1. **Stating the problem:** We need to calculate the area and perimeter of composite figures made up of rectangles, triangles, and semicircles. 2. **Formulas used:** - Area of rectangle: $A = l \times b$ - Area of triangle: $A = \frac{1}{2} \times base \times height$ - Area of semicircle: $A = \frac{\pi r^2}{2}$ - Perimeter (O) is the sum of all outer sides. 3. **Example calculation for the top-left figure:** - Rectangle area: $6.0 \text{ cm} \times 3.0 \text{ cm} = 18.00 \text{ cm}^2$ - Semicircle area: $\frac{3.14 \times (3.0 \text{ cm})^2}{2} = \frac{3.14 \times 9}{2} = 14.13 \text{ cm}^2$ - Shaded area: $18.00 - 14.13 = 3.87 \text{ cm}^2$ 4. **Calculating area and perimeter of the center-left composite figure:** - Rectangle area: $8.9 \times 6.1 = 54.29 \text{ cm}^2$ - Triangle area: $\frac{1}{2} \times 4.0 \times 5.2 = 10.4 \text{ cm}^2$ - Total area: $54.29 + 10.4 = 64.69 \text{ cm}^2$ - Perimeter (O) is sum of all outer sides: $O = 8.9 + 6.1 + 4.0 + 5.2 + \text{other sides depending on shape}$ 5. **Calculating area of bottom-left figures:** - a) Square area: $4 \times 4 = 16 \text{ cm}^2$ - Triangle area (equilateral with height 3): $\frac{1}{2} \times 4 \times 3 = 6 \text{ cm}^2$ - Total area a): $16 + 6 = 22 \text{ cm}^2$ - b) Rectangle area: $8 \times 4 = 32 \text{ cm}^2$ - Triangle area (isosceles with height 4): $\frac{1}{2} \times 8 \times 4 = 16 \text{ cm}^2$ - Total area b): $32 + 16 = 48 \text{ cm}^2$ 6. **Identifying geometric figures:** - Rectangles, triangles (equilateral and isosceles), semicircles, and squares. 7. **Calculating area of the bottom-center figure:** - Two triangles side-by-side with total base $12.4$ cm and height $4.0$ cm. - Area of one triangle: $\frac{1}{2} \times \frac{12.4}{2} \times 4.0 = \frac{1}{2} \times 6.2 \times 4.0 = 12.4 \text{ cm}^2$ - Total area: $12.4 \times 2 = 24.8 \text{ cm}^2$ This approach can be applied to all composite figures by breaking them down into basic shapes, calculating each area, and summing them up.