Edexcel C2 — Question 3 6 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks6
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Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeProving angle relationships
DifficultyStandard +0.8 This is a multi-step geometric proof requiring identification of equilateral triangles, calculation of sector areas, and algebraic manipulation to reach a specific form. It demands more geometric insight than typical C2 circle questions and involves coordinating multiple area calculations, placing it moderately above average difficulty.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta
\includegraphics{figure_1} Figure 1 shows a circle of radius \(r\) and centre \(O\) in which \(AD\) is a diameter. The points \(B\) and \(C\) lie on the circle such that \(OB\) and \(OC\) are arcs of circles of radius \(r\) with centres \(A\) and \(D\) respectively. Show that the area of the shaded region \(OBC\) is \(\frac{1}{6}r^2(3\sqrt{3} - \pi)\). [6]
AnswerMarks Guidance
area of segment \(= (\frac{1}{3} \times r^2 \times \frac{\pi}{3}) - (\frac{1}{2} \times r^2 \times \sin \frac{\pi}{3})\)B1 M2
\(= \frac{1}{6}r^2\pi - \frac{1}{4}r^2\sqrt{3}\)A1
shaded area \(= \frac{1}{6}r^2\pi - 2(\frac{1}{6}r^2\pi - \frac{1}{4}r^2\sqrt{3})\)M1
\(= \frac{1}{6}r^2\pi - \frac{1}{3}r^2\pi + \frac{1}{2}r^2\sqrt{3}\)
\(= \frac{1}{2}r^2\sqrt{3} - \frac{1}{6}r^2\pi = \frac{1}{6}r^2(3\sqrt{3} - \pi)\)A1 (6) 
area of segment $= (\frac{1}{3} \times r^2 \times \frac{\pi}{3}) - (\frac{1}{2} \times r^2 \times \sin \frac{\pi}{3})$ | B1 M2 |
$= \frac{1}{6}r^2\pi - \frac{1}{4}r^2\sqrt{3}$ | A1 |

shaded area $= \frac{1}{6}r^2\pi - 2(\frac{1}{6}r^2\pi - \frac{1}{4}r^2\sqrt{3})$ | M1 |

$= \frac{1}{6}r^2\pi - \frac{1}{3}r^2\pi + \frac{1}{2}r^2\sqrt{3}$ | |

$= \frac{1}{2}r^2\sqrt{3} - \frac{1}{6}r^2\pi = \frac{1}{6}r^2(3\sqrt{3} - \pi)$ | A1 | (6)
\includegraphics{figure_1}

Figure 1 shows a circle of radius $r$ and centre $O$ in which $AD$ is a diameter.

The points $B$ and $C$ lie on the circle such that $OB$ and $OC$ are arcs of circles of radius $r$ with centres $A$ and $D$ respectively.

Show that the area of the shaded region $OBC$ is $\frac{1}{6}r^2(3\sqrt{3} - \pi)$. [6]

\hfill \mbox{\textit{Edexcel C2  Q3 [6]}}
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