Sum of the areas of two squares is `260 m^(2)`. If the difference of their perimeters is 24 m then find the sides of the two squares.

Explanation

Let the sides of the two squares be a metres and b metres.
Then, their areas are `(a^(2))m^(2)" and "(b^(2))m^(2)` respectively.
And, their perimeters are (4a) m and (4b) m respectively.
`:." "4a-4b=24implies4(a-b)=24`
`impliesa-b=6impliesb=(a-6)`
Sum of their areas `=260 m^(2).`
`:." "a^(2)+b^(2)=260`
`implies" "a^(2)+(a-6)^(2)=260" "["using (i)"]`
` implies" "2a^(2)=12a-224=0impliesa^(2)-6a-112=0`
`implies" "a^(2)-14a+8a-112=0`
`implies" "a(a-14)+8(a-14)=0implies(a-14)(a+8)=0`
`implies" "a-14=0" or "a+8=0impliesa=14" or "a=-8`
`implies" "a=14" "[because" side of a square cannot be negative"].`
`:." "a=14" and "b=(14-6)=8.`
Hence, the sides of the square are 14 m and 8 m respectively.