| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Compound shape area |
| Difficulty | Standard +0.3 This is a straightforward compound area problem requiring basic geometry (equilateral triangle properties, midpoint theorem), arc length formula, and sector area formula. All steps are routine applications of standard C2 techniques with no novel insight required. The 'show that' in part (i) guides students, and part (ii) is simple subtraction. Slightly above average due to multiple steps and coordinate geometry setup, but well within typical C2 scope. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Length of \(OD\) is 6 cm | B1 | For stating or using the correct value of \(r\) |
| Angle \(DOE\) is \(\frac{1}{3}\pi\) / \(1.047°\) / \(60°\) / \(\frac{1}{6}\) of circle | B1 | For stating or using the correct angle |
| Hence arc length \(DE\) is \(2\pi\) cm (allow 6.28 cm) | B1 | For correct use of \(s = r\theta\) or equiv in degrees |
| Area is \(\frac{1}{2}\times6^2\times\frac{1}{3}\pi = 6\pi\) cm\(^2\) \(\left(\text{or }\frac{60}{360}\times\pi\times6^2\right)\) | B1 | 4 For obtaining the given answer \(6\pi\) correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Area of small triangle is \(\frac{1}{2}\times6^2\times\frac{1}{2}\sqrt{3} = 9\sqrt{3}\) | *M1 | For use of \(\Delta = \frac{1}{2}ab\sin C\), or equivalent |
| A1 | For correct value \(9\sqrt{3}\), or equiv | |
| Area of segment is \(6\pi - 9\sqrt{3}\) | M1dep*M | For relevant use of (sector \(-\) triangle) |
| Hence shaded area is \((18\sqrt{3}-6\pi)\) cm\(^2\) | A1 | 4 For correct answer \(18\sqrt{3}-6\pi\), or exact equiv |
| Alternative: Attempt area of big triangle / rhombus / segment using \(\Delta = \frac{1}{2}ab\sin C\) | *M1 | |
| Correct area | A1 | |
| Relevant subtraction | M1dep*M | |
| For correct answer \(18\sqrt{3}-6\pi\) | A1 |
# Question 7:
## Part (i):
| Length of $OD$ is 6 cm | B1 | For stating or using the correct value of $r$ |
|---|---|---|
| Angle $DOE$ is $\frac{1}{3}\pi$ / $1.047°$ / $60°$ / $\frac{1}{6}$ of circle | B1 | For stating or using the correct angle |
| Hence arc length $DE$ is $2\pi$ cm (allow 6.28 cm) | B1 | For correct use of $s = r\theta$ or equiv in degrees |
| Area is $\frac{1}{2}\times6^2\times\frac{1}{3}\pi = 6\pi$ cm$^2$ $\left(\text{or }\frac{60}{360}\times\pi\times6^2\right)$ | B1 | **4** For obtaining the given answer $6\pi$ correctly |
## Part (ii):
| Area of small triangle is $\frac{1}{2}\times6^2\times\frac{1}{2}\sqrt{3} = 9\sqrt{3}$ | *M1 | For use of $\Delta = \frac{1}{2}ab\sin C$, or equivalent |
|---|---|---|
| | A1 | For correct value $9\sqrt{3}$, or equiv |
| Area of segment is $6\pi - 9\sqrt{3}$ | M1dep*M | For relevant use of (sector $-$ triangle) |
| Hence shaded area is $(18\sqrt{3}-6\pi)$ cm$^2$ | A1 | **4** For correct answer $18\sqrt{3}-6\pi$, or exact equiv |
| **Alternative:** Attempt area of big triangle / rhombus / segment using $\Delta = \frac{1}{2}ab\sin C$ | *M1 | |
| Correct area | A1 | |
| Relevant subtraction | M1dep*M | |
| For correct answer $18\sqrt{3}-6\pi$ | A1 | |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{608720b6-5b18-45e9-8838-c94b347ab3b7-3_563_639_1379_753}
The diagram shows an equilateral triangle $A B C$ with sides of length 12 cm . The mid-point of $B C$ is $O$, and a circular arc with centre $O$ joins $D$ and $E$, the mid-points of $A B$ and $A C$.\\
(i) Find the length of the arc $D E$, and show that the area of the sector $O D E$ is $6 \pi \mathrm {~cm} ^ { 2 }$.\\
(ii) Find the exact area of the shaded region.
\hfill \mbox{\textit{OCR C2 2005 Q7 [8]}}