sum of areas of two squares is 468m sq. If the difference of their perimeter is 24 m, find the sides of the two squares
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let X and Y are the side of square
according to question
X^2+Y^2=468 --------(1)
and difference in perimeter =4X-4Y=24
4 (X-Y)=24
X-Y=6
X=6+Y --------(2)
put (2) in equation (1)
now we find
(6+Y)^2+Y^2=468
Y^2+12Y+36+Y^2=468
2Y^2+12Y=432
Y^2+6Y-216=0
now solve this quadratic equation
Y=12,-18
but Y is distance so Y=-18 not possible
hence Y=12
and X=6+Y=18
according to question
X^2+Y^2=468 --------(1)
and difference in perimeter =4X-4Y=24
4 (X-Y)=24
X-Y=6
X=6+Y --------(2)
put (2) in equation (1)
now we find
(6+Y)^2+Y^2=468
Y^2+12Y+36+Y^2=468
2Y^2+12Y=432
Y^2+6Y-216=0
now solve this quadratic equation
Y=12,-18
but Y is distance so Y=-18 not possible
hence Y=12
and X=6+Y=18
abhi178:
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