| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Tangent and sector - single tangent line |
| Difficulty | Standard +0.3 This is a straightforward application of basic trigonometry and sector area formulas. Part (i) requires using tan(π/4)=1 to find AD=8cm from the right triangle. Part (ii) involves recognizing that angle DAE=π/4 (since BD is tangent), then calculating shaded area as triangle ABC minus sector ADE using standard formulas. All steps are routine with no novel insight required, 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 |
|---|---|
| 3(i) | 3π |
| Answer | Marks |
|---|---|
| 5 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10 5 | M1 | Angles used must be correct |
| (AD =) 6.47 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 5 | B1 | Angles used must be correct |
| Answer | Marks |
|---|---|
| 5 | M1 |
| (AD =) 6.47 | A1 |
| Answer | Marks |
|---|---|
| 3(ii) | 1( )2×their π π |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 5 | M1 | 19.7(4) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 5 2 10 10 | M1 | Or e.g. ½theirAD× 82 −theirAD2 . |
| Answer | Marks |
|---|---|
| (Shaded area =) 35.0 or 34.9 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(i) ---
3(i) | 3π
Angle EAD = Angle ACD = or 54° or 0.942 soi
10
π
or Angle DAC = or 36° or 0.628 soi
5 | B1
3π π
AD = 8sin( ) or 8cos( )
10 5 | M1 | Angles used must be correct
(AD =) 6.47 | A1
Alternative method for question 3(i)
3π
8sin
AB= 8 or AB= 10 or 1 1. ( 01 )
π π
tan sin
5 5 | B1 | Angles used must be correct
AD=11.0 ( 1 ) sin π oe
5 | M1
(AD =) 6.47 | A1
3
--- 3(ii) ---
3(ii) | 1( )2×their π π
Area sector = theirAD −
2 2 5 | M1 | 19.7(4)
1 π 1 3π 3π
Area ∆ADC= ×8×theirAD×sin or ×8cos ×8sin
2 5 2 10 10 | M1 | Or e.g. ½theirAD× 82 −theirAD2 .
15.2(2)
(Shaded area =) 35.0 or 34.9 | A1
3
Question | Answer | Marks | Guidance\includegraphics{figure_3}
The diagram shows triangle $ABC$ which is right-angled at $A$. Angle $ABC = \frac{1}{4}\pi$ radians and $AC = 8$ cm.
The points $D$ and $E$ lie on $BC$ and $BA$ respectively. The sector $ADE$ is part of a circle with centre $A$ and is such that $BDC$ is the tangent to the arc $DE$ at $D$.
\begin{enumerate}[label=(\roman*)]
\item Find the length of $AD$. [3]
\item Find the area of the shaded region. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2019 Q3 [6]}}