Question

11. The diameter of a circular park is 80 metres. A4 m wide road runs outside around it. Find the cost of constructing
the road at 10 per sq metre.​

Answer 1

Answer :-

Rs. 10560.

Solution :-

Diameter of the circular park i.e Inner diameter = 80 m

Radius of the circular park i.e Inner radius ( r ) = 80/2 = 40 m

A 4 m wide road runs around the circular park

i.e Thickness of road = 4 m

Outer radius ( R ) = 40 + 4 = 44

Area of the road = Area of the ring

= π * (R + r)(R - r)

= π * (44 + 40)(44 - 40)

= π * 84 * 4

= 22/7 * 336

= 22 * 48

= 1056 m²

Cost of constricting 1 m² = Rs. 10

Cos of constructing Area 1056 m² = 10 * 1056 = Rs. 10560

Therefore, the cost of constructing the road is Rs. 10560.

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

$\sf Given \begin{cases} \rm{Diameter \: of \: Park \: = \: 80 \: m} \\ \rm{Radius \: of \: Park \: (r) \: = \: 40 \: m} \\ \rm{Breadth \: of \: Road \: (B) \: = \: 4 \: m} \\ \rm{Cost \: of \: Construction \: = \: Rs. \: 10 \: per \: m^2} \\ \rm{Cost \: of \: Constitution \: = \: ?} \end{cases}$

$\implies {\sf{Outer \: Radius \: (R) \: = \: r \: + \: B}} \\ \\ \implies {\sf{Outer \: Radius \: = \: 40 \: + \: 4}} \\ \\ {\sf{R \: = \: 44 \: m}}$

And Area of Road :

$\large \implies {\sf{Area \: = \: \pi \: \times \: (R \: + \: r)(R \: - \: r) }} \\ \\ \implies {\sf{Area \: = \: \pi (44 \: + \: 40)(44 \: - \:4)}} \\ \\ \implies {\sf{Area \: = \: \pi(84)(4)}} \\ \\ \implies {\sf{Area \: = \: \pi \: \times \: 336}} \\ \\ \implies {\sf{Area \: = \: \dfrac{22}{7} \: \times \: 336}} \\ \\ \implies {\sf{Area \: = \: 22 \: \times \: 48}} \\ \\ \implies {\sf{Area \: = \: 1056}}$

Area is 1056 m²

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$\sf{Cost \: of \: Construction \: on \: 1m^2 \: = \: Rs. \: 10} \\ \\ \implies \sf{Cost \: of \: Construction \: at \: 1056 \: m^2 \: = \: 1056 \: \times \: 10} \\ \\ \implies {\sf{Cost \: = \: Rs. \: 10560 }}$