Semicircle Rectangle 4Bd11B

1. **State the problem:** We have a rectangle with length $14$ cm and breadth $8$ cm, and a semicircle attached along the breadth side. The diameter of the semicircle is $8$ cm. 2. **Calculate the total perimeter:** - The perimeter of the combined shape includes: - The two lengths of the rectangle: $2 \times 14 = 28$ cm - One breadth of the rectangle (the bottom side): $8$ cm - The curved edge of the semicircle (half the circumference of a circle with diameter $8$ cm) - The circumference of a full circle with diameter $8$ cm is: $$C = \pi \times d = \pi \times 8 = 8\pi$$ - The semicircle perimeter (curved part) is half of that: $$\frac{8\pi}{2} = 4\pi$$ - Total perimeter $P$ is: $$P = 28 + 8 + 4\pi = 36 + 4\pi$$ - Approximate value using $\pi \approx 3.1416$: $$P \approx 36 + 4 \times 3.1416 = 36 + 12.5664 = 48.5664 \text{ cm}$$ 3. **Calculate the total area:** - Area of the rectangle: $$A_{rect} = length \times breadth = 14 \times 8 = 112 \text{ cm}^2$$ - Area of the semicircle: - Radius $r = \frac{diameter}{2} = \frac{8}{2} = 4$ cm - Area of full circle: $\pi r^2 = \pi \times 4^2 = 16\pi$ - Area of semicircle: $$A_{semi} = \frac{16\pi}{2} = 8\pi$$ - Total area $A$ is: $$A = 112 + 8\pi$$ - Approximate value: $$A \approx 112 + 8 \times 3.1416 = 112 + 25.1328 = 137.1328 \text{ cm}^2$$ **Final answers:** - Total perimeter: $36 + 4\pi \approx 48.57$ cm - Total area: $112 + 8\pi \approx 137.13$ cm$^2$