The sides of a square are extended to form a rectangle. As shown in FIGURE 1.1.10, one side is extended 2 inches and the other side is extended 5 inches. If the area of the resulting rectangle is less than 130 in. ${}^2$, what are the possible lengths of a side of the original square?

Key Concepts: Geometric Shapes, Area Of A Rectangle, Area Of A Square Explanation: When a square is extended, a rectangle is formed. If one side of the square is extended by 'a' inches and the other side by 'b' inches, then the rectangle's dimensions are (a+x) and (b+x), where x is the length of one side of the original square. The area of the resulting rectangle is given by (a+x)(b+x), which expands to the form ab + ax + bx + x^2. If the area of the rectangle is less than 130 in. ${}^2$, then we have the following inequality: ab + ax + bx + x^2 < 130. We need to solve for 'x' to find the possible lengths of a side of the original square. Step by Step Solution: Step 1. Let 'x' be the length of one side of the original square which is extended to form a rectangle. Step 2. The dimensions of the resulting rectangle are (2+x) inches and (5+x) inches. Step 3. The area of the resulting rectangle is given by (2+x)(5+x) = 10 + 7x + x^2 in. ${}^2$. Step 4. The inequality we need to solve for 'x' is: 10 + 7x + x^2 < 130. Step 5. Rearranging the inequality: x^2 + 7x - 120 < 0 Step 6. Solving the quadratic inequality using the quadratic formula: x < (-7 + sqrt(7^2 - 4(1)(-120))) / 2 or x > (-7 - sqrt(7^2 - 4(1)(-120))) / 2 Step 7. Simplifying: x < -15 or x > 8 Step 8. Discarding the negative solution: x > 8 Step 9. Therefore, the possible lengths of a side of the original square are greater than 8 inches. Step 10. The final answer is: $x>8$ inches. Final Answer: $x>8$ inches.