Read and solve the following problems. Write your answer and complete solution in your answer sheet. 1. A rectangle has a length of 7 feet and a width of 4 feet. It is connected with a semi-circle with the same diameter as the width of the rectangle. Find the combined area of the figures. 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?

Sure, let's solve each of the problems step by step. --- ### **Problem 1** **A rectangle has a length of 7 feet and a width of 4 feet. It is connected with a semi-circle with the same diameter as the width of the rectangle. Find the combined area of the figures.** **Solution:** 1. **Calculate the area of the rectangle:** \[ \text{Area}_{\text{rectangle}} = \text{Length} \times \text{Width} = 7\,\text{ft} \times 4\,\text{ft} = 28\,\text{sq ft} \] 2. **Determine the area of the semi-circle:** - The diameter of the semi-circle is the same as the width of the rectangle, which is 4 feet. - Radius \( r \) of the semi-circle: \[ r = \frac{\text{Diameter}}{2} = \frac{4\,\text{ft}}{2} = 2\,\text{ft} \] - Area of a full circle: \[ \text{Area}_{\text{circle}} = \pi r^2 = \pi (2\,\text{ft})^2 = 4\pi\,\text{sq ft} \] - Since it's a semi-circle: \[ \text{Area}_{\text{semi-circle}} = \frac{1}{2} \times \text{Area}_{\text{circle}} = \frac{1}{2} \times 4\pi\,\text{sq ft} = 2\pi\,\text{sq ft} \] 3. **Combine both areas:** \[ \text{Total Combined Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{semi-circle}} = 28\,\text{sq ft} + 2\pi\,\text{sq ft} \] \[ \text{Total Combined Area} \approx 28 + 6.283 = 34.283\,\text{sq ft} \] **Final Answer:** The combined area of the figures is approximately **34.28 square feet**. --- ### **Problem 2** **A circular picture whose diameter is 4 decimeters is framed in a rectangular board with dimensions of 6 decimeters by 4 decimeters. What is the area of the board that can be seen?** **Solution:** 1. **Calculate the area of the rectangular board:** \[ \text{Area}_{\text{board}} = \text{Length} \times \text{Width} = 6\,\text{dm} \times 4\,\text{dm} = 24\,\text{sq dm} \] 2. **Determine the area of the circular picture:** - Diameter of the circle is 4 decimeters. - Radius \( r \) of the circle: \[ r = \frac{\text{Diameter}}{2} = \frac{4\,\text{dm}}{2} = 2\,\text{dm} \] - Area of the circle: \[ \text{Area}_{\text{circle}} = \pi r^2 = \pi (2\,\text{dm})^2 = 4\pi\,\text{sq dm} \] \[ \text{Area}_{\text{circle}} \approx 12.566\,\text{sq dm} \] 3. **Calculate the visible area of the board:** \[ \text{Visible Area} = \text{Area}_{\text{board}} - \text{Area}_{\text{circle}} = 24\,\text{sq dm} - 4\pi\,\text{sq dm} \] \[ \text{Visible Area} \approx 24 - 12.566 = 11.434\,\text{sq dm} \] **Final Answer:** The area of the board that can be seen is approximately **11.43 square decimeters**. --- ### **Problem 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?** **Solution:** 1. **Determine the dimensions of the cardboard:** - **Length of the cardboard:** \[ \text{Length}_{\text{cardboard}} = \text{Length}_{\text{photo}} + 2 \times \text{Margin} = 20\,\text{cm} + 2 \times 2\,\text{cm} = 24\,\text{cm} \] - **Width of the cardboard:** \[ \text{Width}_{\text{cardboard}} = \text{Width}_{\text{photo}} + 2 \times \text{Margin} = 12\,\text{cm} + 2 \times 2\,\text{cm} = 16\,\text{cm} \] 2. **Calculate the area of the cardboard:** \[ \text{Area}_{\text{cardboard}} = \text{Length}_{\text{cardboard}} \times \text{Width}_{\text{cardboard}} = 24\,\text{cm} \times 16\,\text{cm} = 384\,\text{sq cm} \] 3. **Calculate the area of the photograph:** \[ \text{Area}_{\text{photo}} = \text{Length}_{\text{photo}} \times \text{Width}_{\text{photo}} = 20\,\text{cm} \times 12\,\text{cm} = 240\,\text{sq cm} \] 4. **Determine the uncovered area of the cardboard:** \[ \text{Uncovered Area} = \text{Area}_{\text{cardboard}} - \text{Area}_{\text{photo}} = 384\,\text{sq cm} - 240\,\text{sq cm} = 144\,\text{sq cm} \] **Final Answer:** The area of the cardboard that is not covered by the photograph is **144 square centimeters**. ---