| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Triangle and sector combined - area/perimeter with given values |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard formulas for sector area (A = ½r²θ) and triangle area (A = ½ab sin C). Part (i) is given as 'show that', part (ii) requires simple algebraic manipulation, and part (iii) combines arc length with the cosine rule. All techniques are routine C2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB^2 = 36 \Rightarrow AB = 6\) | M1, A1 2 | Use \((\frac{1}{3})r^2 \theta = 16.2\). Confirm \(AB = 6\) cm (or verify \(\frac{1}{2} \times 6^2 \times 0.9 = 16.2\)). |
| Answer | Marks | Guidance |
|---|---|---|
| \(AC = 13.8\) cm | M1*, M1dep*, A1 3 | Use \(\Delta = \frac{1}{2}bc \sin A\), or equiv. Equate attempt at area to 32.4. Obtain \(AC = 13.8\) cm, or better. |
| Answer | Marks | Guidance |
|---|---|---|
| Hence \(BC = 11.1\) cm | M1, A1∨, A1 | Attempt use of correct cosine formula in \(\triangle ABC\). Correct unsimplified equation, from their \(AC\). Obtain \(BC = 11.1\) cm, or anything that rounds to this. |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 24.3\) cm | B1, M1, A1 6 | State \(BD = 5.4\) cm (seen anywhere in question). Attempt perimeter of region \(BCD\). Obtain 24.3 cm, or anything that rounds to this. |
(i) $\frac{1}{3} \times AB^2 \times 0.9 = 16.2$
$AB^2 = 36 \Rightarrow AB = 6$ | M1, A1 2 | Use $(\frac{1}{3})r^2 \theta = 16.2$. Confirm $AB = 6$ cm (or verify $\frac{1}{2} \times 6^2 \times 0.9 = 16.2$).
(ii) $\frac{1}{3} \times 6 \times AC \times \sin 0.9 = 32.4$
$AC = 13.8$ cm | M1*, M1dep*, A1 3 | Use $\Delta = \frac{1}{2}bc \sin A$, or equiv. Equate attempt at area to 32.4. Obtain $AC = 13.8$ cm, or better.
(iii) $BC^2 = 6^2 + 13.8^2 - 2 \times 6 \times 13.8 \times \cos 0.9$
Hence $BC = 11.1$ cm | M1, A1∨, A1 | Attempt use of correct cosine formula in $\triangle ABC$. Correct unsimplified equation, from their $AC$. Obtain $BC = 11.1$ cm, or anything that rounds to this.
$BD = 6 \times 0.9 = 5.4$ cm
Hence perimeter $= 11.1 + 5.4 + (13.8 - 6)$
$= 24.3$ cm | B1, M1, A1 6 | State $BD = 5.4$ cm (seen anywhere in question). Attempt perimeter of region $BCD$. Obtain 24.3 cm, or anything that rounds to this.8\\
\includegraphics[max width=\textwidth, alt={}, center]{e429080f-8634-46bc-b451-7b13b871e518-3_300_744_1046_703}
The diagram shows a triangle $A B C$, where angle $B A C$ is 0.9 radians. $B A D$ is a sector of the circle with centre A and radius AB .\\
(i) The area of the sector $B A D$ is $16.2 \mathrm {~cm} ^ { 2 }$. Show that the length of $A B$ is 6 cm .\\
(ii) The area of triangle $A B C$ is twice the area of sector $B A D$. Find the length of $A C$.\\
(iii) Find the perimeter of the region $B C D$.
\hfill \mbox{\textit{OCR C2 2007 Q8 [11]}}