A rectangular park 60 , m long and 40 , m wid | vedclass

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(B) Let the width of the road be $x \, m$.Total area of the park $= 60 	imes 40 = 2400 \, m^2$.Area of the lawn $= 2109 \, m^2$.Area of the road $= \

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(B) Let the width of the road be $x \, m$.Total area of the park $= 60 	imes 40 = 2400 \, m^2$.Area of the lawn $= 2109 \, m^2$.Area of the road $= 	ext{Total area} - 	ext{Area of the lawn} = 2400 - 2109 = 291 \, m^2$.The area of the two crossroads is given by the formula: $x(L + W - x) = 	ext{Area of road}$, where $L = 60 \, m$ and $W = 40 \, m$.$x(60 + 40 - x) = 291$$x(100 - x) = 291$$100x - x^2 = 291$$x^2 - 100x + 291 = 0$Solving the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{100 \pm \sqrt{(-100)^2 - 4(1)(291)}}{2(1)}$$x = \frac{100 \pm \sqrt{10000 - 1164}}{2}$$x = \frac{100 \pm \sqrt{8836}}{2}$$x = \frac{100 \pm 94}{2}$$x_1 = \frac{194}{2} = 97$ (not possible as width cannot exceed park dimensions)$x_2 = \frac{6}{2} = 3$Therefore, the width of the road is $3 \, m$.

$A$ rectangular park $60 \, m$ long and $40 \, m$ wide has two concrete crossroads running in the middle of the park, and the rest of the park has been used as a lawn. If the area of the lawn is $2109 \, m^2$, then what is the width of the road? (in $m$)

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