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Sum of the areas of two squares is $468$ m$^2$. If the difference of their perimeters is $24$ m, find the lengths of the sides of the two squares.
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Sol. Let the lengths of the sides of the two squares be '$x$' m and '$y$' m s.t. $x > y$
A.T.Q.
$x^2 + y^2 = 468$ -----(1) (1/2 Mark)
$4x - 4y = 24$
$\Rightarrow x-y=6$ -----(2) (1/2 Mark)
From (1) and (2), we get
$y^2 + 6y - 216 = 0$
$\Rightarrow y = 12$ and $y = -18$ (1 Mark)
But side of a square is always positive,
So, $y = 12$
and $x = 18$ (1 Mark)
Hence, the lengths of the sides of two squares are $12$ m and $18$ m.
A.T.Q.
$x^2 + y^2 = 468$ -----(1) (1/2 Mark)
$4x - 4y = 24$
$\Rightarrow x-y=6$ -----(2) (1/2 Mark)
From (1) and (2), we get
$y^2 + 6y - 216 = 0$
$\Rightarrow y = 12$ and $y = -18$ (1 Mark)
But side of a square is always positive,
So, $y = 12$
and $x = 18$ (1 Mark)
Hence, the lengths of the sides of two squares are $12$ m and $18$ m.