## Question 3:

**Part 3(a):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $L$ is symmetrical about $AD$ | B1 | Any equivalent statement about the symmetry |

**Part 3(b):**

| Answer | Mark | Guidance |
|--------|------|----------|
| Mass ratios: $ABDF = 4a^2 \times M$, $BCD = a^2 \times 3M$, $DEF = a^2 \times 3M$, $L = 10a^2 \times M$ | B1 | Correct mass ratios |
| Distances from $BE$: $-a$, $+\frac{a}{3}$, $-\frac{2a}{3}$, $x$ | B1 | Distance ratios from any horizontal or vertical axis |
| Moments equation set up | M1 | Must be dimensionally correct |
| $-a \times 4a^2M + \frac{a}{3} \times 3a^2M - \frac{2a}{3} \times 3a^2M = 10a^2M \times x$ giving $(-4a + a - 2a = 10x)$ | A1 | Correct unsimplified equation |
| $x = -\frac{5a}{10} = -\frac{a}{2}$ | A1 | Correct horizontal or vertical distance from $D$ |
| Use symmetry and Pythagoras | M1 | Use of Pythagoras with their distance |
| Distance from $D = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \frac{\sqrt{2}}{2}a$ | A1* | Obtain given answer from correct working |

**Part 3(c):**

| Answer | Mark | Guidance |
|--------|------|----------|
| Trig ratio of a relevant angle | M1 | Trig ratio of $\theta$ or $90°-\theta$ or equivalent |
| $\tan\theta = \frac{1}{3}$ or $\cos\theta = \dfrac{\frac{10}{4}a^2 + 4a^2 - \frac{2}{4}a^2}{2 \times \frac{\sqrt{10}}{2}a \times 2a} = \dfrac{6}{2\sqrt{10}}$ | A1ft | Correct unsimplified expression using their $\frac{a}{2}$ |
| $\theta = 18.4°$ | A1 | Correct angle; accept 0.322 radians |

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