12 
Identity
Let the sides of first and second square be X and Y .
Area of first square = (X)²
And,
Area of second square = (Y)²
According to question,
(X)² + (Y)² = 468 m² ------------(1).
Perimeter of first square = 4 × X
and,
Perimeter of second square = 4 × Y
According to question,
4X - 4Y = 24 -----------(2)
From equation (2) we get,
4X - 4Y = 24
4(X-Y) = 24
X - Y = 24/4
X - Y = 6
X = 6+Y ---------(3)
Putting the value of X in equation (1)
(X)² + (Y)² = 468
(6+Y)² + (Y)² = 468
(6)² + (Y)² + 2 × 6 × Y + (Y)² = 468
36 + Y² + 12Y + Y² = 468
2Y² + 12Y - 468 +36 = 0
2Y² + 12Y -432 = 0
2( Y² + 6Y - 216) = 0
Y² + 6Y - 216 = 0
Y² + 18Y - 12Y -216 = 0
Y(Y+18) - 12(Y+18) = 0
(Y+18) (Y-12) = 0
(Y+18) = 0 Or (Y-12) = 0
Y = -18 OR Y = 12
Putting Y = 12 in EQUATION (3)
X = 6+Y = 6+12 = 18
Side of first square = X = 18 m
and,
Side of second square = Y = 12 m.
Ashram, Delhi 
Certified
