Question

The diameter of a circular park is 42 m. A 3.5 m wide path lies along the periphery just outside the park . Find the cost of constructing a pavement on the path at the rate of $20per m²

Answer 1

STEP-BY-STEP EXPLANATION:

.

$\sf Diameter \: of \: Circular \: park = 42m$

$\sf Radius \: of \: Circular \: park = \frac{d}{2}$

$\longmapsto \sf Radius \: of \: Circular \: park = \frac {42m}{2}$

$\longmapsto \sf Radius \: of \: Circular \: park \: (r)= 21m$

$\sf Radius \: of \: Outer \: Circle = 21m + 3.5m$

$\longmapsto\sf Radius \: of \: Outer \: Circle \: (R)= 24.5m$

$\sf Area \: of \: Path = Area_{(Outer \: Circle)}-Area_{(Circular \: park)}$

$\longmapsto \sf Area \: of \: Path = \pi {R}^{2} - \pi {r}^{2}$

$\longmapsto \sf Area \: of \: Path = \pi( {R}^{2} - {r}^{2} )$

$\bf{[ {a}^{2} - {b}^{2} = (a - b)(a + b) ]}$

$\longmapsto \sf Area \: of \: Path = \pi( {R}- {r} ) ({R} + {r})$

$\longmapsto \sf Area \: of \: Path = \pi(24.5m - 21m ) (24.5m + 21m)$

$\longmapsto \sf Area \: of \: Path = \pi(3.5 ) (45.5m)$

$\longmapsto \sf Area \: of \: Path = \frac{22}{ \cancel{7}} ( \cancel{3.5 }) (45.5m)$

$\longmapsto \sf Area \: of \: Path = {22}(0.5m ) (45.5m)$

$\longmapsto \bf Area \: of \: Path = 500.5 {m}^{2}$

$\sf Cost \: of \: pavement \: of \: path = \pounds \: 20 \: per \: {m}^{2}$

$\longmapsto \sf Total \: Cost \: of \: pavement \: of \: path = \frac{\pounds \: 20} { \cancel{{1m}^{2}}} \times 500.5 \cancel{{m}^{2}}$

$\longmapsto \sf Total \: Cost \: of \: pavement \: of \: path = {\pounds \: 20} \times 500.5$

$\longmapsto \bf Total \: Cost \: of \: pavement \: of \: path = \pounds \: 10010$

REQUIRED ANSWER,

.

• $\bf Total \: Cost \: of \: pavement \: of \: path = \pounds \: 10010$

Answer 2

Answer:

Diagram :

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$\begin{gathered}\end{gathered}$

Given :

• The diameter of a circular park is 42 m.
• A 3.5 m wide path lies along the periphery just outside the park.

$\begin{gathered}\end{gathered}$

To Find :

• The cost of constructing a pavement on the path at the rate of $20per m².

$\begin{gathered}\end{gathered}$

Using Formulas :

$\longrightarrow\small{\underline{\boxed{\sf{Radius = \dfrac{Diameter}{2}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{Area \: of \: Inner \: circle = \pi{r}^{2}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{Area \: of \: outer\: circle = \pi{R}^{2}}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{Area_{(parth)} = Area_{(outer\: circle)} - Area_{(inner \: circle)}}}}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{Cost = Area \: of \: path \times constructing \: rate \: per \: {m}^{2}}}}}}}$

$\begin{gathered}\end{gathered}$

Solution :

⚘ Finding the area of park without path :-

$\longrightarrow{\sf{Radius = \dfrac{Diameter}{2}}}$

$\longrightarrow{\sf{Radius = \dfrac{42}{2}}}$

$\longrightarrow{\sf{Radius = \cancel\dfrac{42}{2}}}$

$\longrightarrow{ \underline{\boxed{\sf{\red{Radius = 21 \: m}}}}}$

∴ The radius of park without path is 21 m.

