1. **Problem 29:** Given numbers 40 and $p$ in ratio 4:3, with prime factors shown as: - $40 = 2^3 \times 5$ - $p = 3^1 \times n$ (i) Find $p$: - Since ratio is $\frac{40}{p} = \frac{4}{3}$, solve for $p$: $$p = \frac{3}{4} \times 40 = 30$$ - Prime factorize 30: $$30 = 2 \times 3 \times 5$$ - Given $p = 3^1 \times n$, then $n = 2 \times 5 = 10$ (ii) Find $r$: - From the Venn diagram, $r$ is the common prime factor power of 2 in 40 and $p$. - $40$ has $2^3$, $p$ has no 2 in prime factors, so $r = 0$ (iii) Find $n$: - From above, $n = 10$ 2. **Problem 30:** Find the area of the shaded part (circle inside trapezium): - Trapezium has parallel sides $a=20$ cm, $b=32$ cm, and non-parallel sides $10$ cm each. - Use Pythagoras to find height $h$: $$h = \sqrt{10^2 - \left(\frac{32-20}{2}\right)^2} = \sqrt{100 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \text{ cm}$$ - Area of trapezium: $$A_{trap} = \frac{1}{2} (a+b) h = \frac{1}{2} (20+32) \times 8 = 208 \text{ cm}^2$$ - Diameter of circle = height = 8 cm, so radius $r=4$ cm. - Area of circle: $$A_{circle} = \pi r^2 = \pi \times 4^2 = 16\pi \approx 50.27 \text{ cm}^2$$ - Area of shaded part (trapezium minus circle): $$208 - 16\pi \approx 208 - 50.27 = 157.73 \text{ cm}^2$$ **Final answers:** (i) $p = 30$ (ii) $r = 0$ (iii) $n = 10$ Area of shaded part $\approx 157.73$ cm$^2$