Question
A rectangular field is $50m$ by $40m$. It has two roads through its centre, running parallel to its sides. The widths of the longer and shorter roads are $1.8m$ and $2.5m$ respectively. Find the area of the roads and the area of the remaining portion of the field.

Answer


Let $A B C D$ be the rectangular field and $KLMN$ and $PQRS$ the two rectangular roads with width $1.8 m$ and $2.5 m$ , respectively.
Length of the rectangular field $C D=50 cm$ and breadth of the rectangular field $B C=40 m$
Area of the rectangular field $A B C D=50 m \times 40 m=2000 m^2$
Area of the road $KLMN =40 m \times 2.5 m=100 m^2$
Area of the road $P Q R S=50 m \times 1.8 m=90 m^2$
Clearly area of EFGH is common to the two roads.
Thus, Area of $EFGH =2.5 m \times 1.8 m=4.5 m^2$
Hence, Area of the roads $=$ Area $($ KLMN $)+$ Area(PQRS) - Area(EFGH)
$=\left(100 m^2+90 m^2\right)-4.5 m^2$
$=185.5 m^2$
Area of the remaining portion of the field $=$ Area of the rectangular field $A B C D-$ Area of the roads
$=(2000-185.5) m^2$
$=1814.5 m^2$
 

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