CAIE P1 2008 November — Question 6 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2008
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeTangent meets circle problems
DifficultyStandard +0.3 This is a straightforward multi-part question testing standard circle geometry and arc length formulas. Part (i) uses s=rθ directly, (ii) applies basic trigonometry in a right-angled triangle, and (iii) combines sector and triangle areas. All steps are routine applications of well-known formulas with no novel insight required, making it slightly easier than average.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

6 \includegraphics[max width=\textwidth, alt={}, center]{08729aab-586b-4210-94c9-77b1f6b1d873-3_597_417_274_865} In the diagram, the circle has centre \(O\) and radius 5 cm . The points \(P\) and \(Q\) lie on the circle, and the arc length \(P Q\) is 9 cm . The tangents to the circle at \(P\) and \(Q\) meet at the point \(T\). Calculate
  1. angle \(P O Q\) in radians,
  2. the length of \(P T\),
  3. the area of the shaded region.

(i)
AnswerMarks Guidance
Using \(s = r\theta, 9 = 5\theta \rightarrow \theta = 1.8 \text{ rad.}\)M1, A1 Use of formula. co
(ii)
AnswerMarks Guidance
Uses \(POT\). Halves the angle. Uses tangent in \(POT\).M1, M1, A1 Realises the need to have. Use of tangent – even if angle not halved. co
\(PT = 5\tan 9° = 6.30 \text{ cm (not 6.31)}\)
(iii)
AnswerMarks Guidance
Area of sector = \(\frac{1}{2} \times 5^2 \times 1.8 (22.5)\)M1 Use of \(A = \frac{1}{2}r^2\theta\) with 1.8 or 0.9.
Area of \(POT = \frac{1}{2} \times 5 \times 6.30 (15.75)\)M1 Use of \(\frac{1}{2}bh\) and (2 triangles – sector)
Shaded area = 2 triangles – sector \(\rightarrow 9.00\) (allow 8.95 to 9.05)A1 co
**(i)**

Using $s = r\theta, 9 = 5\theta \rightarrow \theta = 1.8 \text{ rad.}$ | M1, A1 | Use of formula. co

**(ii)**

Uses $POT$. Halves the angle. Uses tangent in $POT$. | M1, M1, A1 | Realises the need to have. Use of tangent – even if angle not halved. co

$PT = 5\tan 9° = 6.30 \text{ cm (not 6.31)}$

**(iii)**

Area of sector = $\frac{1}{2} \times 5^2 \times 1.8 (22.5)$ | M1 | Use of $A = \frac{1}{2}r^2\theta$ with 1.8 or 0.9.

Area of $POT = \frac{1}{2} \times 5 \times 6.30 (15.75)$ | M1 | Use of $\frac{1}{2}bh$ and (2 triangles – sector)

Shaded area = 2 triangles – sector $\rightarrow 9.00$ (allow 8.95 to 9.05) | A1 | co

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6\\
\includegraphics[max width=\textwidth, alt={}, center]{08729aab-586b-4210-94c9-77b1f6b1d873-3_597_417_274_865}

In the diagram, the circle has centre $O$ and radius 5 cm . The points $P$ and $Q$ lie on the circle, and the arc length $P Q$ is 9 cm . The tangents to the circle at $P$ and $Q$ meet at the point $T$. Calculate\\
(i) angle $P O Q$ in radians,\\
(ii) the length of $P T$,\\
(iii) the area of the shaded region.

\hfill \mbox{\textit{CAIE P1 2008 Q6 [8]}}