Step 1: Address Question 28 (Part 1). The question "a circle of radius 17 cm. Find the distance between the chords." is incomplete. To find the distance between chords, their lengths or other properties must be provided. Step 2: Address Question 28 (Part 2). Find the square metres of canvas required for a conical tent with height $h = 3.5$ m and base radius $r = 12$ m. The canvas required is the lateral surface area of the cone, given by $A = \pi r l$, where $l$ is the slant height. First, calculate the slant height $l$: $$l = \sqrt{r^2 + h^2}$$ $$l = \sqrt{(12)^2 + (3.5)^2}$$ $$l = \sqrt{144 + 12.25}$$ $$l = \sqrt{156.25}$$ $$l = 12.5 \text{ m}$$ Now, calculate the lateral surface area: $$A = \pi \times 12 \text{ m} \times 12.5 \text{ m}$$ $$A = 150\pi \text{ m}^2$$ Using $\pi \approx 3.14159$: $$A \approx 150 \times 3.14159 \text{ m}^2$$ $$A \approx 471.2385 \text{ m}^2$$ The canvas required is $\boxed{150\pi \text{ m}^2 \text{ or approximately } 471.24 \text{ m}^2}$. Step 3: Address Question 29 (Part 1). The surface areas of two spheres are in the ratio $1:4$. Find the ratio of their volumes. Let the radii of the two spheres be $r_1$ and $r_2$. The surface area of a sphere is $A = 4\pi r^2$. Given $\frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{1}{4}$. $$\frac{r_1^2}{r_2^2} = \frac{1}{4}$$ $$\left(\frac{r_1}{r_2}\right)^2 = \frac{1}{4}$$ Taking the square root of both sides: $$\frac{r_1}{r_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$ The volume of a sphere is $V = \frac{4}{3}\pi r^3$. The ratio of their volumes is: $$\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3$$ Substitute the ratio of radii: $$\frac{V_1}{V_2} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$$ The ratio of their volumes is $\boxed{1:8}$. Step 4: Address Question 29 (Part 2). In the given figure, $AB \parallel CD$. Find the value of $x, y$ and $z$. Given: • Line $AB$ is parallel to line $CD$. • $\angle AEF = 75^\circ$. • $\angle EGD = 125^\circ$. • $x = \angle EFG$, $y = \angle FGE$, $z = \angle FEG$. 1. Find $x$: Since $AB \parallel CD$, and EF is a transversal, $\angle AEF$ and $\angle EFG$ are alternate interior angles. Therefore, $\angle EFG = \angle AEF$. $$x = 75^\circ$$ 2. Find $y$: Since $AB \parallel CD$, and EG is a transversal, $\angle EGD$ and $\angle FGE$ are angles on a straight line with the angle adjacent to $\angle EGD$ on line CD. The angle $\angle FGE$ and $\angle EGD$ are not directly related as alternate interior or corresponding. However, $\angle FGE$ and $\angle EGC$ are supplementary if G is on CD. The angle $\angle FGE$ is part of the triangle. The angle $\angle EGC$ is $180^\circ - \angle EGD$ (angles on a straight line). $\angle EGC = 180^\circ - 125^\circ = 55^\circ$. So, $y = \angle FGE = 55^\circ$. 3. Find $z$: The sum of angles in a triangle is $180^\circ$. In $\triangle EFG$: $$\angle FEG + \angle EFG + \angle FGE = 180^\circ$$ $$z + x + y = 180^\circ$$ $$z + 75^\circ + 55^\circ = 180^\circ$$ $$z + 130^\circ = 180^\circ$$ $$z = 180^\circ - 130^\circ$$ $$z = 50^\circ$$ The values are $\boxed{x = 75^\circ, y = 55^\circ, z = 50^\circ}$. Step 5: Address Question 30.