| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector with attached triangle |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard formulas (sector area, cosine rule) with no conceptual challenges. Part (a) is direct formula application, part (b) requires cosine rule in a triangle, and part (c) combines arc length with perimeter calculation. All techniques are routine for P1 level, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts \(\frac{1}{2}r^2\theta\) with \(r = 6\) and any allowable angle \(\theta\) | M1 | Allowable angles: \(0.7\), \(\pi - 0.7\) (awrt 2.4), \(2\pi - 0.7\) (awrt 5.6) |
| \(\frac{1}{2} \times 6^2 \times (2\pi - 0.7)\) or \(\pi \times 6^2 - \frac{1}{2} \times 6^2 \times 0.7\) | M1 | Correct method for area of sector \(ABCOA\) |
| \(= 100.5 \text{ cm}^2\) (awrt) | A1 | Units not required |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{\sin \angle ADO}{6} = \frac{\sin 0.7}{5} \Rightarrow \sin \angle ADO = 0.77\ldots\) | M1 A1 | Sine rule in correct positions; note \(\angle ADO = 0.77\) is A0 |
| \(\angle ADO = 2.258\) (awrt) | A1 | Or \(129.4°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Arc length \(ABC = 6 \times (2\pi - 0.7) = 33.50\) | M1 | May be implied by sight of \(6 \times\) awrt 5.6 or awrt 33.5/33.6 |
| \(\frac{\sin(\pi - 0.7 - ``2.258``)}{OD} = \frac{\sin 0.7}{5} \Rightarrow OD = 1.42\) | M1 | Angle \(OAD\) must use correct method \((\pi - 0.7 - ``2.258``)\); cosine rule alternative valid |
| Perimeter \(= ``33.50`` + 5 + 6 - ``1.42``\) | ddM1 | Both previous M's must be scored |
| \(= 43.1 \text{ cm}\) | A1 | cso and cao; units not required |
## Question 7:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts $\frac{1}{2}r^2\theta$ with $r = 6$ and any allowable angle $\theta$ | M1 | Allowable angles: $0.7$, $\pi - 0.7$ (awrt 2.4), $2\pi - 0.7$ (awrt 5.6) |
| $\frac{1}{2} \times 6^2 \times (2\pi - 0.7)$ or $\pi \times 6^2 - \frac{1}{2} \times 6^2 \times 0.7$ | M1 | Correct method for area of sector $ABCOA$ |
| $= 100.5 \text{ cm}^2$ (awrt) | A1 | Units not required |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{\sin \angle ADO}{6} = \frac{\sin 0.7}{5} \Rightarrow \sin \angle ADO = 0.77\ldots$ | M1 A1 | Sine rule in correct positions; note $\angle ADO = 0.77$ is A0 |
| $\angle ADO = 2.258$ (awrt) | A1 | Or $129.4°$ |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Arc length $ABC = 6 \times (2\pi - 0.7) = 33.50$ | M1 | May be implied by sight of $6 \times$ awrt 5.6 or awrt 33.5/33.6 |
| $\frac{\sin(\pi - 0.7 - ``2.258``)}{OD} = \frac{\sin 0.7}{5} \Rightarrow OD = 1.42$ | M1 | Angle $OAD$ must use correct method $(\pi - 0.7 - ``2.258``)$; cosine rule alternative valid |
| Perimeter $= ``33.50`` + 5 + 6 - ``1.42``$ | ddM1 | Both previous M's must be scored |
| $= 43.1 \text{ cm}$ | A1 | cso and cao; units not required |
**Alternative for $OD$ via cosine rule:**
$$OD^2 = 6^2 + 5^2 - 2\times6\times5\cos(\pi - 0.7 - ``2.258``) \Rightarrow OD$$
---7.
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\caption{Figure 2}
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The shape $A B C D A$ consists of a sector $A B C O A$ of a circle, centre $O$, joined to a triangle $A O D$, as shown in Figure 2.
The point $D$ lies on $O C$.\\
The radius of the circle is 6 cm , length $A D$ is 5 cm and angle $A O D$ is 0.7 radians.
\begin{enumerate}[label=(\alph*)]
\item Find the area of the sector $A B C O A$, giving your answer to one decimal place.
Given angle $A D O$ is obtuse,
\item find the size of angle $A D O$, giving your answer to 3 decimal places.
\item Hence find the perimeter of shape $A B C D A$, giving your answer to one decimal place.\\
\href{http://www.dynamicpapers.com}{www.dynamicpapers.com}
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\hfill \mbox{\textit{Edexcel P1 2019 Q7 [10]}}