CAIE P1 2013 June — Question 4 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeSector and arc length
DifficultyModerate -0.3 This is a straightforward application of circle geometry with arc length and sector area formulas. Part (i) uses inverse tangent to find an angle (routine trigonometry), while parts (ii) and (iii) apply standard formulas for arc length and sector area. The setup is clear, all necessary information is given, and the methods are direct applications of standard techniques with no novel problem-solving required. Slightly easier than average due to its procedural nature.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

4 \includegraphics[max width=\textwidth, alt={}, center]{fe4c3555-5736-48c4-b61a-9f6b9a1ee46e-2_645_652_1023_744} The diagram shows a square \(A B C D\) of side 10 cm . The mid-point of \(A D\) is \(O\) and \(B X C\) is an arc of a circle with centre \(O\).
  1. Show that angle \(B O C\) is 0.9273 radians, correct to 4 decimal places.
  2. Find the perimeter of the shaded region.
  3. Find the area of the shaded region.

AnswerMarks Guidance
(i) \(BOC = 2\tan^{-1}\frac{1}{2} = 0.9273\)M1 A1 Correct trigonometry. (ans given)
(ii) \(OB = \sqrt{(10^2 + 5^2)} \text{ or } 11.2 = r\)
AnswerMarks Guidance
Arc \(BXC = \sqrt{125} \times 0.9273\) → Perimeter = 20.4 cmB1 M1 A1 Use of trig (or Pyth) for the \(OB = \sqrt{125}\). Use of \(s = r\theta\) with \(\theta\) in rads, \(r \neq 10\)
(iii) Area \(= \frac{1}{2}r^2\theta\) → \(\frac{1}{2}.10.10 \rightarrow 7.96 \text{ cm}^2\)M1 A1 Correct formula used with rads, \(r \neq 10\). Allow 7.95 or 7.96
[3][2]
(i) $BOC = 2\tan^{-1}\frac{1}{2} = 0.9273$ | M1 A1 | Correct trigonometry. (ans given)

(ii) $OB = \sqrt{(10^2 + 5^2)} \text{ or } 11.2 = r$

Arc $BXC = \sqrt{125} \times 0.9273$ → Perimeter = 20.4 cm | B1 M1 A1 | Use of trig (or Pyth) for the $OB = \sqrt{125}$. Use of $s = r\theta$ with $\theta$ in rads, $r \neq 10$

(iii) Area $= \frac{1}{2}r^2\theta$ → $\frac{1}{2}.10.10 \rightarrow 7.96 \text{ cm}^2$ | M1 A1 | Correct formula used with rads, $r \neq 10$. Allow 7.95 or 7.96

[3] | [2] |

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4\\
\includegraphics[max width=\textwidth, alt={}, center]{fe4c3555-5736-48c4-b61a-9f6b9a1ee46e-2_645_652_1023_744}

The diagram shows a square $A B C D$ of side 10 cm . The mid-point of $A D$ is $O$ and $B X C$ is an arc of a circle with centre $O$.\\
(i) Show that angle $B O C$ is 0.9273 radians, correct to 4 decimal places.\\
(ii) Find the perimeter of the shaded region.\\
(iii) Find the area of the shaded region.

\hfill \mbox{\textit{CAIE P1 2013 Q4 [7]}}