- Define variables \(s\) and \(L=s+5\) and use the area relationship \(s^2 + L^2 = 1025\).
- Simplify to the quadratic equation \(s^2 + 5s - 500 = 0\).
- Factor the equation as \((s+25)(s-20)=0\) and identify the valid physical solution.
- The side length of the smaller window is \(\boxed{20}\).
Explanation
- Problem Analysis and Setup
We are given two square windows with side lengths \(s\) and \(L\), respectively, where \(s\) is the side length of the smaller square and \(L\) is the side length of the larger square. We know that the side of the larger window is \(5 \text{ in.}\) longer than the side of the smaller window, which gives us the relation:
\[L = s + 5\]
The area of a square with side length \(x\) is given by \(x^2\). The total area of the two windows is given as \(1025 \text{ in.}^2\), leading to the following equation:
\[s^2 + L^2 = 1025\]
- Deriving the Quadratic Equation
Substitute the expression for \(L\) into the area equation:
\[s^2 + (s + 5)^2 = 1025\]
Expand the squared term \((s+5)^2\):
\[s^2 + (s^2 + 10s + 25) = 1025\]
\[2s^2 + 10s + 25 = 1025\]
Subtract 1025 from both sides to form a quadratic equation equal to zero:
\[2s^2 + 10s - 1000 = 0\]
Simplify the equation by dividing all terms by 2:
\[s^2 + 5s - 500 = 0\]
- Solving the Equation
To solve the quadratic equation \(s^2 + 5s - 500 = 0\), we look for two numbers that multiply to \(-500\) and add up to \(5\). These numbers are \(25\) and \(-20\). Thus, we can factor the quadratic as:
\[(s + 25)(s - 20) = 0\]
This gives us two possible values for \(s\):
\[s = -25 \quad \text{or} \quad s = 20\]
Since the side length of a square must be a positive value, we discard the negative solution. Therefore, the side length of the smaller window is \(20 \text{ in.}\).
- Conclusion and Verification
The side length of the smaller window is \(20 \text{ in.}\). We can verify this: if \(s = 20\), then \(L = 20 + 5 = 25\). The sum of their areas is \(20^2 + 25^2 = 400 + 625 = 1025 \text{ in.}^2\), which matches the given total area. The final answer is \(\boxed{20}\).
Examples
This problem demonstrates how to set up and solve quadratic equations based on geometric constraints. In real-life scenarios, this type of calculation is useful when you have limited material and need to optimize dimensions, such as deciding the sizes of two picture frames to fit within a specific total glass area, or when dividing a fixed amount of space between two separate projects. Understanding the relationship between sides and areas allows for precise planning in construction, design, and manufacturing.