Question

The sum of the areas of two square is 244 and the their difference between their perimeter is 8 cm.find the ratio of their diagonal

Answer 1

Let the side length of two squares be x and y respectively.
Given, Perimeter difference = 8
So
$4x - 4y = 8 \\ x - y = 2 \\ x = y + 2 - - - (1)$
Sum of areas = 244
So
${x}^{2} + {y}^{2} = 244 \\ substituting \: \: from \: \: (1) \\ {(y + 2)}^{2} + {y}^{2} = 244 \\ {y}^{2} + 2y - 120 = 0 \\ (y + 12)(y - 10) = 0 \\ considering \: \: positive \: \: value \: \\ y = 10 \\ from \: \: (1) \\ x = 10 + 2 = 12 \\ \\ ratio \: \: of \: \: diagonals = \sqrt{2} x : \sqrt{2} y \\ = x:y = 12:10 = 6:5$

Answer 2

The sides of the square are 10 cm and 12 cm

Let the sides of square be x and y cm.

Sum of their areas would be x² + y² = 244

The difference in their perimeter would be 4x-4y = 8

=> x-y = 2

x = y+2

Substituting this value of x in the first equation:

(y+2)² + y² = 244

2y² + 4 + 4y = 244

y² + 2y + 2 = 122

y² + 2y -120 = 0

y² -10y + 12 y-120 =0

y(y-10) +12(y-10) = 0

(y+12)(y-10) = 0

y = 10, -12

Discarding the negative value,

y =10

x = y+2, x = 12