Basic relations:
\[ \text{Area: } A = x \cdot y \qquad (1) \qquad \text{Diagonal: } L^2 = x^2 + y^2 \qquad (2) \]
Solving (1) for \(y\) : \[y = A/x\] and substitute in (2) \[L^2 = x^2 + (A/x)^2\] Rewrite the above equation as \[x^4 - L^2 x^2 + A^2 = 0\] discriminant \(\Delta = L^4 - 4A^2\)
Existence condition: \(L \ge \sqrt{2A}\) (otherwise rectangle impossible)
Given \(A = 15\) , \(L = 12\) :
\[ \Delta = 12^4 - 4\cdot15^2 = 20736 - 900 = 19836 \] \[ x = \sqrt{\frac{L^2 + \sqrt{\Delta}}{2}} = \sqrt{\frac{144 + \sqrt{19836}}{2}} \approx 11.066 \] \[ y = \frac{A}{x} = \frac{15}{11.066} \approx 1.356 \]
→ length ≈ 11.066, width ≈ 1.356
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