Question

diameter of a circular field is 70 m. the area of the same field will be


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Answer 1

Answer:

Use curved surface area of Circle

Answer 2

Given : Diameter of Circular Feild is 70 m .

Need To Find : Area of Circular Field .

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❍ Finding Radius of of Circular Field for Area of Field using Formula is given by :

$\underline {\dag \frak{ As, \: We \:Know \:that \:}}$

⠀⠀⠀⠀⠀ $\implies {\underline {\sf{ Radius _{(Circle)} = \dfrac{d}{2} \:units .}}}$

⠀⠀⠀⠀⠀Here d is the Diameter of Circular Feild in metres.

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀ $\::\implies \sf{ Radius _{(Circle)} = \dfrac{\cancel {70}}{\cancel{2}}.}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  Radius = 35\: m}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore\:{ \mathrm {  Radius \:of\:Circular \:Field \:is\:35\: m}}}$

⠀⠀⠀⠀⠀ Finding Area of the Circular Field using Formula is given by :

$\underline {\dag \frak{ As, \: We \:Know \:that \:}}$

⠀⠀⠀⠀⠀ $\implies {\underline {\sf{ Area _{(Circle)} = \pi r^{2} \:sq.units .}}}$

⠀⠀⠀⠀⠀ Here r is the Radius of Circular Field in metres and $\pi =\dfrac{22}{7}$.

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀ $:\implies \sf{ Area _{(Circle)} = \dfrac{22}{7} \times 35^{2} }$

⠀⠀⠀⠀⠀ $:\implies \sf{ Area _{(Circle)} = \dfrac{22}{7} \times 35 \times 35 }$

⠀⠀⠀⠀⠀ $:\implies \sf{ Area _{(Circle)} = \dfrac{22}{\cancel {7}} \times \cancel {35}\times 35 }$

⠀⠀⠀⠀⠀ $:\implies \sf{ Area _{(Circle)} = 22 \times 5 \times 35 }$

⠀⠀⠀⠀⠀ $:\implies \sf{ Area _{(Circle)} = 110 \times 35 }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  Area = 3850\: m^{2}}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore\:{ \mathrm {  Area\:of\:Circular \:Field \:is\bf{\:3,850\: m^{2}}}}}$

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More To Know :

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$\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}$

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