Twenty-nine times the area of a square is one square metre less than six times the area of the second square and nine times its side exceeds the perimeter of other square by 1 metre. The difference in the sides of these squares is (M.A.T., 2010)

### Concept & Algebraic Substitution Translate the word problem into a system of two equations. Use substitution to solve the resulting quadratic equation. Let the side of the first square be $x$ and the second square be $y$. $$ \text{Area} = \text{side}^2, \text{Perimeter} = 4 \times \text{side} $$ ### Step-by-Step Solution 1. Let the side of the first square be $x$ and the second square be $y$. 2. "Twenty-nine times the area of a square is one less than six times the area of the second": $29x^2 = 6y^2 - 1$ (Equation 1) 3. "Nine times its side exceeds the perimeter of other square by 1": $9x = 4y + 1$ (Equation 2) 4. Solve Equation 2 for $y$: $4y = 9x - 1 \Rightarrow y = \frac{9x - 1}{4}$ 5. Substitute $y$ into Equation 1: $29x^2 = 6\left(\frac{9x - 1}{4}\right)^2 - 1$ $29x^2 = \frac{6(81x^2 - 18x + 1)}{16} - 1$ $29x^2 = \frac{3(81x^2 - 18x + 1)}{8} - 1$ 6. Multiply the entire equation by 8 to remove the fraction: $232x^2 = 3(81x^2 - 18x + 1) - 8$ $232x^2 = 243x^2 - 54x + 3 - 8$ $232x^2 = 243x^2 - 54x - 5$ 7. Rearrange into a standard quadratic equation: $11x^2 - 54x - 5 = 0$ 8. Factor the quadratic equation: $11x^2 - 55x + x - 5 = 0$ $11x(x - 5) + 1(x - 5) = 0$ $(11x + 1)(x - 5) = 0$ 9. Since a side length must be positive, $x = 5$ m. 10. Find $y$ using $y = \frac{9(5) - 1}{4} = \frac{44}{4} = 11$ m. 11. The difference in their sides is $y - x = 11 - 5 = 6$ m. ### Exam Strategy & Shortcut Because solving quadratics with large coefficients is tedious, check integer answers. If $x=5$, $y=11$. Difference = 6. Test it in eq1: $29(25) = 725$. $6(121) - 1 = 726 - 1 = 725$. It balances perfectly. Guessing and checking logical integer roots based on the options (like diff = 6) can save minutes. ### Common Pitfall A standard trap is misinterpreting "perimeter of other square" as just the side of the other square, leading to $9x = y + 1$ which creates an unsolvable quadratic mess. ### Final Answer Therefore, the correct answer is **6 m**.