Question

if the areas of two circles are in the ratio 4 ratio 9 then the ratio of the perimeter of the semicircle is​

Answer 1

Answer:

I hope this is the right answer

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that

• Radius of first circle be x units

• Radius of second circle be y units

We know that

Area of circle of radius 'r' is given by

$\red{\rm :\longmapsto\:\boxed{\tt{ \: \: Area_{Circle} \: = \: \pi \: {r}^{2} \: \: }}}$

So, According to statement

$\rm :\longmapsto\:Area_{first \: Circle} : Area_{second \: Circle} = 4 : 9$

$\rm :\longmapsto\:\dfrac{Area_{first \: Circle}}{Area_{second \: Circle}} = \dfrac{4}{9}$

$\rm :\longmapsto\:\dfrac{\pi \: {x}^{2} }{\pi \: {y}^{2} } = \dfrac{4}{9}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {y}^{2} } = \dfrac{ {2}^{2} }{ {3}^{2} }$

$\bf :\longmapsto\:\dfrac{x}{y} = \dfrac{2}{3} - - - - (1)$

Now, we know that

Perimeter of a circle of radius 'r' is given by

$\rm :\longmapsto\:\boxed{\tt{ \: \: Perimeter_{Circle} \: = \: 2 \: \pi \: r \: \: }}$

So,

$\rm :\longmapsto\:Perimeter_{first \: Circle} : Perimeter_{second \: Circle}$

$\rm \:  =  \: \dfrac{Perimeter_{first \: Circle}}{Perimeter_{second \: Circle}}$

$\rm \:  =  \: \dfrac{2 \: \pi \: x}{2 \: \pi \: y}$

$\rm \:  =  \: \dfrac{ \: x}{ \: y}$

$\rm \:  =  \: \dfrac{ \: 2 \: }{ \: 3 \: }$

$\rm \:  =  \: 2 : 3$

Hence,

$\red{\boxed{\tt{ Perimeter_{first \: Circle} : Perimeter_{second \: Circle} = 2 : 3 \: }}}$

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