sum of the areas if two squares is 468m^2 . if the difference of their perimeters is 24m , find the sides if two squares
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let sides of two squares are x and y
a/q,x^2+y^2=468 ..........(1)
and 4x-4y=24
x-y=6
x=6+y
putting the value of x in equation 1
(6+y)^2+y^2=468
36+y^2+12y+y^2 =468
2y^2+12y-432=0
2y^2+36y-24y-432=0
2y(y+18)-24(y+18)=0
(2y-24)(y+18)=0
then 2y-24=0
y=12
and y=-18
side can not -ve so y=12 then x=18
a/q,x^2+y^2=468 ..........(1)
and 4x-4y=24
x-y=6
x=6+y
putting the value of x in equation 1
(6+y)^2+y^2=468
36+y^2+12y+y^2 =468
2y^2+12y-432=0
2y^2+36y-24y-432=0
2y(y+18)-24(y+18)=0
(2y-24)(y+18)=0
then 2y-24=0
y=12
and y=-18
side can not -ve so y=12 then x=18
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