OCR C2 — Question 6 8 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSine and Cosine Rules
TypeTriangle with circular sector
DifficultyStandard +0.3 This is a straightforward C2 question involving basic trigonometry (isosceles triangle, sine rule/area formula) and sector area calculations. Part (i) requires finding the area using standard formulas, while part (ii) involves subtracting circular sectors from a triangle—both are routine techniques with clear methods. The 'show that' format in part (ii) provides the target answer, making it easier than an open-ended question.
Spec1.05b Sine and cosine rules: including ambiguous case1.05c 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

\includegraphics{figure_6} The diagram shows triangle \(ABC\) in which \(AC = 8\) cm and \(\angle BAC = \angle BCA = 30°\).
  1. Find the area of triangle \(ABC\) in the form \(k\sqrt{3}\). [4]
The point \(M\) is the mid-point of \(AC\) and the points \(N\) and \(O\) lie on \(AB\) and \(BC\) such that \(MN\) and \(MO\) are arcs of circles with centres \(A\) and \(C\) respectively.
  1. Show that the area of the shaded region \(BNMO\) is \(\frac{8}{3}(2\sqrt{3} - \pi)\) cm\(^2\). [4]

\includegraphics{figure_6}

The diagram shows triangle $ABC$ in which $AC = 8$ cm and $\angle BAC = \angle BCA = 30°$.

\begin{enumerate}[label=(\roman*)]
\item Find the area of triangle $ABC$ in the form $k\sqrt{3}$. [4]
\end{enumerate}

The point $M$ is the mid-point of $AC$ and the points $N$ and $O$ lie on $AB$ and $BC$ such that $MN$ and $MO$ are arcs of circles with centres $A$ and $C$ respectively.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that the area of the shaded region $BNMO$ is $\frac{8}{3}(2\sqrt{3} - \pi)$ cm$^2$. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR C2  Q6 [8]}}