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Mensuration - I (Perimeter and Area of Rectilinear Figures) — Maths STD 7 — Question

Maharashtra BoardEnglish MediumSTD 7MathsMensuration - I (Perimeter and Area of Rectilinear Figures)4 Marks
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 $PQRS =50 m \times 1.8 m=90 m^2$
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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