Question

Sum ols area of square 244 cm² and
diffrence their perimeter is 8cm
the ratio of their diagonal.
the
find​

Answer 1

Let the side length of two squares be x and y respectively.

Given, Perimeter difference = 8

So

$\sf {4x - 4y = 8}\\\\\sf {x - y = 2}\\\\\sf {x = y + 2 --- (1)}$

Sum of areas = 244  cm²

So

$\sf {x^{2} + y^{2} = 244}\\\\\sf {Substituting \: \: from \: \: (1)} \\\sf {(y + 2)}^{2} + {y}^{2} = 244 \\y^{2} + 2y - 120 = 0 \\\\(y + 12)(y - 10) = 0 \\ considering \: \: positive \: \: value \: \\ y = 10 \\ from \: \: (1) \\ x = 10 + 2 = 12 \\ \\ Ratio \: \: of \: \: diagonals = \sqrt{2} x : \sqrt{2} y \\ = x:y = 12:10 = 6:5$

Answer 2

Let the side length of two squares be x and y respectively.

Given, Perimeter difference = 8

So

\begin{gathered}\sf {4x - 4y = 8}\\\\\sf {x - y = 2}\\\\\sf {x = y + 2 --- (1)}\end{gathered}

Sum of areas = 244  cm²

So

\begin{gathered}\sf {x^{2} + y^{2} = 244}\\\\\sf {Substituting \: \: from \: \: (1)} \\\sf {(y + 2)}^{2} + {y}^{2} = 244 \\y^{2} + 2y - 120 = 0 \\\\(y + 12)(y - 10) = 0 \\ considering \: \: positive \: \: value \: \\ y = 10 \\ from \: \: (1) \\ x = 10 + 2 = 12 \\ \\ Ratio \: \: of \: \: diagonals = \sqrt{2} x : \sqrt{2} y \\ = x:y = 12:10 = 6:5\end{gathered}