To find the perimeter of the figure, we need to sum the lengths of all its outer edges. Let's identify the lengths of the exterior sides: 1. The left vertical side of rectangle B: $5 \text{ cm}$ 2. The bottom horizontal side of rectangle B: $12 \text{ cm}$ 3. The right vertical side of rectangle B: $5 \text{ cm}$ 4. The top horizontal side of rectangle B that is exposed (not covered by A): $9 \text{ cm}$ 5. The left vertical side of rectangle A: $3 \text{ cm}$ 6. The top horizontal side of rectangle A: $3 \text{ cm}$ 7. The right vertical side of rectangle A: $3 \text{ cm}$ Now, we sum these lengths to find the total perimeter: $$P = 5 \text{ cm} + 12 \text{ cm} + 5 \text{ cm} + 9 \text{ cm} + 3 \text{ cm} + 3 \text{ cm} + 3 \text{ cm}$$ $$P = 40 \text{ cm}$$ Alternatively, we can calculate the perimeter of each rectangle and subtract the overlapping internal boundary. Perimeter of rectangle B: $2 \times (12 \text{ cm} + 5 \text{ cm}) = 2 \times 17 \text{ cm} = 34 \text{ cm}$ Perimeter of rectangle A: $2 \times (3 \text{ cm} + 3 \text{ cm}) = 2 \times 6 \text{ cm} = 12 \text{ cm}$ The length of the common boundary (where A and B meet) is $3 \text{ cm}$. This boundary is counted twice when summing the individual perimeters, so we subtract $2 \times 3 \text{ cm}$ from the sum. Total perimeter = (Perimeter of B) + (Perimeter of A) - $2 \times (\text{common boundary})$ $$P = 34 \text{ cm} + 12 \text{ cm} - 2 \times 3 \text{ cm}$$ $$P = 46 \text{ cm} - 6 \text{ cm}$$ $$P = 40 \text{ cm}$$ Both methods yield the same result. Comparing with the given options: (a) $69 \text{ cm}$ (b) $46 \text{ cm}$ (c) $43 \text{ cm}$ (d) $40 \text{ cm}$ (e) $37 \text{ cm}$ The calculated perimeter matches option (d). The final answer is $\boxed{\text{40cm}}$.