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A rectangular field has 80 m length and 48 m breadth.three roads each of 2 m pass through the field such that two roads are parallel to the breadth and the

third road parallel to the length.find the area covered by the three roads



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Area of road 1(parallel to length) = 2×80=160
area of road 2 (parallel to breadth) = 2×48 = 96
area of road 3 (parallel to breadth) = 2×48 = 96
common areas(a and b)=2×(2×2) = 8m

area covered by road = 160+96+96-8=344 m²

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answer given is 348m^2
it can't be 348!! it should be 344.....there are 2 intersections...
Length of two roads are 48 m each
       Area of these two roads= 2*48*2= 192 m²
Length of one road is 80 m
       Area of this road = 80*2=160 m²
       Common area of these roads= 2*2*2= 8 m²
Area covered by three roads =192 + 160 - 8 = 344 m²
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