CAIE P1 2009 June — Question 5 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2009
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeOptimization with sectors
DifficultyStandard +0.3 This is a straightforward sector problem requiring students to set up an equation relating perimeter to arc length (part i), then use the given area to find the radius and calculate the complementary region (part ii). The algebra is routine and the 'show that' guides students to the answer, making this slightly easier than average.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

5 \includegraphics[max width=\textwidth, alt={}, center]{3b527397-7781-41e9-8218-57277cc977bf-2_385_403_1866_872} The diagram shows a circle with centre \(O\). The circle is divided into two regions, \(R _ { 1 }\) and \(R _ { 2 }\), by the radii \(O A\) and \(O B\), where angle \(A O B = \theta\) radians. The perimeter of the region \(R _ { 1 }\) is equal to the length of the major \(\operatorname { arc } A B\).
  1. Show that \(\theta = \pi - 1\).
  2. Given that the area of region \(R _ { 1 }\) is \(30 \mathrm {~cm} ^ { 2 }\), find the area of region \(R _ { 2 }\), correct to 3 significant figures.

AnswerMarks Guidance
(i) Perimeter of \(R_1 = r + r + r\theta\)B1 co answer was given
Major arc length \(= 2\pi r - r\theta\)B1
Equated and solved \(\rightarrow \theta = \pi - 1\)B1 [3]
(ii) \(\frac{1}{2}r^2\theta\) with \(\theta = \pi - 1\), equated to 30M1 Use of correct formula once
\(\rightarrow r^2 = 60/(\pi - 1)\) (\(r = 5.29\))A1 Any correct form for \(r\) or \(r^2\)
\(R_2 = \frac{1}{2}r^2(\pi + 1) \rightarrow 58.0\)M1 A1 [4] Attempt at \(r\) (or \(r^2\)) and at area of \(R_2\). co (could be full circle – sector)
or [Reflex \(= \pi + 1\).
\(R_2 = 30(\pi + 1) ÷ (\pi - 1)\)M1 A1 M1 A1]
(i) Perimeter of $R_1 = r + r + r\theta$ | B1 | co answer was given
Major arc length $= 2\pi r - r\theta$ | B1 |
Equated and solved $\rightarrow \theta = \pi - 1$ | B1 [3] |

(ii) $\frac{1}{2}r^2\theta$ with $\theta = \pi - 1$, equated to 30 | M1 | Use of correct formula once
$\rightarrow r^2 = 60/(\pi - 1)$ ($r = 5.29$) | A1 | Any correct form for $r$ or $r^2$
$R_2 = \frac{1}{2}r^2(\pi + 1) \rightarrow 58.0$ | M1 A1 [4] | Attempt at $r$ (or $r^2$) and at area of $R_2$. co (could be full circle – sector)
or [Reflex $= \pi + 1$. | |
$R_2 = 30(\pi + 1) ÷ (\pi - 1)$ | M1 A1 | M1 A1]
5\\
\includegraphics[max width=\textwidth, alt={}, center]{3b527397-7781-41e9-8218-57277cc977bf-2_385_403_1866_872}

The diagram shows a circle with centre $O$. The circle is divided into two regions, $R _ { 1 }$ and $R _ { 2 }$, by the radii $O A$ and $O B$, where angle $A O B = \theta$ radians. The perimeter of the region $R _ { 1 }$ is equal to the length of the major $\operatorname { arc } A B$.\\
(i) Show that $\theta = \pi - 1$.\\
(ii) Given that the area of region $R _ { 1 }$ is $30 \mathrm {~cm} ^ { 2 }$, find the area of region $R _ { 2 }$, correct to 3 significant figures.

\hfill \mbox{\textit{CAIE P1 2009 Q5 [7]}}