Question

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There is a inner circular cemented ground of radius 7m. It is surrounded a grassy circular plot of radius 14 m. Finally, both of these are surrounded by plot of dimension 63 m × 22 m.
Find the ratio of areas of all these three.


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Answer 1

Answer :-

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

Here the concept of Area of Rectangle and Area of Circle has been used. Here firstly we will calculate the area of the Inner Circle. Then we will subtract it from the Area of Bigger circle to find the area occupied by bigger circle. Then, we will subtract the Area of Bigger circle from the Area of Rectangle.

Let's do it  !!

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★ Formula Used :-

$\: \\ \: \large{\boxed{\boxed{\sf{Area \: of \: Rectangle \: = \: \bf{Length(L) \: \times \: Breadth(B)}}}}}$

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

$\\ \large{\sf{Area \: occupied \: by \: Bigger \: Circular \: Part \: = \: \bf{Area \: of \: Bigger \: Circular \: - \: Area \: of \: Inner \: Circle}}}$

$\\ \large{\sf{Area \: occupied \: by \: Rectangular \: Part \: = \: \bf{Area \: of \: Rectangle \: - \: Area \: of \: Bigger \: Circle\: }}}$

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★ Solution :-

Given,

» Dimensions of Rectangular plot = 63 m × 22 m

» Radius of Bigger Circular plot = 14 m

» Radius of the smaller circular part = 7 m

~ For Area of Inner Circular Plot where r = 7  :-

$\: \\\qquad \: \large{\sf{:\Rightarrow \: \: \: Area \: of \: Circle_{(inner \: circular\: part)} \: = \: \bf{\pi \: \times \:r^{2}}}}$

$\: \\ \qquad \: \large{\sf{:\Rightarrow \: \: \: Area \: of \: Circle_{(inner \: circular\: part)} \: = \: \bf{\dfrac{22}{\cancel{7}} \: \times \: (7)^{\cancel{2}} \: \: = \: \: \underline{\underline{154 \: m^{2}}}}}}$

$\: \: \large{\boxed{\boxed{\tt{Area \:\; Occupied \:\; by \;\: Inner \:\; Circle \:\; = \: \bf{154 \: m^{2}}}}}}$

~ For Area occupied by Bigger Circular Plot :-

$\: \\ \qquad \: \sf{:\Rightarrow \: \: \: Area \: of \: Circle_{(bigger \: circular\: part)} \: = \: \bf{\pi \: \times \: r^{2}}}$

$\: \\ \qquad \: \sf{:\Rightarrow \: \: \: Area \: of \: Circle_{(bigger \: circular\: part)} \: = \: \dfrac{22}{7} \: \times \: (14)^{2} \: \: = \: \: 616 \: m^{2}}$

Then area occupied by Bigger Circular plot is given as :-

$\: \\ \: \large{\sf{\longrightarrow \: \: \: Area \: occupied \: by \: Bigger \: Circular \: Part \: = \: \bf{616 \: m^{2} \: - \: 154 \: m^{2} \: \: = \: \: \underline{\underline{462 \: m^{2}}}}}}$

$\: \: \large{\boxed{\boxed{\tt{Area \:\; Occupied \:\; by \;\: Outer \:\; Circle \:\; = \: \bf{462 \: m^{2}}}}}}$

~ For the Area occupied by Rectangular Plot  :-

$\: \\ \qquad \: \sf{:\Rightarrow \: \: \: Area \: of \: Rectangular \: Part \: = \: Length(L) \: \times \: Breadth(B)}$

$\: \\ \qquad \: \sf{:\Rightarrow \: \: \: Area \: of \: Rectangular \: Part \: = \: 63 \: m \: \times \: 22 \: m \: \: = \: \: 1386 \: m^{2}}$

Then, area occupied by rectangular portion :-

$\large{\sf{\longrightarrow \: \: \: Area \: occupied \: by \: Rectangle \: = \: \bf{Area \: of \: Rectangle\: - \: Area \: of \: Bigger \: Circle}}}$

$\: \\ \: \large{\sf{\longrightarrow \: \: \: Area \: occupied \: by \: Rectangular \: Part \: = \: \bf{1386 \: m^{2} \: - \: 616 \: m^{2} \: \: = \: \: \underline{\underline{770 \: m^{2}}}}}}$

$\: \: \large{\boxed{\boxed{\tt{Area \:\; Occupied \:\; by \;\: Rectangular \:\; Portion \:\; = \: \bf{770 \: m^{2}}}}}}$

~ For the Ratio of areas of all three plots :-

$\large{\sf{:\implies\:\:\:Ratio\:of\:all\:plots\:=\:\bf{Area\:of\:Inner\:Circle\::\:Area\:of\:Outer\:Circle\::\:Area\:of\:Rect.}}}$

