Question

a circular grassy plot of Land of diameter 42 metre has a path of wide 3.5 running around it on the outside find the cost of travelling the path at rupees 4 per square metre​

Answer 1

Answer :-

$\: \: \\ \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Areas of Circles has been used. We are given that a grassy circular plot is surrounded by a path 3.5 wide running around it. If we can find out the radius from the diameter of the grassy circular plot, then we can add 3.5 m to it and find tha radius of whole circular land with path and plot. Then we can subtract, this from the area of circular plot and then find out the area of path. Finally we can multiply it with the rate.

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★ Formula Used :-

$\: \\ \large{\boxed{\boxed{\sf{Area \: \: of \: \: Circle \: \: = \: \: \bf{\pi r^{2}}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{Radius \: \: of \: \: Circle \: \: = \: \: \bf{\dfrac{Diameter}{2}}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{Area \: of \: the \: Path \: = \: \bf{Area \: of \: whole \: ground_{(plot \: + \: path)} \: - \: Area \: of \: Plot}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{Total \: cost \: Gravelling \: = \: \bf{Rate_{(in \: per \: m^{2})} \: \times \: Area \: of \: Path}}}}}$

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★ Question :-

A circular grassy plot of Land of diameter 42 metre has a path of wide 3.5 running around it on the outside. Find the cost of gravelling the path at rupees 4 per square metre.

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★ Solution :-

Given,

» Diameter of Grassy Plot = 42 m

» Width of the path = 3.5 m

» Rate of Gravelling = 4 per m²

Then, according to the question :-

~ Area of Circular Grassy Plot :-

$\: \\ \sf{\longrightarrow \: \: \: Radius \: \: of \: \: Circle, (r) \: \: = \: \bf{\dfrac{Diameter}{2}}}$

$\: \\ \sf{\longrightarrow \: \: \: Radius \: \: of \: \: Circle, (r) \: \: = \: \: \bf{\dfrac{42}{2} \: = \: \underline{21 \: m}}}$

Then,

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Area \: \: of \: \: Circle \: \: = \: \: \bf{\pi r^{2}}}}$

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Area \: \: of \: \: Circle \: \: \: = \: \: \bf{\dfrac{22}{7} \: \times \: (21 \: m)^{2}}}}$

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Area \: \: of \: \: Circle \: \: = \: \: \bf{22 \: \times \: 3 \: \times \: 21 \: \: m^{2} \: \: = \: \: \underline{1386 \: \: m^{2}}}}}$

$\: \\ \large{\boxed{\boxed{\tt{Area \: \: of \: \: Grassy \: \: Plot \: \: = \: \: \bf{1386 \: \: m^{2}}}}}}$

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~ Area of The Whole Land (Plot + Path) :-

❐ Radius of the whole land, r' = 21 m + 3.5 m

= 24.5 m = r'

(since, width of path is 3.5 m and radius of plot is 21 m)

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Area \: \: of \: \: Circle_{(for \: r')} \: \: = \: \: \bf{\dfrac{22}{7} \: \times \: (24.5 \: m)^{2}}}}$

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Area \: \: of \: \: Circle \: \: = \: \: \bf{22 \: \times \: 3.5 \: \times \: 24.5\: \: m^{2} \: \: = \: \: \underline{1886.5\: \: m^{2}}}}}$

$\: \\ \large{\boxed{\boxed{\tt{Area \: \: of \: \: Whole \: \: Land_{(plot \: + \: path)}\: \: = \: \: \bf{1886.5\: \: m^{2}}}}}}$

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~ For Area of the Path :-

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Area \: of \: the \: Path \: = \: \bf{Area \: of \: ground_{(plot \: + \: path)} \: - \: Area \: of \: Plot}}}$

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Area \: of \: the \: Path \: = \: \bf{1886.5 \; m^{2} \: - \: 1386 \; m^{2} \: \: = \: \: \underline{500.5 \: \: m^{2}}}}}$

$\: \\ \large{\boxed{\boxed{\tt{Area \: \: of \: \: Circular \: \: Path \: \: = \: \: \bf{500.5\: \: m^{2}}}}}}$

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~ Rate of Gravelling the Path :-

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Total \: cost \: Gravelling \: = \: \bf{Rate_{(in \: per \: m^{2})} \: \times \: Area \: of \: Path}}}$

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Total \: cost \: Gravelling \: = \: \bf{Rs. \: 4 \: per \: m^{2}\: \times \: 500.5 \; m^{2} \: \: = \: \: \underline{Rs.\: \: 2002}}}}$

$\: \\ \large{\boxed{\boxed{\tt{Total \: \: Cost \: \: of \: \: Gravelling \: \: the \: \: Path \: \: = \: \: \bf{Rs. \: 2002 \: \:}}}}}$

$\: \: \\ \large{\underline{\underline{\sf{\leadsto \: \: \: Thus, \: cost \: of \: Gravelling \: the \: path \: is \: \: \boxed{\bf{Rs. \: \: 2002}}}}}}$

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$\: \qquad \large{\underline{\underline{\rm{\mapsto \: \: \: \: Let's \: \: Understand \: \: More \: \: :-}}}}$

⌬ Area of Square = (Side)²

⌬ Area of Rectangle = Length × Breadth

⌬ Area of Triangle = ½ × Base × Height

⌬ Area of Parallelogram = Base × Height

⌬ Perimeter of Circle = 2πr

⌬ Perimeter of Square = 4 × Side

⌬ Perimeter of Rectangle = 2 × (Length + Breadth)

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Answer 2

Given,

A circular grassy plot of Land of diameter 42 metre has a path of wide 3.5 running around it on the outside.

To find :

Find the cost of gravelling the path at rupees 4 per square metre​.

Solution :

Diameter of circular land = 42 m

Radius of circular land = 42/2 m

Radius of circular land = 21 m

Now we know,

Area of circle = πr²

⇒ Area of land = 22/7 * 21²

⇒ Area of land = 22/7 * 21 * 21

⇒ Area of land = 22 * 3 * 21

⇒ Area of land = 1386 m²

Now with path, radius of land = 21 + 3.5 = 24.5 m

⇒ Area of land with path = 22/7 * (24.5)²

⇒ Area of land with path = 22/7 * 600.25

⇒ Area of land with path = 1886.5 m²

Now area of path :

⇒ Area of path = Area of land with path - Area of land

⇒ Area of path = 1886.5 - 1386

⇒ Area of path = 500.5 m²

Now cost of gravelling :

⇒ Cost of gravelling = Area * Rate

⇒ Cost of gravelling = 500.5 * 4

⇒ Cost of gravelling = Rs. 2002

Therefore,

Cost of gravelling the path = Rs. 2002