Question

sum of the area of two squares is 468m sq. If the difference of their perimeters is 24 m. find the side of two squares​

Answer 1

A-B=6

A²+B²=468

hence

A=18 and B=12.

Answer 2

Answer:

Let the side of the first square be 'a' m and that of the second be  

′ A′ m.

Area of the first square =a^2 sq m.

Area of the second square =A^2 sq m.

Their perimeters would be 4a and 4A respectively.

Given 4A−4a=24

A−a=6 --(1)

A^2 +a ^2 =468 --(2)

From (1), A=a+6

Substituting for A in (2), we get

(a+6)^2 +a^2 =468

a^2 +12a+36+a^2 =468

2a^2 +12a+36=468

a^2 +6a+18=234

a^2 +6a−216=0

a^2 +18a−12a−216=0

a(a+18)−12(a+18)=0

(a−12)(a+18)=0

a=12,−18

So, the side of the first square is 12 m. and the side of the second square is 18 m.