1. **State the problem:** Kim has 64 cm of wire to make a square and a rectangle. The square has side 6 cm. The rectangle's length is 4 cm more than its breadth. We need to find the difference between the areas of the square and the rectangle. 2. **Write down what is given:** - Perimeter of square wire used: $4 \times 6 = 24$ cm - Total wire length: 64 cm - Wire left for rectangle: $64 - 24 = 40$ cm 3. **Find the rectangle's dimensions:** Let the breadth be $b$ cm. Then length $l = b + 4$ cm. 4. **Use the perimeter formula for the rectangle:** $$2(l + b) = 40$$ Substitute $l = b + 4$: $$2(b + 4 + b) = 40$$ $$2(2b + 4) = 40$$ $$4b + 8 = 40$$ 5. **Solve for $b$:** $$4b = 40 - 8$$ $$4b = 32$$ $$b = \frac{32}{4}$$ $$b = 8$$ 6. **Find length $l$:** $$l = b + 4 = 8 + 4 = 12$$ 7. **Calculate areas:** - Area of square: $6 \times 6 = 36$ cm$^2$ - Area of rectangle: $l \times b = 12 \times 8 = 96$ cm$^2$ 8. **Find the difference between areas:** $$|96 - 36| = 60$$ **Final answer:** The difference between the areas of the square and the rectangle is 60 cm$^2$.