Question

Question 11
Sum of the areas of two squares is 468$m^2$. If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer 1

Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively and their areas will be $x^{2}and{y}^2$ respectively. It is given that
4x - 4y = 24
x - y = 6
x = y + 6
$Also,x^2+y^2=468$
$\Rightarrow (6+y{)}^2+y^2=468$
$\Rightarrow 36+y^2+12y+y^2=468$
$\Rightarrow 2y^2+12y+432=0$
$\Rightarrow y^2+6y-216=0$
$\Rightarrow y^2+18y-12y-216=0$
$\Rightarrow y(y+18)-12(y+18)=0$
$\Rightarrow (y+18)(y-12)=0$
$\Rightarrow y=-18,12$
However, side of a square cannot be negative. Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.