1. **State the problem:** We need to find the area of a rectangle where the length is 5 inches longer than the width, and the perimeter is 34 inches. 2. **Define variables:** Let the width be $w$ inches. Then the length is $w + 5$ inches. 3. **Write the perimeter formula:** The perimeter $P$ of a rectangle is given by: $$P = 2(\text{length} + \text{width})$$ 4. **Substitute known values:** $$34 = 2((w + 5) + w)$$ 5. **Simplify inside the parentheses:** $$34 = 2(2w + 5)$$ 6. **Distribute the 2:** $$34 = 4w + 10$$ 7. **Isolate $w$:** $$34 - 10 = 4w$$ $$24 = 4w$$ 8. **Divide both sides by 4:** $$\cancel{4}w = \frac{24}{\cancel{4}}$$ $$w = 6$$ 9. **Find the length:** $$\text{length} = w + 5 = 6 + 5 = 11$$ 10. **Calculate the area:** Area $A$ is: $$A = \text{length} \times \text{width} = 11 \times 6 = 66$$ **Final answer:** The area of the rectangle is $66$ square inches.