Calculate the length and width of a rectangle when you know its area \(A\) and the diagonal length \(L\). Let \(x\) be the length and \(y\) be the width. The basic formulas are:
\[ \text{Area: } A = x \cdot y \] \[ \text{Diagonal: } L = \sqrt{x^2 + y^2} \]
To solve for \(x\) and \(y\), first express \(y\) in terms of \(x\): \[ y = \frac{A}{x} \] Substitute into the diagonal formula: \[ L^2 = x^2 + \left(\frac{A}{x}\right)^2 \] Multiply both sides by \(x^2\) to obtain a quartic equation: \[ x^4 - L^2 x^2 + A^2 = 0 \] The discriminant is: \[ \Delta = L^4 - 4A^2 \] The rectangle exists if: \[ L \ge \sqrt{2A} \]
Suppose a rectangle has area \(A = 15\) and diagonal \(L = 12\). Compute \(x\) and \(y\): \[ \Delta = 12^4 - 4(15^2) = 20736 - 900 = 19836 \] \[ x = \sqrt{\frac{L^2 + \sqrt{\Delta}}{2}} = \sqrt{\frac{144 + \sqrt{19836}}{2}} \approx 11.066 \] \[ y = \frac{A}{x} = \frac{15}{11.066} \approx 1.356 \]
Enter the area \(A\) and diagonal \(L\) of the rectangle to find the dimensions.