Stacey has a square piece of cloth. She cuts 3 inches off of the length of the square and 3 inches off of the width. The area of the smaller square is $\frac{1}{4}$ the area of the original square. What was the side length of the original square?
A. 1 in.
B. 2 in.
C. 6 in.
D. 12 in.
Answer (2)
Let s be the side length of the original square, then the area of the original square is s 2 .
The side length of the smaller square is s − 3 , and its area is ( s − 3 ) 2 .
Set up the equation ( s − 3 ) 2 = 4 1 s 2 and simplify it to s 2 − 8 s + 12 = 0 .
Solve the quadratic equation to find s = 2 or s = 6 . Since 3"> s > 3 , the side length of the original square is 6 inches.
Explanation
Define the variable Let s be the side length of the original square. The area of the original square is s 2 .
Find the side length and area of the smaller square The side length of the smaller square is s − 3 . The area of the smaller square is ( s − 3 ) 2 .
Set up the equation The area of the smaller square is 4 1 the area of the original square, so we have the equation ( s − 3 ) 2 = 4 1 s 2 .
Expand and simplify the equation Expanding the left side, we get s 2 − 6 s + 9 = 4 1 s 2 . Multiplying both sides by 4, we have 4 s 2 − 24 s + 36 = s 2 .
Simplify the quadratic equation Rearranging the terms, we get 3 s 2 − 24 s + 36 = 0 . Dividing by 3, we have s 2 − 8 s + 12 = 0 .
Solve the quadratic equation We can factor the quadratic equation as ( s − 2 ) ( s − 6 ) = 0 . Thus, the possible values for s are s = 2 or s = 6 .
Determine the correct solution However, since we cut 3 inches off each side, the side length of the original square must be greater than 3. Therefore, s = 2 is not a valid solution. So, the side length of the original square is s = 6 inches.
Verify the solution We can check our answer: If the original square has side length 6, its area is 6 2 = 36 . The smaller square has side length 6 − 3 = 3 , so its area is 3 2 = 9 . Since 9 = 4 1 × 36 , our answer is correct.
Examples
Understanding how areas change when dimensions are altered is crucial in many real-world applications. For example, architects use these principles when designing buildings to ensure that scaling dimensions proportionally affects the usable space as intended. Similarly, landscapers apply this knowledge when planning gardens or patios, adjusting sizes to meet specific area requirements while maintaining aesthetic balance. This problem demonstrates a fundamental concept in geometry that directly impacts design and spatial planning.
The side length of the original square is 6 inches, as confirmed by solving the quadratic equation formed by the relationship between the areas of the original and smaller squares and ensuring the length is valid after cutting. The valid solution to the quadratic equation yields 6 inches while also verifying that this dimension satisfies the condition set by the problem.
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