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The sum of the areas of two squares is $640$ m$^2$. If the difference in their perimeters is $64$ m, find the sides of the two squares.
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Let the sides of the two squares be $x$ m and $y$ m $(x > y)$
$x^2 + y^2 = 640$ (I) (1 Mark)
and $4x – 4y = 64 \Rightarrow y = x - 16$ (II) (1 Mark)
$\therefore x^2 + (x – 16)^2 = 640$ (III) (1 Mark)
$\Rightarrow x^2 - 16x – 192 = 0$ (IV) (1 Mark)
$\Rightarrow (x – 24)(x + 8) = 0$
$\therefore x = 24$ (V) (1/2 Mark)
$x = -8$ (Rejected)
$\Rightarrow y = 24 – 16 = 8$ (VI) (1/2 Mark)
Hence the sides of the two squares are $24$ m and $8$ m
$x^2 + y^2 = 640$ (I) (1 Mark)
and $4x – 4y = 64 \Rightarrow y = x - 16$ (II) (1 Mark)
$\therefore x^2 + (x – 16)^2 = 640$ (III) (1 Mark)
$\Rightarrow x^2 - 16x – 192 = 0$ (IV) (1 Mark)
$\Rightarrow (x – 24)(x + 8) = 0$
$\therefore x = 24$ (V) (1/2 Mark)
$x = -8$ (Rejected)
$\Rightarrow y = 24 – 16 = 8$ (VI) (1/2 Mark)
Hence the sides of the two squares are $24$ m and $8$ m