SMART 1.3. Mr Paul intends to bulld an all-round rectangular flower garden bed made of wood sheets with the following dimensions. \[ \begin{array}{l} \text { Length }=1500 \mathrm{~cm} \\ \text { breadth }=7,5 \mathrm{~m} \end{array} \] 1.3.1. Calculate the Perimeter of the garden (2) 13.2. Determine the area (in square metres) You may use the formula: \( A=\lambda \times b \) (3) [16] HOMEWORK 1. The dimensions of the cylindrical water tank are given below: Diameter \( =28 \mathrm{~m} \) \[ \text { Height }=7 \mathrm{~m} \] Note: \( 1 m^{3}=1000 \) litres. 1.1. Explain the term radius. 1.2. Calculate the radus of the water tank (2) 1.3. The lid of the water tank must be painted On the outside for it to be visible. Calculate the area of the lid \( \left(i_{n} n^{2}\right) \) if the diameter is 28 m . The following formula may be (3) used: Area of a circle \( =\pi \times \) radius \( { }^{2} \) using \[ \pi=3,142 \] 14. A one lite tin of paint covers an area of \( 15 \mathrm{~m}^{2} \) calcuate the size of the area that can be covered with 17 lities of paint (2) 15. The inside of the tank must be sealed with water proof Paink. The lid does not get water proofed on the inside. Calculate the nherior Sufface mrea in \( m^{2} \) by using the forml. Insele area \( =(2 \times \pi \times r \times h)+\left(\pi \times r^{2}\right) \) (4)

**SMART** 1.3. Mr Paul’s garden dimensions: - Length: \(1500 \text{ cm} = 15 \text{ m}\) - Breadth: \(7.5 \text{ m}\) **1.3.1. Perimeter Calculation** The formula for the perimeter of a rectangle is \[ P = 2(l + b) \] Substitute \(l = 15 \text{ m}\) and \(b = 7.5 \text{ m}\): \[ P = 2(15 + 7.5) = 2(22.5) = 45 \text{ m} \] **1.3.2. Area Calculation** The area is calculated using the formula: \[ A = l \times b \] Substitute \(l = 15 \text{ m}\) and \(b = 7.5 \text{ m}\): \[ A = 15 \times 7.5 = 112.5 \text{ m}^2 \] --- **HOMEWORK** Dimensions of the cylindrical water tank: - Diameter: \(28 \text{ m}\) - Height: \(7 \text{ m}\) - Note: \(1 \text{ m}^3 = 1000\) litres. **1.1. Explanation of Radius** The radius is the distance from the center of a circle to any point on its circumference. It is exactly half of the diameter. **1.2. Calculation of the Radius** Given the diameter \(= 28 \text{ m}\), the radius \(r\) is: \[ r = \frac{28}{2} = 14 \text{ m} \] **1.3. Area of the Lid** The lid of the water tank is a circle. The formula for the area of a circle is: \[ \text{Area} = \pi r^2 \] Using \(r = 14 \text{ m}\) and \(\pi = 3.142\): \[ \text{Area} = 3.142 \times (14)^2 = 3.142 \times 196 = 616.232 \text{ m}^2 \] **14. Paint Coverage Calculation** One litre of paint covers \(15 \text{ m}^2\). Therefore, 17 litres will cover: \[ 17 \times 15 = 255 \text{ m}^2 \] **15. Interior Surface Area Calculation** The inside of the tank (excluding the lid) is given by the lateral (curved) surface area plus the area of the bottom. The formula is: \[ \text{Inside Area} = (2 \pi r h) + (\pi r^2) \] Substitute \(r = 14 \text{ m}\) and \(h = 7 \text{ m}\): - Lateral surface area: \[ 2 \pi r h = 2 \times 3.142 \times 14 \times 7 = 2 \times 3.142 \times 98 = 615.832 \text{ m}^2 \] - Area of the bottom: \[ \pi r^2 = 3.142 \times (14)^2 = 3.142 \times 196 = 615.832 \text{ m}^2 \] Thus, the total inside area is: \[ 615.832 + 615.832 = 1231.664 \text{ m}^2 \] Rounded appropriately, the interior surface area is approximately \(1231.66 \text{ m}^2\).