Question

Sum of the areas of two squares is 625 m\(^2\). If the difference of their perimeters is 20 m, find the sides of the two squares.

• A. 20 m, 5 m
• B. 15 m, 10 m
• C. 20 m, 15 m
• D. 25 m, 5 m

Hint:

When solving word problems involving squares, remember that the area of a square is \(side^2\) and the perimeter is \(4 \times side\).

Answer 1

The Correct Option is C

Let the sides of the two squares be \(x\) and \(y\). Then, the areas of the squares are \(x^2\) and \(y^2\), and we are given that: \[ x^2 + y^2 = 625 \] The perimeters of the squares are \(4x\) and \(4y\), and we are given that: \[ 4x - 4y = 20 \quad \Rightarrow \quad x - y = 5 \] Now, we have the system of equations: 1. \(x^2 + y^2 = 625\) 2. \(x - y = 5\) Solving these equations: From the second equation, \(x = y + 5\). Substituting this into the first equation: \[ (y + 5)^2 + y^2 = 625 \quad \Rightarrow \quad y^2 + 10y + 25 + y^2 = 625 \] \[ 2y^2 + 10y - 600 = 0 \quad \Rightarrow \quad y^2 + 5y - 300 = 0 \] Solving this quadratic equation gives \(y = 15\). Substituting \(y = 15\) into \(x = y + 5\), we get \(x = 20\). Thus, the sides of the two squares are 20 m and 15 m.