| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Quadratic applied to similar/geometric figures with surds |
| Difficulty | Moderate -0.5 This is a straightforward two-part question requiring the triangle area formula (1/2 ab sin C) and then the cosine rule. Both are standard C2 techniques with no problem-solving insight needed. The surd manipulation is routine for this level, making it slightly easier than average. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(24\sqrt{3} = \frac{1}{2} \times 3x \times x \sin 60°\) | M1 | Attempt Area \(= \frac{1}{2}ab\sin C\) |
| Uses \(\sin 60° = \frac{\sqrt{3}}{2}\) to reach \(x^2 = k\) | dM1 | Sight of \(x^2 = \frac{16\sqrt{3}}{\sin 60°}\) oe. Or correct simplified intermediate line then correct answer |
| \(x^2 = 32 \Rightarrow x = 4\sqrt{2}\) | A1* | Must see \(x = 4\sqrt{2}\) following \(x^2=32\) or \(x^2=16\times2\) or \(x=\sqrt{32}\). Show that question |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(BC^2 = (12\sqrt{2})^2 + (4\sqrt{2})^2 - 2(12\sqrt{2})(4\sqrt{2})\cos 60°\) | M1 | Uses cosine rule. Condone missing brackets. Can score for \(BC^2=(3x)^2+(x)^2-2(3x)(x)\cos60°\) |
| \(BC^2 = 224\) | A1 | May be implied by \(BC = \sqrt{224}\) or \(4\sqrt{14}\) |
| \(BC = 4\sqrt{14}\) | A1 | If candidate uses decimals: \(BC^2=(5.66)^2+(16.97)^2-2(5.66)(16.97)\cos60° \Rightarrow BC=4\sqrt{14}\) award M1,A1,A0 |
## Question 4:
**(a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $24\sqrt{3} = \frac{1}{2} \times 3x \times x \sin 60°$ | M1 | Attempt Area $= \frac{1}{2}ab\sin C$ |
| Uses $\sin 60° = \frac{\sqrt{3}}{2}$ to reach $x^2 = k$ | dM1 | Sight of $x^2 = \frac{16\sqrt{3}}{\sin 60°}$ oe. Or correct simplified intermediate line then correct answer |
| $x^2 = 32 \Rightarrow x = 4\sqrt{2}$ | A1* | Must see $x = 4\sqrt{2}$ following $x^2=32$ or $x^2=16\times2$ or $x=\sqrt{32}$. Show that question |
**(b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $BC^2 = (12\sqrt{2})^2 + (4\sqrt{2})^2 - 2(12\sqrt{2})(4\sqrt{2})\cos 60°$ | M1 | Uses cosine rule. Condone missing brackets. Can score for $BC^2=(3x)^2+(x)^2-2(3x)(x)\cos60°$ |
| $BC^2 = 224$ | A1 | May be implied by $BC = \sqrt{224}$ or $4\sqrt{14}$ |
| $BC = 4\sqrt{14}$ | A1 | If candidate uses decimals: $BC^2=(5.66)^2+(16.97)^2-2(5.66)(16.97)\cos60° \Rightarrow BC=4\sqrt{14}$ award M1,A1,A0 |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb1becd5-96c1-426d-9b85-4bbc4a61af27-08_287_689_255_625}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a triangle $A B C$ with $A B = 3 x \mathrm {~cm} , A C = x \mathrm {~cm}$ and angle $C A B = 60 ^ { \circ }$
Given that the area of triangle $A B C = 24 \sqrt { 3 }$
\begin{enumerate}[label=(\alph*)]
\item show that $x = 4 \sqrt { 2 }$
\item Hence find the exact length of $B C$, giving your answer as a simplified surd.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q4 [6]}}