| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector area calculation |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring direct application of standard sector formulas (perimeter = 2r + rθ, area = ½r²θ). Given the radius and perimeter, students solve a simple linear equation for θ, then substitute into the area formula. No problem-solving insight needed, just routine formula recall and arithmetic. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\text{perimeter} = 2r + r\theta\) with \(16 + 8\theta = 23.2\), \(8\theta = 7.2\), \(\theta = 0.9\) rads | B1* | State or imply that arc length \(8\theta\), or equiv in degrees is \(\frac{5}{360} \times 16\pi\). Allow B1 by implication for \(^{25}/_{8}r\) or equivalent in degrees. |
| Equate attempt at perimeter to 23.2 and attempt to solve for \(\theta\) | M1d* | Need to get as far as attempting \(\theta\). Must include 2 radii and correct expression for arc length, either in radians or degrees. M0 if using chord length. |
| Obtain \(\theta = 0.9\) rads | A1 | 3 |
| (ii) \(\frac{1}{2} \times 8^2 \times 0.9 = 28.8\) | M1 | Attempt area of sector using \((\frac{1}{2}) r^2\theta\). Condone omission of \(\frac{1}{2}\), but no other error. Allow if incorrect angle from part (i), as long as clearly intended to be in radians. Allow equivalent method using fractions of the area. Allow working in degrees as long as it is a valid method. Allow M1 if using 0.9π (even if 0.9 was answer to (i)), as long as clearly attempting \((\frac{1}{2}) r^2\theta\) with error on angle rather then \((\frac{1}{2}) \pi r^2\theta\). |
| Obtain 28.8 | A1 | 2 |
**(i)** $\text{perimeter} = 2r + r\theta$ with $16 + 8\theta = 23.2$, $8\theta = 7.2$, $\theta = 0.9$ rads | B1* | State or imply that arc length $8\theta$, or equiv in degrees is $\frac{5}{360} \times 16\pi$. Allow B1 by implication for $^{25}/_{8}r$ or equivalent in degrees.
**Equate attempt at perimeter to 23.2 and attempt to solve for $\theta$** | M1d* | Need to get as far as attempting $\theta$. Must include 2 radii and correct expression for arc length, either in radians or degrees. M0 if using chord length.
**Obtain $\theta = 0.9$ rads** | A1 | 3 | Obtaining 0.9 and then giving final answer as 0.9π is A0 – do not isw as this shows lack of understanding. Finding $\theta$ in degrees (51.6°) and then converting to radians can get A1 as long as final answer is 0.9 (and not eg 0.9006 from premature approximation).
**(ii)** $\frac{1}{2} \times 8^2 \times 0.9 = 28.8$ | M1 | Attempt area of sector using $(\frac{1}{2}) r^2\theta$. Condone omission of $\frac{1}{2}$, but no other error. Allow if incorrect angle from part (i), as long as clearly intended to be in radians. Allow equivalent method using fractions of the area. Allow working in degrees as long as it is a valid method. Allow M1 if using 0.9π (even if 0.9 was answer to (i)), as long as clearly attempting $(\frac{1}{2}) r^2\theta$ with error on angle rather then $(\frac{1}{2}) \pi r^2\theta$.
**Obtain 28.8** | A1 | 2 | Or any exact equiv. If 0.9 obtained incorrectly in part (i), full credit can still be gained in part (ii). Condone minor inaccuracies from working in degrees, as long as final answer is given as 28.8 exactly.
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{4d03f4e3-ae6c-4a0d-ae4d-d89258d2919a-2_515_501_1439_822}
The diagram shows a sector $A O B$ of a circle, centre $O$ and radius 8 cm . The perimeter of the sector is 23.2 cm .\\
(i) Find angle $A O B$ in radians.\\
(ii) Find the area of the sector.
\hfill \mbox{\textit{OCR C2 2011 Q3 [5]}}