| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Compound shape area |
| Difficulty | Standard +0.3 This is a straightforward application of arc length formula (s = rθ) to find the angle, followed by decomposing the shaded region into a right triangle and sector. The geometry is clearly defined with perpendicular and parallel lines making calculations routine. Slightly above average difficulty due to the two-part geometric decomposition, but all techniques are standard P1 level. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks |
|---|---|
| \(0.8\) oe | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(BD = 5\sin(\text{their } 0.8)\) | M1 | Expect 3.58(7). Methods using degrees acceptable |
| \(DC = 5 - 5\cos(\text{their } 0.8)\) | M1 | Expect 1.51(6) |
| Sector \(= \frac{1}{2} \times 5^2 \times \text{their } 0.8\) OR Seg \(= \frac{1}{2} \times 5^2 \times [\text{their } 0.8 - \sin(\text{their } 0.8)]\) | M1 | Expect 10 for sector. Expect 1.03(3) for segment |
| Trap \(= \frac{1}{2}(5 + \text{their}DC) \times \text{their}BD\) oe OR \(\Delta BDC = \frac{1}{2}\text{their}BD \times \text{their}CD\) | M1 | OR (for last 2 marks) if \(X\) is on \(AB\) and \(XC\) is parallel to \(BD\) |
| Shaded area \(= 11.69 - 10\) OR \(2.71(9) - 1.03(3) = 1.69\) cao | A1 | \(BDCX - (\text{sector} - \Delta AXC) = 5.43(8) - [10 - 6.24(9)] = 1.69\) cao M1A1 |
## Question 3(i):
| $0.8$ oe | B1 | |
|---|---|---|
## Question 3(ii):
| $BD = 5\sin(\text{their } 0.8)$ | M1 | Expect 3.58(7). Methods using degrees acceptable |
|---|---|---|
| $DC = 5 - 5\cos(\text{their } 0.8)$ | M1 | Expect 1.51(6) |
| Sector $= \frac{1}{2} \times 5^2 \times \text{their } 0.8$ OR Seg $= \frac{1}{2} \times 5^2 \times [\text{their } 0.8 - \sin(\text{their } 0.8)]$ | M1 | Expect 10 for sector. Expect 1.03(3) for segment |
| Trap $= \frac{1}{2}(5 + \text{their}DC) \times \text{their}BD$ oe OR $\Delta BDC = \frac{1}{2}\text{their}BD \times \text{their}CD$ | M1 | OR (for last 2 marks) if $X$ is on $AB$ and $XC$ is parallel to $BD$ |
| Shaded area $= 11.69 - 10$ OR $2.71(9) - 1.03(3) = 1.69$ cao | A1 | $BDCX - (\text{sector} - \Delta AXC) = 5.43(8) - [10 - 6.24(9)] = 1.69$ cao M1A1 |
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\includegraphics[max width=\textwidth, alt={}, center]{d5d94eb8-7f41-4dff-b503-8be4f20e21b7-04_467_401_260_872}
The diagram shows an arc $B C$ of a circle with centre $A$ and radius 5 cm . The length of the arc $B C$ is 4 cm . The point $D$ is such that the line $B D$ is perpendicular to $B A$ and $D C$ is parallel to $B A$.\\
(i) Find angle $B A C$ in radians.\\
(ii) Find the area of the shaded region $B D C$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q3 [6]}}