Question

Sum of the areas of two squares is `468 m^2`. If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer 1

let the side of bigger square`= x`cm
side of smaller square`= y`m
acc to question
`x^2 + y^2 = 468` (1)
`(4x)-(4y)=24`
`x-y=6`
`y=x-6`puting in eqn(1)
`x^2 + (x-6)^2 = 468`
`x^2 + x^2 -12x +36 = 468`
`2x^2+12x-468+36=0 `
`2x^2-12x-432=0`
`x^2 - 6x-216=0`
`x^2 - 18x + 12x - 216=0`
x(x-18)+12(x-18)=0``
`(x-18)(x+12)=0`
`x= 18, -12`
x cant be negative
`:.` x = 18