3. A cylindrical tin with one end open has diameter 140 mm and height 50 mm . calculate the i. Area of the base ii. Curved surface area iii. Total surface area

Sure, let's solve each part step by step for the cylindrical tin with one end open. **Given:** - **Diameter (D)** = 140 mm - **Radius (r)** = Diameter/2 = 140 mm / 2 = 70 mm - **Height (h)** = 50 mm --- ### **i. Area of the Base** The base of the cylinder is a circle. The area \( A \) of a circle is given by: \[ A = \pi r^2 \] **Calculation:** \[ A = \pi \times (70 \, \text{mm})^2 = \pi \times 4900 \, \text{mm}^2 = 4900\pi \, \text{mm}^2 \] **Approximate Value (using \( \pi \approx 3.14 \)):** \[ A \approx 4900 \times 3.14 = 15,\!386 \, \text{mm}^2 \] **Answer:** \[ \boxed{\text{i. Area of the base} = 4900\pi \, \text{mm}^2 \approx 15,\!386 \, \text{mm}^2} \] --- ### **ii. Curved Surface Area (CSA)** The curved surface area (also known as the lateral area) of a cylinder is calculated by: \[ \text{CSA} = 2\pi r h \] **Calculation:** \[ \text{CSA} = 2 \times \pi \times 70 \, \text{mm} \times 50 \, \text{mm} = 7000\pi \, \text{mm}^2 \] **Approximate Value:** \[ \text{CSA} \approx 7000 \times 3.14 = 21,\!980 \, \text{mm}^2 \] **Answer:** \[ \boxed{\text{ii. Curved Surface Area} = 7000\pi \, \text{mm}^2 \approx 21,\!980 \, \text{mm}^2} \] --- ### **iii. Total Surface Area (TSA)** Since the cylinder has one end open, the total surface area is the sum of the curved surface area and the area of the single base. \[ \text{TSA} = \text{CSA} + \text{Area of Base} = 7000\pi + 4900\pi = 11,\!900\pi \, \text{mm}^2 \] **Approximate Value:** \[ \text{TSA} \approx 11,\!900 \times 3.14 = 37,\!366 \, \text{mm}^2 \] **Answer:** \[ \boxed{\text{iii. Total Surface Area} = 11,\!900\pi \, \text{mm}^2 \approx 37,\!366 \, \text{mm}^2} \] --- **Summary of Results:** 1. **Area of the base:** \( 4900\pi \, \text{mm}^2 \) (≈ 15,386 mm²) 2. **Curved Surface Area:** \( 7000\pi \, \text{mm}^2 \) (≈ 21,980 mm²) 3. **Total Surface Area:** \( 11,\!900\pi \, \text{mm}^2 \) (≈ 37,366 mm²)