Through a rectangular field of length 90m and breadth 60m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3m, Find:
(i) the area covered by the roads
(ii) the cost of constructing the roads at the rate of ₹110 per m²
R.E.F. Image.
Given length of field = 90 m
width of field = 60 m.
Area of field =90×60
Area =5400m
2
length of road GEFH = 90 m.
length of road ABCD = 60 m.
width of each road = 3m [Given].
Area of roads =ar(ABCD)+ar.(GEFH)−ar(LIJK)
=90×3+60×3−3×3
=270+180−9
Area of roads =441m
2
.
∴ Area covered by roads =441m
2
.
cost of construction of roads =441×10
Total cost =4410Rs.
✲Question:-
❖Given:–
❅To find:–
✹Solution:-
[tex] \sf \: ↦Area \: of \: roads = Area \: of \: EFGH+ \: Area \: of \: IJKL-Area \: of \: PQRS[/tex]
EFGH is a rectangle with
EFGH is a rectangle withLength =3mBreadth = 60m↦Area of EFGH= length ×breadth
↦3m×60m
↦180m
Area of EFGH is 180m²
IJKL is a rectangle with
Area of IJKL=length ×breadth
↠90m×3m
↠270m
Area of IJKL is 270 m²
PQRS is a square with
Area of PQRS =side²
↠3m×3m
↠9m²
Area of PQRS is 9m²
↦Area of roads = Area of EFGH+ Area of IJKL-Area of PQRS
[tex]↠180 + 270 – 9[/tex]
[tex]↠450 – 9[/tex]
[tex]↠441 {m}^{2} [/tex]
[tex] \therefore \: Area \: covered \: by \: \: roads = 441 {m}^{2} [/tex]
2)the cost of constructing the roads at the rate of ₹110 per m
=> Cost of constructing 1m² road = Rs.110
[tex]\sf \: so \: cost \: of \: constructing \:441 {m }^{2} \: road = ₹110×441[/tex]
[tex] \implies \: ₹48,510[/tex]
∴The cost of constructing the road is Rs.48510
ㅤ✩。:•.───── ❁❁ ─────.•:。✩