| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| 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 shape problem requiring standard formulas for sector area and arc length, with simple subtraction of a triangle area. The calculations are routine with given values, requiring no geometric insight beyond recognizing the shapes involved. Slightly easier than average due to all necessary information being provided directly. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| sector \(= \frac{1}{2} \times 8^2 \times 1.2\ (= 38.4)\) | M1* | Must be correct formula including \(\frac{1}{2}\); M0 if \(1.2\pi\) used not \(1.2\); M0 if \(\frac{1}{2}r^2\theta\) used with \(\theta\) in degrees |
| \(\frac{1}{2} \times 2.6 \times 5.2 \times \sin 1.2\ (= 6.3)\) | M1* | Must be correct formula including \(\frac{1}{2}\); angle could be in radians (\(1.2\) not \(1.2\pi\)) or degrees (\(68.8°\)); must have sides of \(2.6\) and \(5.2\) |
| \(38.4 - 6.3 = 32.1\) | M1d* | Attempt area of sector \(-\) area of triangle; need area of sector \(>\) area of triangle |
| A1 | Obtain \(32.1\) or better; allow final answers rounding to \(32.10\) if \(> 3\)sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(8 \times 1.2 = 9.6\) | M1* | Attempt use of \(r\theta\); allow if \(8 \times 1.2\) seen even if incorrectly evaluated |
| \(CD^2 = 2.6^2 + 5.2^2 - 2 \times 2.6 \times 5.2 \times \cos 1.2\) | M1* | Must be correct cosine rule; M0 if \(1.2\pi\) used not \(1.2\); allow if incorrectly evaluated |
| \(CD = 4.90\) or \(\sqrt{24}\) | A1 | Allow any answer in range \([4.89, 4.90]\) |
| perimeter \(= 2.8 + 4.9 + 5.4 + 9.6\) | M1d* | \((8-5.2)+(8-2.6)+\) their \(AB +\) their \(CD\) (not their \(CD^2\)) |
| \(= 22.7\) | A1 | Accept any answer in range \([22.69, 22.70]\) if \(>3\)sf |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| sector $= \frac{1}{2} \times 8^2 \times 1.2\ (= 38.4)$ | M1* | Must be correct formula including $\frac{1}{2}$; M0 if $1.2\pi$ used not $1.2$; M0 if $\frac{1}{2}r^2\theta$ used with $\theta$ in degrees |
| $\frac{1}{2} \times 2.6 \times 5.2 \times \sin 1.2\ (= 6.3)$ | M1* | Must be correct formula including $\frac{1}{2}$; angle could be in radians ($1.2$ not $1.2\pi$) or degrees ($68.8°$); must have sides of $2.6$ and $5.2$ |
| $38.4 - 6.3 = 32.1$ | M1d* | Attempt area of sector $-$ area of triangle; need area of sector $>$ area of triangle |
| | A1 | Obtain $32.1$ or better; allow final answers rounding to $32.10$ if $> 3$sf |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $8 \times 1.2 = 9.6$ | M1* | Attempt use of $r\theta$; allow if $8 \times 1.2$ seen even if incorrectly evaluated |
| $CD^2 = 2.6^2 + 5.2^2 - 2 \times 2.6 \times 5.2 \times \cos 1.2$ | M1* | Must be correct cosine rule; M0 if $1.2\pi$ used not $1.2$; allow if incorrectly evaluated |
| $CD = 4.90$ or $\sqrt{24}$ | A1 | Allow any answer in range $[4.89, 4.90]$ |
| perimeter $= 2.8 + 4.9 + 5.4 + 9.6$ | M1d* | $(8-5.2)+(8-2.6)+$ their $AB +$ their $CD$ (not their $CD^2$) |
| $= 22.7$ | A1 | Accept any answer in range $[22.69, 22.70]$ if $>3$sf |
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\includegraphics[max width=\textwidth, alt={}, center]{6dd10d03-5fe2-4a70-b5a2-03347dff0360-2_576_599_1062_733}
The diagram shows a sector $A O B$ of a circle with centre $O$ and radius 8 cm . The angle $A O B$ is 1.2 radians. The points $C$ and $D$ lie on $O A$ and $O B$ respectively such that $O C = 5.2 \mathrm {~cm}$ and $O D = 2.6 \mathrm {~cm} . C D$ is a straight line.\\
(i) Find the area of the shaded region $A C D B$.\\
(ii) Find the perimeter of the shaded region $A C D B$.
\hfill \mbox{\textit{OCR C2 2015 Q3 [9]}}