We solve word problems involving the side lengths and areas of figures using quadratic equations.
📘 Example ①
A rectangle has a length that is 3cm greater than its width and an area of 40cm². Find its width.
Answer:Let the width be x cm: x(x+3) = 40 → x² + 3x − 40 = 0 → (x+8)(x−5) = 0 → x = 5 (positive). Width = 5cm
A rectangle has a length that is 3cm greater than its width and an area of 40cm². Find its width.
Answer:Let the width be x cm: x(x+3) = 40 → x² + 3x − 40 = 0 → (x+8)(x−5) = 0 → x = 5 (positive). Width = 5cm
💡 Key Points
- Lengths are positive → discard negative solutions
- Area = length × width (for a rectangle)
Practice Problem
- A rectangle has a difference of 2cm between its length and width and an area of 35cm². Find the lengths of its sides.
Answer and Explanation
- Answer:5cm and 7cm. x(x+2)=35 → x²+2x−35=0 → (x+7)(x−5)=0 → x=5.