Question

Sum of the areas of two squares is 468m.if the diffrences of their perimetre is 24m,find the sides of two squares

Answer 1

I'll go step by step so that it's easy for you to understand :

Let the length of side of smaller square be x m
      so,perimeter of smaller square=4x m
Now, perimeter of the bigger square is 24m MORE THAN the perimeter of the smaller one
       so,perimeter of bigger square=(4x+24)m
Length of larger side of the bigger square=[4x+24/4]m = (x+6)m   {side=p/4}

We already know that the sum of areas of those two squares is 468 m^2 
=(x+6)^2+x^2=468
={x^2+12x+36}+x^2=468
=2{x^2+6x+18}=468
=x^2+6x+18=468/2=234
=x^2+6x+18-234=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)=0 or (x-12)=0
x=-18 or x=12

The measure of side of square is NON-NEGATIVE quantity
Thus, measure of side of smaller square=12m and the bigger one=18m(=12m+6m)

Hope I helped you!!..ask me if you have any problem in any of those steps..!