Two perpendicular cross roads of equal width run through the middle of a rectangular field of length 80 m and breadth 60 m. If the area of the cross roads is 675 square meters, then the width of each road is:

Let the width of each road be $w$ meters.

The horizontal road runs along the length of the field with dimensions 80 m × $w$ m.

Area of horizontal road $= 80w$ m²

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The vertical road runs along the breadth of the field with dimensions 60 m × $w$ m.

Area of vertical road $= 60w$ m²

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The two roads intersect at the middle, forming a square of side $w$ m.

Area of intersection $= w^2$ m²

When calculating the total area of the cross roads, the intersection is counted twice (once in each road), so it must be subtracted once.

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Total area of cross roads $= 80w + 60w - w^2$

Given that the total area is 675 m²:

$80w + 60w - w^2 = 675$

$140w - w^2 = 675$

$w^2 - 140w + 675 = 0$

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Using the quadratic formula:

$w = \dfrac{140 \pm \sqrt{140^2 - 4(1)(675)}}{2}$

$w = \dfrac{140 \pm \sqrt{19600 - 2700}}{2}$

$w = \dfrac{140 \pm \sqrt{16900}}{2}$

$w = \dfrac{140 \pm 130}{2}$

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Two solutions:

$w = \dfrac{140 + 130}{2} = \dfrac{270}{2} = 135$ m

$w = \dfrac{140 - 130}{2} = \dfrac{10}{2} = 5$ m

Since the width cannot be 135 m (larger than the field dimensions), the width of each road is 5 m.