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⚘ Finding the area of park with path :-

$\longrightarrow{\sf{Radius = 21 + 3.5}}$

$\longrightarrow{\sf{Radius = 24.5 \: m}}$

$\longrightarrow{ \underline{\boxed{\sf{\red{Radius = 24.5\: m}}}}}$

∴ The area of park with path is 24.5 m.

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⚘ Finding area of Inner circle :-

$\longrightarrow{\sf{Area \: of \: Inner \: circle = \pi{r}^{2}}}$

${\longrightarrow{\sf{Area \: of \: Inner \: circle = \dfrac{22}{7} \times {(21)}^{2}}}}$

${\longrightarrow{\sf{Area \: of \: Inner \: circle = \dfrac{22}{7} \times {(21 \times 21)}}}}$

${\longrightarrow{\sf{Area \: of \: Inner \: circle = \dfrac{22}{7} \times {441}}}}$

${\longrightarrow{\sf{Area \: of \: Inner \: circle = \dfrac{22 \times 441}{7}}}}$

${\longrightarrow{\sf{Area \: of \: Inner \: circle = \dfrac{9702}{7}}}}$

${\longrightarrow{\sf{Area \: of \: Inner \: circle = \cancel{\dfrac{9702}{7}}}}}$

${\longrightarrow{\sf{Area \: of \: Inner \: circle = 1386 \: {m}^{2}}}}$

${\longrightarrow{ \underline{\boxed{\sf{\red{Area \: of \: Inner \: circle = 1386 \: {m}^{2}}}}}}}$

∴ The area of inner circle is 1386 m².

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⚘ Finding the area of outer circle :-

$\longrightarrow{\sf{Area \: of \: outer\: circle = \pi{R}^{2}}}$

${\longrightarrow{\sf{Area \: of \: outer\: circle = \dfrac{22}{7} \times {(24.5)}^{2}}}}$

${\longrightarrow{\sf{Area \: of \: outer\: circle = \dfrac{22}{7} \times {(24.5 \times 24.5)}}}}$

${\longrightarrow{\sf{Area \: of \: outer\: circle = \dfrac{22}{7} \times {600.25}}}}$

${\longrightarrow{\sf{Area \: of \: outer\: circle = \dfrac{22 \times 600.25}{7}}}}$

${\longrightarrow{\sf{Area \: of \: outer\: circle = \dfrac{13205.5}{7}}}}$

${\longrightarrow{\sf{Area \: of \: outer\: circle = \cancel{\dfrac{13205.5}{7}}}}}$

${\longrightarrow{\sf{Area \: of \: outer\: circle = 1886.5 \: {m}^{2}}}}$

${\longrightarrow{ \underline{\boxed{\sf{\red{Area \: of \: outer\: circle = 1886.5\: {m}^{2}}}}}}}$

∴ The area of outer circle is 1886.5 m².

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⚘ Finding the area of path :-

${\longrightarrow{\small{\underline{\boxed{\sf{Area_{(parth)} = Area_{(outer\: circle)} - Area_{(inner \: circle)}}}}}}}$

${\longrightarrow{\sf{Area_{(parth)} = 1886.5 - 1386}}}$

${\longrightarrow{\sf{Area_{(parth)} =500.5 \: {m}^{2}}}}$

${\longrightarrow{\underline{\boxed{\sf{\red{Area \: of \: parth =500.5 \: {m}^{2}}}}}}}$

∴ The area of path is 505.5 m².

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⚘ Now finding the cost of constructing a pavement on the path :-

${\longrightarrow{\sf{Cost = Area \: of \: path \times constructing \: rate \: per \: {m}^{2}}}}$

${\longrightarrow{\sf{Cost = 500.5 \times 20}}}$

${\longrightarrow{\sf{Cost = \ 10010}}}$

${\longrightarrow{\underline{\boxed{\sf{\red{Cost = \ 10010}}}}}}$

∴ The cost of constructing a pavement on the path is $10010.

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