$\: \\ \: \large{\sf{:\implies \: \: \: Ratio \: of \: areas \: of \: all \: three \: plots \: = \: \: \bf{154 \: m^{2}\:\: : \:\: 462 \: m^{2} \:\: : \:\: 770 \: m^{2}}}}$

Dividing all the terms by a common factor of 154 m², we get,

$\: \\ \: \large{\sf{:\implies \: \: \: Ratio \: of \: areas \: of \: all \: three \: plots \: = \: \: \bf{\cancel{154}\:\:(1)\:\: \cancel{m^{2}}\:\: : \:\: \cancel{462}\: \:(3) \:\: \cancel{m^{2}} \:\: : \:\: \cancel{770}\:\:(5)\:\: \cancel{m^{2}}}}}$

$\: \\ \:\large{\sf{:\implies\:\:\:Ratio\:of \: areas\:of\:all\:three\:plots\:=\:\: \bf{\underline{\underline{1\:\:\:\::\:\:\:3\:\:\:\::\:\:\:\:5}}}}}$

$\:\:\\\large{\underline{\underline{\sf{:\implies\:\:\:Ratio\:of\:areas\:of\:all\: three\:plots\: is \:\: \boxed{\bf{1 \:\:\:\::\:\:\:3\:\:\:\::\:\:\:\:5}}}}}}$

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$\: \: \qquad \large{\underbrace{\underbrace{\rm{Evaluation\:from\:the\:figure\: :-}}}}$

• DEFG is the rectangle.

Then, are occupied by DEFG will be :-

Area of DEFG - Area of Outer Circle

• There is a Outer Circle.

Area of outer circle is given as :-

Area of Outer Circle - Area of Inner Circle

Answer 2

Answer :

• Ratio of the areas of all the three figures is 1 : 3 : 5.

Explanation :

Given :

• Radius of the inner circle, r = 7 m.
• Radius of the outer fircle, R = 14 m.
• Dimensions of the plot = 63 m × 22 m.

To find :

• Ratio of the areas of all the three figures.

Knowledge required :

• Formula for area of a circle :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀A = πr²

Where,

⠀⠀⠀⠀⠀⠀⠀⠀⠀● A = Area of the circle.

⠀⠀⠀⠀⠀⠀⠀⠀⠀● r = Radius of the circle.

• Formula for area of the rectangle :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀A = lb

Where,

⠀⠀⠀⠀⠀⠀⠀⠀⠀● A = Area of the rectangle.

⠀⠀⠀⠀⠀⠀⠀⠀⠀● l = Length of the rectangle.

⠀⠀⠀⠀⠀⠀⠀⠀⠀● b = Breadth of the rectangle.

• The value of π is 22/7.

Solution :

• Area of the inner circle :

Let the area of the inner circle be α.

Now by using the formula area of a circle and substituting the values in it, we get :

⠀=> α = πr²

⠀=> α = 22/7 × 7²

⠀=> α = 22/7 × 49

⠀=> α = 22 × 7

⠀=> α = 154

⠀⠀⠀⠀⠀⠀⠀⠀⠀∴ α = 154 m²

• Area of the outer circle :

Let the area of the outer circle be β.

Now by using the formula area of a circle and substituting the values in it, we get :

⠀=> A = πr²

⠀=> β = 22/7 × 14²

⠀=> β = 22/7 × 196

⠀=> β = 22 × 28

⠀=> β = 616

∴ β = 616 m²

Since some of the area of the outer circle is occupied by the inner circle , we will consider the area of the outer circle without the inner circle.

Area of the outer circle excluding the inner circle :

⠀=> β = 616 - 154

⠀=> β = 462

⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀∴ β = 462 m²

• Area of the rectangular plot :

Let the area of the rectangular plot be γ.

Now by using the formula for area of a Rectangle and substituting the values in it, we get :

⠀=> A = lb

⠀=> γ = 63 × 22

⠀=> γ = 1386

⠀⠀⠀⠀⠀⠀⠀⠀⠀∴ β = 1386 m²

Since some of the area of the Rectangular plot is occupied by the outer, we will consider the area of the Rectangle without the outer circle.

Area of the Rectangular plot excluding the outer circle :

⠀=> γ = 1386 - 616

⠀=> γ = 770

⠀⠀⠀⠀⠀⠀⠀⠀⠀∴ β = 770 m²

Now we have to find ratio of the areas i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀α : β : γ

By substituting the value of α,β and γ, we get :

⠀=> α : β : γ

⠀=> α : β : γ = 154 : 462 : 770

⠀=> α : β : γ = 1 : 3 : 5

⠀ ∴ α : β : γ = 1 : 3 : 5

Therefore,

• Ratio of the areas of all the three figures is 1 : 3 : 5.