Through a rectangular field of length 80 m and breadth 60 m ,two roads are constructed which are parallel to the sides and cut each other at right angles through the center of the field. If the width of each road is 3m, find:
a)The area covered by the roads
b)the cost of constructing the roads at the rate of Rs 90 per m sq.
c)Cost of leveling the field at the rate of Rs 90 per m sq.
Answers
This is a problem of area and length from mathematics.
In this case,
Area of road parallel to length =90 \mathrm {m} \times 3 \mathrm{m}=270 \mathrm {sq} \mathrm{m}=90m×3m=270sqm
Area of road parallel to breadth =60 m \times 3 m=180 \mathrm {sq} m.=60m×3m=180sqm.
But,
Four plots of length and breadth 3 m is counted
twice.
So, total area will be =270 \mathrm {sq} \mathrm{m}+180 \mathrm {sq} \mathrm{m}-3 \times 3 \mathrm {sq} \mathrm{m}=270sqm+180sqm−3×3sqm
=270 \mathrm {sq} m+180 \mathrm {sq} \mathrm{m}-9 \mathrm {sq} \mathrm{m}=270sqm+180sqm−9sqm
=441 \mathrm {sq} \mathrm{m}=441sqm
So, the area of pavement required to fill up the area of the road is
441 sq m.
Given:-
Length of Rectangular field = 90m
Breadth of Rectangular field = 60m
Width of road at centre = 3m.
Need To Find:-
Area of Roads
cost of constructing the roads at the rate of ₹110 per m²
Where:-
L denotes Length
B denotes Breadth
Explanation:-
[Take a look at Attachment] In this Question, We are provided with the length of Rectangular field that is 90m and Breadth that is 60m and there is small road at centre of Rectangular field of 3m that is cuts at right angles through centre of fields.
Solution:-
PQ = 3m ; QR = 60m
EF = 90m ; FG = 3m
KL = 3m
⠀⠀
Area of Rectangular Road = QR × PQ
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 60 × 3
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 180m²
Hence, Area of Rectangular road is 180m2
∴Hence,AreaofRectangularroadis180m2
⠀⠀
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Now, We have to find the area Rectangular road EFGH.
Area of Rectangular road EFGH = L × B
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ = 90 × 3
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 270m²
Hence, Area of Rectangular road is 270m^2
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∴Hence, Area of Rectangular road is 270m2