Question

What will be the area of a circle whose diameter is 1.8 m and value of π=3.14?

Answer 1

Answer: area =  2.5434 m²

Step-by-step explanation:

Answer 2

$\frak{ \dag \: Given }\begin{cases} \frak{Diameter \: of \: a \: circle = 1.8m} \\ \\ \\ \frak{Value \: of \: \pi = 3.14} \end{cases}$

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To find: Area of circle?

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Here, In this question We are asked to find the area of circle.

In order to calculate the area, Firstly we will find radius and then we will put it in "Area of circle" formula. i.e πr².

$\begin{gathered}\dag\;{\underline{\frak{As\;we\;know\;that,}}}\\\end{gathered}$

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$\begin{gathered}\star\;{\boxed{\sf{\pink{Radius_{\;(circle)} = \dfrac{Diameter}{2}}}}}\\\end{gathered}$

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$\begin{gathered}:\implies\sf \frac{1.8}{2} \\ \\ \\ :\implies\sf 0.9m\\ \\ \\:\implies{\underline{\boxed{\frak{\purple{r = 0.9\:m}}}}}\;\bigstar\\ \end{gathered}$

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$\therefore\:{\underline{\sf{Radius\:of\:circle\:is\: {\textsf{\textbf{0.9\:cm}}}.}}}$

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$\begin{gathered}\dag\;{\underline{\frak{Now,\:Finding\:area\:of\:circle,}}}\\\end{gathered}$

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$\begin{gathered}\star\;{\boxed{\sf{\pink{Area_{\;(circle)} = \pi r^2 }}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf Area_{\;(Circle)} = 3.14 \times 0.9 \times 0.9\\ \\ \\ :\implies\sf Area_{\;(Circle)} = 3.14\times 0.81\\ \\ \\:\implies{\underline{\boxed{\frak{\purple{Area_{\;(Circle)} = 2.5434\:cm^2}}}}}\;\bigstar\\ \\\end{gathered}$

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$\therefore\:{\underline{\sf{Area\:of\:circle\:is\: \bf{2.5434\:cm^2}.}}}$

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$\begin{gathered}\qquad\qquad\boxed{\underline{\underline{\pink{\bigstar \: \bf\:More\:to\:know\:\bigstar}}}} \\\end{gathered}$

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$\begin{gathered}\boxed{\begin{minipage}{6.2 cm}\bigstar \:\underline{\textbf{Formulae Related to Cylinder :}}\\\\\sf {\textcircled{\footnotesize\textsf{1}}} \:Area\:of\:Base\:and\:top =\pi r^2 \\\\ \sf {\textcircled{\footnotesize\textsf{2}}} \:\:Curved \: Surface \: Area =2 \pi rh\\\\\sf{\textcircled{\footnotesize\textsf{3}}} \:\:Total \: Surface \: Area = 2 \pi r(h + r)\\ \\{\textcircled{\footnotesize\textsf{4}}} \: \:Volume=\pi r^2h\end{minipage}}\end{gathered}$