CAIE P1 2006 June — Question 7 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2006
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeTangent and sector - two tangents from external point
DifficultyStandard +0.3 This is a standard tangent-circle geometry problem requiring basic trigonometry (tan θ = 15/8 to find angle OAT, then doubling for AOB), arc length formula (rθ), and sector area formula (½r²θ). All techniques are routine applications with clear structure and no novel insight required, making it slightly easier than average.
Spec1.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05g Exact trigonometric values: for standard angles

7 \includegraphics[max width=\textwidth, alt={}, center]{cbcb15b4-1870-4dfd-b6e9-839aa4601511-3_545_759_269_694} The diagram shows a circle with centre \(O\) and radius 8 cm . Points \(A\) and \(B\) lie on the circle. The tangents at \(A\) and \(B\) meet at the point \(T\), and \(A T = B T = 15 \mathrm {~cm}\).
  1. Show that angle \(A O B\) is 2.16 radians, correct to 3 significant figures.
  2. Find the perimeter of the shaded region.
  3. Find the area of the shaded region.

AnswerMarks Guidance
(i) \(\tan(\frac{x}{5}) = \frac{15}{8} = 1.875\)M1 Uses correct 90° triangle and sine. Realises the need to ÷ 2
\(\frac{1}{2}x = 1.081\)A1
→ \(x = 2.16\)A1 co
(ii) \(P = 15 + 15 + r\theta = 30 + 17.3\)M1 For \(r\theta\) only – \(\theta\) must be in radians.
→ \(47.3\)A1 co.
(iii) Sector area \(= \frac{1}{2}r^2\theta = 69.1\)M1 For use of \(\frac{1}{2}r^2\theta\).
Area of \(AOB T = 2 \times \frac{1}{2} \times 8 \times 15 = 120\)M1 For use of 2 triangles or equivalent.
Shaded area \(= 120 - 69.1\)
→ \(50.8\) or \(50.9\)A1 co.
Total: [3] [2] [3]
**(i)** $\tan(\frac{x}{5}) = \frac{15}{8} = 1.875$ | M1 | Uses correct 90° triangle and sine. Realises the need to ÷ 2
$\frac{1}{2}x = 1.081$ | A1
→ $x = 2.16$ | A1 | co

**(ii)** $P = 15 + 15 + r\theta = 30 + 17.3$ | M1 | For $r\theta$ only – $\theta$ must be in radians.
→ $47.3$ | A1 | co.

**(iii)** Sector area $= \frac{1}{2}r^2\theta = 69.1$ | M1 | For use of $\frac{1}{2}r^2\theta$.
Area of $AOB T = 2 \times \frac{1}{2} \times 8 \times 15 = 120$ | M1 | For use of 2 triangles or equivalent.
Shaded area $= 120 - 69.1$ | 
→ $50.8$ or $50.9$ | A1 | co.

**Total: [3] [2] [3]**

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7\\
\includegraphics[max width=\textwidth, alt={}, center]{cbcb15b4-1870-4dfd-b6e9-839aa4601511-3_545_759_269_694}

The diagram shows a circle with centre $O$ and radius 8 cm . Points $A$ and $B$ lie on the circle. The tangents at $A$ and $B$ meet at the point $T$, and $A T = B T = 15 \mathrm {~cm}$.\\
(i) Show that angle $A O B$ is 2.16 radians, correct to 3 significant figures.\\
(ii) Find the perimeter of the shaded region.\\
(iii) Find the area of the shaded region.

\hfill \mbox{\textit{CAIE P1 2006 Q7 [8]}}