| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Tangent and sector - two tangents from external point |
| Difficulty | Standard +0.3 This is a standard tangent-sector problem requiring routine application of arc length, tangent properties, and area formulas. Part (i) involves expressing AT in terms of r and θ using tan θ, then adding arc length. Part (ii) requires calculating area by subtracting triangle from sector. While multi-step, it follows a well-established template with no novel insight required, making it slightly easier than average. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Arc length \(AB = 2r\theta\) | B1 | |
| \(\tan\theta = \dfrac{AT}{r}\) or \(\dfrac{BT}{r} \rightarrow AT\) or \(BT = r\tan\theta\) | B1 | Accept or \(\sqrt{\left(\dfrac{r}{\cos\theta}\right)^2 - r^2}\) or \(\dfrac{r\sin\theta}{\sin\left(\frac{\pi}{2}-\theta\right)}\). NOT \((90-\theta)\) |
| \(P = 2r\theta + 2r\tan\theta\) | B1FT | OE, FT for *their* arc length \(+ 2 \times\) *their* \(AT\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area \(\Delta AOT = \frac{1}{2} \times 5 \times 5 \tan 1.2\) or Area \(AOBT = 2 \times \frac{1}{2} \times 5 \times 5 \tan 1.2\) | B1 | |
| Sector area \(= \frac{1}{2} \times 25 \times 2.4\) (or 1.2) | \*M1 | Use of \(\frac{1}{2}r^2\theta\) with \(\theta = 1.2\) or \(2.4\) |
| Shaded area \(= 2\) triangles \(-\) sector | DM1 | Subtraction of sector, using 2.4 where appropriate, from 2 triangles |
| Area \(= 34.3\) (cm\(^2\)) | A1 | AWRT |
| Alternative: Area of \(\Delta ABT = \frac{1}{2} \times (5 \times \tan 1.2)^2 \times \sin(\pi - 2.4)\) (\(= 55.86\)) | B1 | |
| Segment area \(= \frac{1}{2} \times 25 \times (2.4 - \sin 2.4)\) (\(= 21.56\)) | \*M1 | Use of \(\frac{1}{2}r^2(\theta - \sin\theta)\) with \(\theta = 1.2\) or \(2.4\) |
| Shaded area \(=\) triangle \(-\) segment | DM1 | Subtraction of segment from \(\Delta ABT\), using 2.4 where appropriate |
| Area \(= 34.3\) (cm\(^2\)) | A1 | AWRT |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Arc length $AB = 2r\theta$ | B1 | |
| $\tan\theta = \dfrac{AT}{r}$ or $\dfrac{BT}{r} \rightarrow AT$ or $BT = r\tan\theta$ | B1 | Accept or $\sqrt{\left(\dfrac{r}{\cos\theta}\right)^2 - r^2}$ or $\dfrac{r\sin\theta}{\sin\left(\frac{\pi}{2}-\theta\right)}$. NOT $(90-\theta)$ |
| $P = 2r\theta + 2r\tan\theta$ | B1FT | OE, FT for *their* arc length $+ 2 \times$ *their* $AT$ |
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area $\Delta AOT = \frac{1}{2} \times 5 \times 5 \tan 1.2$ **or** Area $AOBT = 2 \times \frac{1}{2} \times 5 \times 5 \tan 1.2$ | B1 | |
| Sector area $= \frac{1}{2} \times 25 \times 2.4$ (or 1.2) | \*M1 | Use of $\frac{1}{2}r^2\theta$ with $\theta = 1.2$ or $2.4$ |
| Shaded area $= 2$ triangles $-$ sector | DM1 | Subtraction of sector, using 2.4 where appropriate, from 2 triangles |
| Area $= 34.3$ (cm$^2$) | A1 | AWRT |
| **Alternative:** Area of $\Delta ABT = \frac{1}{2} \times (5 \times \tan 1.2)^2 \times \sin(\pi - 2.4)$ ($= 55.86$) | B1 | |
| Segment area $= \frac{1}{2} \times 25 \times (2.4 - \sin 2.4)$ ($= 21.56$) | \*M1 | Use of $\frac{1}{2}r^2(\theta - \sin\theta)$ with $\theta = 1.2$ or $2.4$ |
| Shaded area $=$ triangle $-$ segment | DM1 | Subtraction of segment from $\Delta ABT$, using 2.4 where appropriate |
| Area $= 34.3$ (cm$^2$) | A1 | AWRT |
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\includegraphics[max width=\textwidth, alt={}, center]{567c3d72-c633-4ae0-8605-f63f93d718c4-06_517_768_262_685}
The diagram shows a circle with centre $O$ and radius $r \mathrm {~cm}$. Points $A$ and $B$ lie on the circle and angle $A O B = 2 \theta$ radians. The tangents to the circle at $A$ and $B$ meet at $T$.\\
(i) Express the perimeter of the shaded region in terms of $r$ and $\theta$.\\
(ii) In the case where $r = 5$ and $\theta = 1.2$, find the area of the shaded region.\\
\hfill \mbox{\textit{CAIE P1 2019 Q4 [7]}}