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The side of a square exceeds the side of another square by $4$ cm and the sum of the areas of the two squares is $400$ cm$^2$. Find the sides of the squares.
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Let the side of first square = $x$ cm (1/2 Mark)
$\therefore$ Side of second square = $(x + 4)$ cm (1/2 Mark)
$x^2 + (x + 4)^2 = 400$ (1 Mark)
$x^2 + x^2 + 8x + 16 = 400 \Rightarrow 2x^2 + 8x - 384 = 0 \Rightarrow x^2 + 4x - 192 = 0$ (1 Mark)
$(x + 16)(x - 12) = 0$ (1/2 Mark)
$x = 12$ (as $x \neq -16$) (1/2 Mark)
Side of squares = $12$cm and $16$cm (1 Mark)
$\therefore$ Side of second square = $(x + 4)$ cm (1/2 Mark)
$x^2 + (x + 4)^2 = 400$ (1 Mark)
$x^2 + x^2 + 8x + 16 = 400 \Rightarrow 2x^2 + 8x - 384 = 0 \Rightarrow x^2 + 4x - 192 = 0$ (1 Mark)
$(x + 16)(x - 12) = 0$ (1/2 Mark)
$x = 12$ (as $x \neq -16$) (1/2 Mark)
Side of squares = $12$cm and $16$cm (1 Mark)