Question

Sum of the areas of two squares is 468 m2 . If the difference of their perimeters is 24 m . Find the sides of the two squares.

Answer 1

Step-by-step explanation:

Given = areas of two squares

let the rea of first square =a'

the area of second square =a''

According to the question,

sum of their areas =a' + a" = 468 m (equation 1)

difference of their perimeter = 4a'-4a" = 24m (equation 2)

(perimeter = 4a)

divide the equation second by 2 we get,

2a' - 2a = 12, a' - a" = 6 (equation 3)

now add the two equations

a'+ a" = 468

+a' - a" = 6

= 2a' = 6

= a' = 3

now put the obtained value of a' in equation 2, we get

4a' + 4a" = 12

4*3 + 4a" = 12

12 + 4a" = 12

4a" = 0

a" = 0