Question

Sum of areas of two squares is 117 m\(^2\). If the difference between their perimeters is 12 m, then find the sides of both the squares.

Hint:

When solving problems involving squares, use the properties of areas and perimeters, and set up a system of equations to find the side lengths.

Answer 1

Let the sides of the two squares be \( a \) and \( b \), where \( a>b \). Step 1: Use the sum of areas. The area of a square is \( \text{side}^2 \). We are given that the sum of the areas of the squares is 117 m\(^2\): \[ a^2 + b^2 = 117 \quad \text{(Equation 1)}. \] Step 2: Use the difference of perimeters. The perimeter of a square is \( 4 \times \text{side} \). We are given that the difference between the perimeters is 12 m: \[ 4a - 4b = 12 \quad \implies \quad a - b = 3 \quad \text{(Equation 2)}. \] Step 3: Solve the system of equations. From Equation 2, we have: \[ a = b + 3. \] Substitute \( a = b + 3 \) into Equation 1: \[ (b + 3)^2 + b^2 = 117, \] \[ (b^2 + 6b + 9) + b^2 = 117, \] \[ 2b^2 + 6b + 9 = 117, \] \[ 2b^2 + 6b - 108 = 0. \] Divide the entire equation by 2: \[ b^2 + 3b - 54 = 0. \] Step 4: Solve the quadratic equation. The quadratic equation is \( b^2 + 3b - 54 = 0 \). Use the quadratic formula: \[ b = \frac{-3 \pm \sqrt{3^2 - 4(1)(-54)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 216}}{2} = \frac{-3 \pm \sqrt{225}}{2} = \frac{-3 \pm 15}{2}. \] Thus, the two solutions are: \[ b = \frac{-3 + 15}{2} = 6 \quad \text{or} \quad b = \frac{-3 - 15}{2} = -9. \] Since \( b \) is a side length, we take \( b = 6 \). Step 5: Find \( a \). Substitute \( b = 6 \) into Equation 2: \[ a = 6 + 3 = 9. \] Thus, the sides of the squares are \( a = 9 \) m and \( b = 6 \) m.
Conclusion:
The sides of the two squares are 9 m and 6 m.