. Ignore reference to which triangle |
| Area of sector $= \dfrac{1}{2}\times r^2 \times \dfrac{\pi}{3}$ | M1 | Use of sector formula $\dfrac{1}{2}r^2\theta$ with $\theta = \dfrac{\pi}{3}$, which may be embedded within a segment |
| Area $R$ = Sector + 2 Segments $= \dfrac{1}{2}r^2\times\dfrac{\pi}{3} + 2\left(\dfrac{1}{2}r^2\times\dfrac{\pi}{3} - \dfrac{1}{2}r^2\times\dfrac{\sqrt{3}}{2}\right)$ or equivalent correct methods (Triangle + 3 Segments, 3 Sectors $-$ 2 Triangles, Big Triangle $-$ 3 White bits) | M1A1 | M1: fully correct method (may be implied by final answer awrt $0.705r^2$). A1: correct exact expression — $\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$ must be seen |
| $= \dfrac{1}{2}\pi r^2 - \dfrac{\sqrt{3}}{2}r^2 = r^2\left(\dfrac{1}{2}\pi - \dfrac{\sqrt{3}}{2}\right)$ | A1 | Cso. Allow $\dfrac{r^2}{2}(\pi - \sqrt{3})$ or any exact equivalent with $r^2$ taken out as common factor |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f39ade34-32e2-4b5c-b80a-9663c6a65c87-26_780_871_242_539}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows the design for a logo.\\
The logo is in the shape of an equilateral triangle $A B C$ of side length $2 r \mathrm {~cm}$, where $r$ is a constant.
The points $L , M$ and $N$ are the midpoints of sides $A C , A B$ and $B C$ respectively.\\
The shaded section $R$, of the logo, is bounded by three curves $M N , N L$ and $L M$.
The curve $M N$ is the arc of a circle centre $L$, radius $r \mathrm {~cm}$.\\
The curve $N L$ is the arc of a circle centre $M$, radius $r \mathrm {~cm}$.\\
The curve $L M$ is the arc of a circle centre $N$, radius $r \mathrm {~cm}$.
Find, in $\mathrm { cm } ^ { 2 }$, the area of $R$. Give your answer in the form $k r ^ { 2 }$, where $k$ is an exact constant to be determined.
\hfill \mbox{\textit{Edexcel C12 2017 Q15 [5]}}