## Question 12(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $[6 \times P\hat{A}Q = 2\pi],\ [P\hat{A}Q =]\ 2\pi \div 6$ | M1 | |
| Explaining six sectors around diagram make up a complete circle | A1 | AG |

**Alternative 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Using area or circumference of circle centre $A \div 6$ | M1 | $\frac{400\pi}{6}$ or $\frac{40\pi}{6}$ |
| Justification for dividing by 6 followed by comparison with sector area or arc length | A1 | AG |

**Alternative 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Explain why $\triangle PAQ$ is an equilateral triangle | M1 | Assumption of this scores M0 |
| Using $\triangle PAQ$ equilateral $\therefore P\hat{A}Q = \frac{\pi}{3}$ | A1 | AG |

**Alternative 3:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Using internal angle of regular hexagon $= \frac{2\pi}{3}$, or $F\hat{A}O + O\hat{A}B = \frac{2\pi}{3}$, equilateral triangles | M1 | |
| $P\hat{A}Q = 2\pi - \left(\frac{\pi}{2} + \frac{2\pi}{3} + \frac{\pi}{2}\right) = \frac{\pi}{3}$ | A1 | AG |

**Alternative 4:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin\theta = \frac{20}{40}$, with $\theta$ clearly identified | M1 | |
| $\theta = \frac{\pi}{6},\ 2\theta = \frac{\pi}{3} = F\hat{A}O$ and by similar triangles $= P\hat{A}Q$ | A1 | AG |

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## Question 12(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Each straight section of rope has length 40 cm | B1 | SOI |
| Each curved section round each pipe has length $r\theta = 20\times\frac{\pi}{3}$ | \*M1 | Use of $r\theta$ with $r = 20$ and $\theta$ in radians |
| Total length $= 6\times\bigl((\text{their}\ 40) + k\pi\bigr)$ | DM1 | $6\times(\text{their straight section} + \text{their curved section})$. Their curved section must be from acceptable use of $r\theta$ |
| $240 + 40\pi$ or $366$ (AWRT) (cm) | A1 | Or directly: $(6\times \text{diameter}) + \text{circumference}$ |

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## Question 12(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Triangle area $= \frac{1}{2}\times40\times40\times\sin\!\left(\frac{\pi}{3}\right)$ or $\frac{1}{2}\times40\times20\sqrt{3}$ or $400\sqrt{3}$ or 693 (AWRT) | B1 | |
| Total area of hexagon $= 6\times400\sqrt{3} = 2400\sqrt{3}$ | B1 | Condone $4800\frac{\sqrt{3}}{2}$ |

**Alternative 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Trapezium area $= \frac{1}{2}\times(40+80)\times40\sin\!\left(\frac{\pi}{3}\right)$ or $1200\sqrt{3}$ or 2080 (AWRT) | B1 | |
| Total area of hexagon $= 2\times1200\sqrt{3} = 2400\sqrt{3}$ | B1 | Condone $4800\frac{\sqrt{3}}{2}$ |

**Alternative 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Area of $\triangle ABC = 400\sqrt{3}$ or 693 (AWRT), or $4\times$ area of half $\triangle ABC = 4\times200\sqrt{3}$ or 1390 (AWRT), or area of rectangle $ABDE = 1600\sqrt{3}$ or 2770 (AWRT) | B1 | |
| Total area $= 2\times400\sqrt{3} + 1600\sqrt{3} = 2400\sqrt{3}$, or $[= 4\times200\sqrt{3} + 1600 =]\ 2400\sqrt{3}$ | B1 | Condone $4800\frac{\sqrt{3}}{2}$. If B0B0, SC B1 for sight of 4160 (AWRT) as final answer |

## Question 12(d):

| Answer | Mark | Guidance |
|--------|------|----------|
| Each rectangle area $= 40 \times 20 (= 800)$ | B1 | SOI, e.g. by sight of 4800 |
| Each sector area $= \frac{1}{2}r^2\theta = \frac{1}{2} \times 20^2 \times \frac{\pi}{3} \left[= \frac{200\pi}{3}\right]$ | B1 | SOI. |
| Total area $= 2400\sqrt{3} + 4800 + 400\pi$ or $10\,200$ (cm²) (AWRT) | B1 | Or directly: part **(c)** $+ 6800 +$ area circle radius 20. |
| | **3** | |