## Question 5:

### Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| $\angle AOB = \cos^{-1}\frac{1.5}{4} = ...$ or $\angle AOB = \tan^{-1}\frac{\sqrt{4^2-1.5^2}}{1.5} = ...$ or $\angle AOB = \frac{\pi}{2} - \sin^{-1}\frac{1.5}{4} = ...$ or $\angle AOB = \frac{\pi}{2} - \tan^{-1}\frac{1.5}{\sqrt{4^2-1.5^2}} = ...$ | M1 | Complete and correct method to find angle $AOB$ in radians or degrees. May attempt angle $AOB$ directly using cosine, or angle $BOD$ using sine then subtract from $\frac{\pi}{2}$, or equivalent using tan |
| $= 1.186$ | A1* | Obtains angle $AOB = 1.186$ with no errors |

### Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| $l = 4 \times 1.186\ (= 4.744)$ | M1 | Uses correct arc length formula with $r=4$ and $1.186$. May be implied by awrt 4.74 or 4.75 |
| $BC = 6 - 4\sin 1.186\ (= 6 - 3.708... = 2.29...)$ or $BC = 6 - 4\cos\left(\frac{\pi}{2}-1.186\right)\ (= 2.29...)$ or $BC = 6 - \sqrt{4^2-1.5^2}\ \left(= \frac{12-\sqrt{55}}{2}\right)$ or $BC = 6 - \frac{1.5}{\tan"0.384"}\ (= 2.29...)$ | M1 | Attempts length of $BC$ using appropriate trig function and angle, or Pythagoras. May be implied by awrt 2.3 provided no incorrect working seen |
| Perimeter is $4\times1.186... + (6-4\cos 0.384...) + 1.5 + 6 + 4 = ..$ | ddM1 | Depends on both previous marks. Adds the five lengths of the sides to obtain the perimeter |
| $= \text{awrt } 18.5\ \text{(m)}$ | A1 | Accept awrt 18.5. No units required |

### Part (c):

| Working | Mark | Guidance |
|---------|------|----------|
| Area sector $= \frac{1}{2} \times 4^2 \times 1.186...$ | M1 | Correct attempt at sector area. May be implied by e.g. $\frac{1186}{125}$ or awrt 9.5. Allow equivalent in degrees e.g. $\frac{1}{2}\times 4^2 \times 67.9756...\times\frac{\pi}{180}$ but **not** $\frac{1}{2}\times 4^2 \times 67.9756$ |
| Area $OBCD$: $6\times1.5 - \frac{1}{2}\times\text{"3.708"}\times1.5$ or $(6-\text{"3.708"})\times1.5 + \frac{1}{2}\times\text{"3.708"}\times1.5$ or $\frac{1}{2}(6+6-\text{"3.708"})\times1.5$ or $6\times1.5 - \frac{1}{2}\times1.5\times4\sin1.186$ or $\frac{1}{2}\times6\times1.5 + \frac{1}{2}(6-\text{"3.708"})\times1.5\ (= 6.219...)$ | M1, A1 on EPEN | Correct attempt at area of $OBCD$ — may be found as difference of rectangle and triangle, sum of rectangle and triangle, or as a trapezium. Attempt to find $BC$, if used, must involve trigonometry or Pythagoras |
| $\frac{1}{2}\times4^2\times1.186 + 6.219... =$ | ddM1 | Depends on both previous marks. Correct attempt at full area, adding sector area to area of $OBCD$ |
| $= \text{awrt } 15.7\ \text{(m}^2)$ | A1 | Accept awrt 15.7. No units required |

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