Explanation:

Given:-

\SF \: Diameter = 75 \: mDiameter=75m

\sf \: Radius \: = 37.5 \: mRadius=37.5m

\sf \: Outer \: radius = 37.5 + 3.5 = 41 \: mOuterradius=37.5+3.5=41m

Solution:-

{ \underline{ \boxed{ \sf{Outer \: AREA \: from \: center =\pi {r}^{2} }}}} < /p > < p >

Outerareafromcenter=πr

2

\bullet= \sf \: 3.14 \times {41}^{2}∙=3.14×41

2

\bullet \sf \: = 5278.34 \: {m}^{2}∙=5278.34m

2

{ \underline{ \boxed{ \sf{Inner \: area \: = \: \pi {r}^{2} }}}}

Innerarea=πr

2

\bullet = \sf \: 3.14 \times{37.5}^{2}∙=3.14×37.5

2

\bullet\sf \: = \: 4,415.625 \: m {}^{2}∙=4,415.625m

2

{ \underline{ \boxed{ \sf{Area \: of \: round = Outer \: area \: - \: Inner \:area}}}}

Areaofround=Outerarea−Innerarea

\bullet \sf \: = (5278.34 ) - (4,415.625)∙=(5278.34)−(4,415.625)

\bullet \sf \: = 862.715 \: {m}^{2}∙=862.715m

2

{ \underline{ \boxed{ \sf{Cost \: of \: CONSTRUCTION \: = Rs.2.00 \: {m}^{2} }}}}

Costofconstruction=Rs.2.00m

2

{ \underline{ \boxed{ \sf{cost \: of \: construction \: = Rs.2.00 \times 862.715}}}}

costofconstruction=Rs.2.00×862.715

\bullet \sf \: = Rs. \: 1,725.43∙=Rs.1,725.43

The cost of constructing road is Rs.

1,725.43..