| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Triangle and sector combined - algebraic/general expressions |
| Difficulty | Standard +0.3 This is a standard application of sector and triangle area formulas with straightforward angle-chasing in an isosceles triangle. Part (a) requires expressing areas algebraically (routine manipulation), and part (b) involves direct substitution and calculating arc length—slightly above average due to the combined geometry but no novel insight required. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of correct formula for the area of triangle \(ABC\) | M1 | Use of \(180-2\theta\) scores M0. Condone \(2\pi - 2\theta\) |
| \(\frac{1}{2}r^2\sin(\pi - 2\theta)\) or \(\frac{1}{2}r^2\sin 2\theta\) or \(2 \times \frac{1}{2}r \times r\cos\theta \times \sin\theta\) or \(2 \times \frac{1}{2}r\cos\theta \times r\sin\theta\) | A1 | OE |
| Shaded area = triangle \(-\) sector \(=\) *their* triangle area \(- \frac{1}{2}r^2\theta\) | B1 FT | FT for *their* triangle area \(- \frac{1}{2}r^2\theta\) (Condone use of 180 degrees for triangle area for B1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Arc \(BD = r\theta = 6\) cm | B1 | SOI |
| \(AC = 2r\cos\theta = (2 \times 10\cos 0.6) = 20\cos 0.6 = 16.506\) or \(\sqrt{2r^2 - 2r^2\cos(\pi - 2\theta)}\) or \(\frac{r \times \sin(\pi - 2\theta)}{\sin\theta}\) | \*M1 | Finding \(AC\) or \(\frac{1}{2}AC\) (= 8.25) |
| \(DC = 2r\cos\theta - r\) or \(\sqrt{2r^2 - 2r^2\cos(\pi-2\theta)} - r\) (= 6.506) | DM1 | Subtracting \(r\) from *their* \(AC\) or \(r\cos\theta\) from *their* half \(AC\) (8.25−1.75) |
| Perimeter \(= 10 + 6 + 6.506 = 22.5\) | A1 | AWRT |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of correct formula for the area of triangle $ABC$ | M1 | Use of $180-2\theta$ scores M0. Condone $2\pi - 2\theta$ |
| $\frac{1}{2}r^2\sin(\pi - 2\theta)$ or $\frac{1}{2}r^2\sin 2\theta$ or $2 \times \frac{1}{2}r \times r\cos\theta \times \sin\theta$ or $2 \times \frac{1}{2}r\cos\theta \times r\sin\theta$ | A1 | OE |
| Shaded area = triangle $-$ sector $=$ *their* triangle area $- \frac{1}{2}r^2\theta$ | B1 FT | FT for *their* triangle area $- \frac{1}{2}r^2\theta$ (Condone use of 180 degrees for triangle area for B1) |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Arc $BD = r\theta = 6$ cm | B1 | SOI |
| $AC = 2r\cos\theta = (2 \times 10\cos 0.6) = 20\cos 0.6 = 16.506$ or $\sqrt{2r^2 - 2r^2\cos(\pi - 2\theta)}$ or $\frac{r \times \sin(\pi - 2\theta)}{\sin\theta}$ | \*M1 | Finding $AC$ or $\frac{1}{2}AC$ (= 8.25) |
| $DC = 2r\cos\theta - r$ or $\sqrt{2r^2 - 2r^2\cos(\pi-2\theta)} - r$ (= 6.506) | DM1 | Subtracting $r$ from *their* $AC$ or $r\cos\theta$ from *their* half $AC$ (8.25−1.75) |
| Perimeter $= 10 + 6 + 6.506 = 22.5$ | A1 | AWRT |
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\includegraphics[max width=\textwidth, alt={}, center]{6bcc553c-4938-46ef-bba4-97391b4d58d4-10_348_700_262_721}
In the diagram, $A B C$ is an isosceles triangle with $A B = B C = r \mathrm {~cm}$ and angle $B A C = \theta$ radians. The point $D$ lies on $A C$ and $A B D$ is a sector of a circle with centre $A$.
\begin{enumerate}[label=(\alph*)]
\item Express the area of the shaded region in terms of $r$ and $\theta$.
\item In the case where $r = 10$ and $\theta = 0.6$, find the perimeter of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q8 [7]}}