OCR C2 — Question 8 10 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeSegment area calculation
DifficultyStandard +0.3 This is a standard C2 sector/segment question requiring arc length formula, chord length via cosine rule or triangle geometry, and segment area (sector minus triangle). All techniques are routine applications with clear scaffolding across three parts, making it slightly easier than average.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

8. \includegraphics[max width=\textwidth, alt={}, center]{27703044-8bb3-4809-9454-ae6774fec060-3_501_492_242_607} The diagram shows a circle of radius 12 cm which passes through the points \(P\) and \(Q\). The chord \(P Q\) subtends an angle of \(120 ^ { \circ }\) at the centre of the circle.
  1. Find the exact length of the major arc \(P Q\).
  2. Show that the perimeter of the shaded minor segment is given by \(k ( 2 \pi + 3 \sqrt { 3 } ) \mathrm { cm }\), where \(k\) is an integer to be found.
  3. Find, to 1 decimal place, the area of the shaded minor segment as a percentage of the area of the circle.

Question 8:
Part (i):
AnswerMarks
\(= 12 \times (2\pi - \frac{2\pi}{3}) = 16\pi\) cmM1 A1
Part (ii):
AnswerMarks
chord \(= 2 \times 12 \sin\frac{\pi}{3} = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3}\)M1 A1
\(P = (12 \times \frac{2\pi}{3}) + 12\sqrt{3}\)M1
\(= 8\pi + 12\sqrt{3} = 4(2\pi + 3\sqrt{3})\) cm \(\quad [k=4]\)A1
Part (iii):
AnswerMarks Guidance
area of segment \(= (\frac{1}{2} \times 12^2 \times \frac{2\pi}{3}) - (\frac{1}{2} \times 12^2 \times \sin\frac{2\pi}{3})\)M2
\(= 72(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}) = 88.443\)
as % of area of circle \(= \frac{88.443}{\pi \times 12^2} \times 100\% = 19.6\%\) (1dp)M1 A1 (10)
# Question 8:

## Part (i):
| $= 12 \times (2\pi - \frac{2\pi}{3}) = 16\pi$ cm | M1 A1 | |

## Part (ii):
| chord $= 2 \times 12 \sin\frac{\pi}{3} = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3}$ | M1 A1 | |
| $P = (12 \times \frac{2\pi}{3}) + 12\sqrt{3}$ | M1 | |
| $= 8\pi + 12\sqrt{3} = 4(2\pi + 3\sqrt{3})$ cm $\quad [k=4]$ | A1 | |

## Part (iii):
| area of segment $= (\frac{1}{2} \times 12^2 \times \frac{2\pi}{3}) - (\frac{1}{2} \times 12^2 \times \sin\frac{2\pi}{3})$ | M2 | |
| $= 72(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}) = 88.443$ | | |
| as % of area of circle $= \frac{88.443}{\pi \times 12^2} \times 100\% = 19.6\%$ (1dp) | M1 A1 | **(10)** |

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8.\\
\includegraphics[max width=\textwidth, alt={}, center]{27703044-8bb3-4809-9454-ae6774fec060-3_501_492_242_607}

The diagram shows a circle of radius 12 cm which passes through the points $P$ and $Q$. The chord $P Q$ subtends an angle of $120 ^ { \circ }$ at the centre of the circle.\\
(i) Find the exact length of the major arc $P Q$.\\
(ii) Show that the perimeter of the shaded minor segment is given by $k ( 2 \pi + 3 \sqrt { 3 } ) \mathrm { cm }$, where $k$ is an integer to be found.\\
(iii) Find, to 1 decimal place, the area of the shaded minor segment as a percentage of the area of the circle.\\

\hfill \mbox{\textit{OCR C2  Q8 [10]}}