A rectangular field is 50 m by 40 m. It has two roads through its centre, running parallel to its sides. The width of the longer and the shorter roads are 2 m and 2.5-m-respectively. Find the area of the roads and the area of the remaining portion of the field.

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Solution

Let ABCD be the rectangular field and KLMN and PQRS be the two rectangular roads with width 2.5 m and 2 m, respectively.

Length of the rectangular field CD = 50 cm
Breadth of the rectangular field BC = 40 m
∴ Area of the rectangular field ABCD = 50 m $\times$ 40 m = 2000 m²
Area of road KLMN = 40 m $\times$ 2.5 m = 100 m²
Area of road PQRS = 50 m $\times$ 2 m = 100 m²

Clearly, area of EFGH is common to both the two roads.

∴ Area of EFGH = 2.5 m $\times$ 2 m = 5 m²

∴ Area of the roads = Area (KLMN) + Area (PQRS) − Area (EFGH)
= (100 m² + 100 m²) − 5 m² = 195 m²

Area of the remaining portion of the field = Area of the rectangular field (ABCD) − Area of the roads
= (2000 − 195) m²
= 1805 m²

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