2. A circular picture whose diameter is 4 decimeters is framed in a rectangular board with dimension of 6 decimeter by 4 decimeters. What is the area of the board that can be seen? 3. A photograph measuring 20 cm by 12 cm is mounted on a rectangular cardboard leaving a margin of 2 cm around. What is the area of the cardboard that is not covered by the photograph?

Let's solve each problem step by step. --- ### **Problem 2** **Given:** - **Circular Picture:** - Diameter = 4 decimeters (dm) - Radius \( r = \frac{\text{Diameter}}{2} = 2 \) dm - **Rectangular Board:** - Length = 6 dm - Width = 4 dm **Objective:** Find the area of the **visible** part of the board (i.e., the area of the board not covered by the picture). **Solution:** 1. **Calculate the Area of the Rectangular Board:** \[ \text{Area}_{\text{Board}} = \text{Length} \times \text{Width} = 6 \, \text{dm} \times 4 \, \text{dm} = 24 \, \text{dm}^2 \] 2. **Calculate the Area of the Circular Picture:** \[ \text{Area}_{\text{Circle}} = \pi r^2 = \pi (2 \, \text{dm})^2 = 4\pi \, \text{dm}^2 \] 3. **Determine the Visible Area of the Board:** \[ \text{Visible Area} = \text{Area}_{\text{Board}} - \text{Area}_{\text{Circle}} = 24 \, \text{dm}^2 - 4\pi \, \text{dm}^2 \] 4. **Approximate Numerical Value (Optional):** \[ \text{Visible Area} \approx 24 - 12.566 \approx 11.434 \, \text{dm}^2 \] (Using \( \pi \approx 3.1416 \)) **Answer:** The area of the visible part of the board is \( 24 - 4\pi \) square decimeters, which is approximately **11.434 dm²**. --- ### **Problem 3** **Given:** - **Photograph:** - Length = 20 centimeters (cm) - Width = 12 cm - **Cardboard:** - Margins = 2 cm on all sides **Objective:** Find the area of the cardboard that is **not** covered by the photograph. **Solution:** 1. **Determine the Total Dimensions of the Cardboard:** - **Total Length:** \[ \text{Cardboard Length} = \text{Photograph Length} + 2 \times \text{Margin} = 20 \, \text{cm} + 2 \times 2 \, \text{cm} = 24 \, \text{cm} \] - **Total Width:** \[ \text{Cardboard Width} = \text{Photograph Width} + 2 \times \text{Margin} = 12 \, \text{cm} + 2 \times 2 \, \text{cm} = 16 \, \text{cm} \] 2. **Calculate the Area of the Cardboard:** \[ \text{Area}_{\text{Cardboard}} = \text{Length} \times \text{Width} = 24 \, \text{cm} \times 16 \, \text{cm} = 384 \, \text{cm}^2 \] 3. **Calculate the Area Covered by the Photograph:** \[ \text{Area}_{\text{Photograph}} = 20 \, \text{cm} \times 12 \, \text{cm} = 240 \, \text{cm}^2 \] 4. **Determine the Uncovered Area of the Cardboard:** \[ \text{Uncovered Area} = \text{Area}_{\text{Cardboard}} - \text{Area}_{\text{Photograph}} = 384 \, \text{cm}^2 - 240 \, \text{cm}^2 = 144 \, \text{cm}^2 \] **Answer:** The area of the cardboard not covered by the photograph is **144 cm²**.