CAIE P1 2023 June — Question 4 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeSector and arc length
DifficultyModerate -0.3 This is a straightforward sector problem requiring the formula for sector area to find the angle (θ = 2A/r² = π/3), then using arc length formula (rθ = 8π/3) and chord length (2r sin(θ/2) = 8). The calculation is routine with standard formulas and minimal problem-solving, making it slightly easier than average.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta
4 The diagram shows a sector \(A B C\) of a circle with centre \(A\) and radius 8 cm . The area of the sector is \(\frac { 16 } { 3 } \pi \mathrm {~cm} ^ { 2 }\). The point \(D\) lies on the \(\operatorname { arc } B C\). Find the perimeter of the segment \(B C D\).
Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{1}{2} \times 8^2 \times \theta = \frac{16\pi}{3} \Rightarrow \theta = \frac{\pi}{6}\)B1 SOI OE e.g. \(\frac{2\pi}{12}\), 0.524 (3 s.f.). Use of degrees acceptable throughout provided conversion used in formulae for sector area and arc length.
Arc length \(= 8 \times their\frac{\pi}{6}\) \([= 4.1887\ldots]\)M1 OE FT *their* \(\theta\). Look for \(\frac{4\pi}{3}\).
\([BC =]\ 2 \times 8 \sin\!\left(\frac{1}{2} \times their\frac{\pi}{6}\right)\) \([= 4.1411\ldots]\)M1 Attempt to find \(BC\) or \(BC^2\) (see alt. methods below). FT *their* \(\theta\). Look for \(16\sin\frac{\pi}{12}\) or \(4\sqrt{6} - 4\sqrt{2}\).
Perimeter \(= 8.33\)A1 AWRT Must be combined into one term.
4
Alternative methods for Question 4: 2nd M1 mark
AnswerMarks
AnswerGuidance
ALT 1 \(BC^2 = 8^2 + 8^2 - 2\times8\times8\cos\!\left(their\frac{\pi}{6}\right)\) \([\Rightarrow BC = 4.14\ldots]\)ALT 1: Substitute into correct cosine rule. FT *their* \(\theta\). Look for \(128 - 64\sqrt{3}\).
ALT 2 \(BC^2 = \left(8 - 4\sqrt{3}\right)^2 + 4^2\) \([\Rightarrow BC = 4.14\ldots]\)ALT 2: Find lengths 4 and \(4\sqrt{3}\), then use Pythagoras in the left-hand triangle.
ALT 3 \(\dfrac{BC}{\sin\!\left(\frac{\pi}{6}\right)} = \dfrac{8}{\sin\!\left(\frac{5\pi}{12}\right)}\) \([\Rightarrow BC = 4.14\ldots]\)ALT 3: Substitute into correct sine rule. 
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 8^2 \times \theta = \frac{16\pi}{3} \Rightarrow \theta = \frac{\pi}{6}$ | **B1** | SOI OE e.g. $\frac{2\pi}{12}$, 0.524 (3 s.f.). Use of degrees acceptable throughout provided conversion used in formulae for sector area and arc length. |
| Arc length $= 8 \times their\frac{\pi}{6}$ $[= 4.1887\ldots]$ | **M1** | OE FT *their* $\theta$. Look for $\frac{4\pi}{3}$. |
| $[BC =]\ 2 \times 8 \sin\!\left(\frac{1}{2} \times their\frac{\pi}{6}\right)$ $[= 4.1411\ldots]$ | **M1** | Attempt to find $BC$ or $BC^2$ (see alt. methods below). FT *their* $\theta$. Look for $16\sin\frac{\pi}{12}$ or $4\sqrt{6} - 4\sqrt{2}$. |
| Perimeter $= 8.33$ | **A1** | AWRT Must be combined into one term. |
| | **4** | |

**Alternative methods for Question 4: 2nd M1 mark**

| Answer | Guidance |
|--------|----------|
| **ALT 1** $BC^2 = 8^2 + 8^2 - 2\times8\times8\cos\!\left(their\frac{\pi}{6}\right)$ $[\Rightarrow BC = 4.14\ldots]$ | ALT 1: Substitute into correct cosine rule. FT *their* $\theta$. Look for $128 - 64\sqrt{3}$. |
| **ALT 2** $BC^2 = \left(8 - 4\sqrt{3}\right)^2 + 4^2$ $[\Rightarrow BC = 4.14\ldots]$ | ALT 2: Find lengths 4 and $4\sqrt{3}$, then use Pythagoras in the left-hand triangle. |
| **ALT 3** $\dfrac{BC}{\sin\!\left(\frac{\pi}{6}\right)} = \dfrac{8}{\sin\!\left(\frac{5\pi}{12}\right)}$ $[\Rightarrow BC = 4.14\ldots]$ | ALT 3: Substitute into correct sine rule. |

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4

The diagram shows a sector $A B C$ of a circle with centre $A$ and radius 8 cm . The area of the sector is $\frac { 16 } { 3 } \pi \mathrm {~cm} ^ { 2 }$. The point $D$ lies on the $\operatorname { arc } B C$.

Find the perimeter of the segment $B C D$.\\

\hfill \mbox{\textit{CAIE P1 2023 Q4 [4]}}
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