A square field with side 30, m is surrounded | vedclass

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(D) Let $x$ (in metres) be the width of the path.The side of the inner square is $30\, m$.The side of the outer square is $(30 + 2x)\, m$.The area of

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$2$

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(D) Let $x$ (in metres) be the width of the path.The side of the inner square is $30\, m$.The side of the outer square is $(30 + 2x)\, m$.The area of the path is given by the difference between the area of the outer square and the area of the inner square.Area of path $= (30 + 2x)^2 - 30^2 = 256$.Expanding the equation: $(900 + 120x + 4x^2) - 900 = 256$.$4x^2 + 120x = 256$.Dividing by $4$: $x^2 + 30x = 64$.$x^2 + 30x - 64 = 0$.Factoring the quadratic equation: $(x + 32)(x - 2) = 0$.This gives $x = -32$ or $x = 2$.Since the width cannot be negative,we have $x = 2\, m$.

$A$ square field with side $30\, m$ is surrounded by a path of uniform width. If the area of the path is $256\, m^2$,then its width is ........ $m$.

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