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The total area of two square windows is 1,025 in.[tex]\( ^2 \)[/tex]. Each side of the larger window is 5 in. longer than the sides of the smaller window. How long are the sides of the smaller window?

[tex]\[ A = s^2 \][/tex]

A. 13.5 in.
B. 20 in.
C. 25 in.
D. 31.6 in.

Sagot :

Let's denote the side length of the smaller window as [tex]\( s \)[/tex].

### Step-by-Step Solution:

1. Determine the relationship between the sizes:
- Side length of the smaller window: [tex]\( s \)[/tex]
- Side length of the larger window: [tex]\( s + 5 \)[/tex]

2. Calculate the area of each window:
- Area of the smaller window: [tex]\( s^2 \)[/tex]
- Area of the larger window: [tex]\( (s + 5)^2 \)[/tex]

3. Express the total area of the two windows:
- According to the problem, the total area is 1,025 square inches:
[tex]\[ s^2 + (s + 5)^2 = 1025 \][/tex]

4. Expand the equation:
- Expand [tex]\( (s + 5)^2 \)[/tex]:
[tex]\[ (s + 5)^2 = s^2 + 10s + 25 \][/tex]

5. Substitute and combine like terms:
- Substitute into the equation and combine:
[tex]\[ s^2 + s^2 + 10s + 25 = 1025 \][/tex]
[tex]\[ 2s^2 + 10s + 25 = 1025 \][/tex]

6. Simplify the equation:
- Subtract 1025 from both sides:
[tex]\[ 2s^2 + 10s + 25 - 1025 = 0 \][/tex]
[tex]\[ 2s^2 + 10s - 1000 = 0 \][/tex]

7. Divide the equation by 2:
- To simplify further, divide everything by 2:
[tex]\[ s^2 + 5s - 500 = 0 \][/tex]

8. Solve the quadratic equation using the quadratic formula [tex]\( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
- In this equation, [tex]\( a = 1 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = -500 \)[/tex]:
[tex]\[ s = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-500)}}{2 \cdot 1} \][/tex]
[tex]\[ s = \frac{-5 \pm \sqrt{25 + 2000}}{2} \][/tex]
[tex]\[ s = \frac{-5 \pm \sqrt{2025}}{2} \][/tex]
[tex]\[ s = \frac{-5 \pm 45}{2} \][/tex]

9. Evaluate the solutions:
- We get two potential solutions:
[tex]\[ s = \frac{-5 + 45}{2} = \frac{40}{2} = 20 \][/tex]
[tex]\[ s = \frac{-5 - 45}{2} = \frac{-50}{2} = -25 \][/tex]

10. Select the positive solution:
- Since the side length cannot be negative, the solution is:
[tex]\[ s = 20 \][/tex]

Therefore, the sides of the smaller window are 20 inches long.

So, the correct answer is:
[tex]\[ 20 \text{ in.} \][/tex]