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Mensuration – I (Perimeter and Area of Rectilinear Figures) — MATHS STD 7 — Question

Gujarat BoardEnglish MediumSTD 7MATHSMensuration – I (Perimeter and Area of Rectilinear Figures)5 Marks
Question
Two cross roads, each of width $5m$, run at right angles through the centre of a rectangular park of length $70m$ and breadth $45m$ parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at the rate of $Rs. 105$ per $m^2$.
✓

Answer


Let $A B C D$ be the rectangular park then $EFGH$ and $IJKL$ the two rectangular roads with width $5 m .$
Length of the rectangular park $A D=70 \mathrm{~cm}$
Breadth of the rectangular park $C D=45 \mathrm{~m}$
Area of the rectangular park $=$ Length $\times$ Breadth $=70 \mathrm{~m} \times 45 \mathrm{~m}=3150 \mathrm{~m}^2$
Area of the road $\mathrm{EFGH}=70 \mathrm{~m} \times 5 \mathrm{~m}=350 \mathrm{~m}^2$
Area of the road $\mathrm{JKIL}=45 \mathrm{~m} \times 5 \mathrm{~m}=225 \mathrm{~m}^2$
Clearly area of MNOP is common to the two roads.
Thus, Area of MNOP $=5 \mathrm{~m} \times 5 \mathrm{~m}=25 \mathrm{~m}^2$
Hence,
Area of the roads $=$ Area $(\mathrm{EFGH})+$ Area(JKIL) - Area(MNOP)
$=(350+225) m^2-25 m^2=550 m^2$
Again, it is given that the cost of constructing the roads $=$ Rs. 105 per $m^2$
Therefore,
Cost of constructing $550 \mathrm{~m}^2$ area of the roads $=$ Rs. $(105 \times 550)$
$=\text { Rs. } 57750$

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