Question

If the area of ​​a circle and the perimeter of the circle are equal then what is the radius of the circle?​

Answer 1

Given: If the area of a circle and the perimeter of the circle are equal.

To be Found: what is the radius of the circle?

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${ \underline{ \bf{ \bigstar \: According \: to \: the \: question : }}}$

$\: \: \: \: \: \: \: \dag \bigg( \tt \: perimeter _ {(circle)} = area _ {(circle)} \bigg)$

❒ Let the ratio of the perimeter of the circle and the area of the circle be 1 : 1

${ \underline{ \frak{ \dag \: As \: we \: know \: that : }}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \star \: { \underline{ \boxed{ \frak{Perimeter = 2\pi \: r}}}}$

$\star \: { \underline{ \boxed{ \frak{Area = \pi \: {r}^{2} }}}}$

${ \underline{ \bf{ \dag{Framing \: an \: equation \: we \: get}}}}$

${ : \implies} \sf \: \frac{\pi {r}^{2} }{2\pi \: r} = 1 : \: 1 \\ \\ \\ { : \implies} \sf \frac{ \cancel\pi {r}^{2} }{2 \cancel\pi \: r} = \frac{1}{1} \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf \frac{ \cancel{r}^{2} }{ 2{\cancel{r}}} = \frac{1}{1} \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf \frac{r}{2} = \frac{1}{1} \: \: \: \: \: \: \: \: \: \\ \\ \\{ : \implies} \sf r ( 1) = 2 ( 1 ) \\ \\ \\ { : \implies} \sf { \underline{ \boxed{ \frak{r = 2} } \star}} \: \: \: \: \: \: \:$

Therefore,

$\: \: \: \: \: \: \: \: \: \: \: { \underline{ \rm{ \star \: The \: radius \: of \: the \: circle \: is \: 2cm}}}$

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Additional information:

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