1. **State the problem:** Calculate the total area of a compound shape consisting of a square, an isosceles triangle on top of the square, and two semicircles on either side of the triangle. 2. **Given dimensions:** - Square side length $s = 6$ cm - Triangle height $h = 5.2$ cm - The base of the triangle is the same as the square side, $6$ cm - Two semicircles on either side of the triangle, each with radius $r = \frac{6}{2} = 3$ cm (since the diameter equals the side length of the square) 3. **Formulas used:** - Area of square: $A_{square} = s^2$ - Area of triangle: $A_{triangle} = \frac{1}{2} \times base \times height$ - Area of circle: $A_{circle} = \pi r^2$ - Area of semicircle: $A_{semicircle} = \frac{1}{2} A_{circle} = \frac{1}{2} \pi r^2$ 4. **Calculate each area:** - Square area: $$A_{square} = 6^2 = 36 \text{ cm}^2$$ - Triangle area: $$A_{triangle} = \frac{1}{2} \times 6 \times 5.2 = 3 \times 5.2 = 15.6 \text{ cm}^2$$ - Area of one semicircle: $$A_{semicircle} = \frac{1}{2} \pi (3)^2 = \frac{1}{2} \pi \times 9 = \frac{9\pi}{2} \text{ cm}^2$$ - Area of two semicircles: $$2 \times A_{semicircle} = 2 \times \frac{9\pi}{2} = 9\pi \text{ cm}^2$$ 5. **Sum all areas to get total area:** $$A_{total} = A_{square} + A_{triangle} + 2 \times A_{semicircle} = 36 + 15.6 + 9\pi$$ 6. **Approximate numerical value:** $$9\pi \approx 9 \times 3.1416 = 28.2744$$ $$A_{total} \approx 36 + 15.6 + 28.2744 = 79.8744 \text{ cm}^2$$ **Final answer:** $$\boxed{79.87 \text{ cm}^2}$$