Question

please solve fast as possible

Answer 1

it is ncert question
u can get there

Answer 2

Let the side of larger square be 'x' and that of smaller square be 'y'.

Thus,

${x}^{2} + {y}^{2} = 468 \\ 4x - 4y = 24$
Place the value of x from second equation into first.

$(6 + y)^{2} + {y}^{2} = 468 \\ {y}^{2} + 12y + 36 + {y}^{2} = 468 \\ 2 {y}^{2} + 12y + 36 = 468 \\ {y}^{2} + 6y + 18 = 234 \\ {y}^{2} + 6y -216 = 0 \\ (y - 12)(y + 18) = 0 \\As \: y\neq -18, y = 12$
The side of smaller square is 12cm.

Side of Larger square = 6+y
= 6+12=18cm

The side of larger square is 18cm.