Math, asked by varshinihanagal, 3 months ago

The areas of two circles are in ratio a 9:4, then
what is ratio of their circumferences ?
I need step by step calculation.​

Answers

Answered by snehitha2
16

Answer:

The required ratio of their circumferences is 3 : 2

Step-by-step explanation:

Given :

The areas of two circles are in the ratio of 9 : 4

To find :

the ratio of their circumferences

Solution :

Let r₁ and r₂ be the radii of the two circles.

Area of a circle of radius r = πr²

So,

area of first circle = πr₁²

area of second circle = πr₂²

Their ratio = 9 : 4

\tt \dfrac{\pi r_1 ^2}{\pi r_2 ^2} = \dfrac{9}{4} \\ \tt \dfrac{r_1 ^2}{r_2 ^2} = \dfrac{9}{4} \\ \tt \bigg( \dfrac{r_1}{r_2}\bigg) ^2 = \dfrac{9}{4} \\ \tt \dfrac{r_1}{r_2} = \sqrt{\dfrac{9}{4}} \\ \tt \dfrac{r_1}{r_2} = \sqrt{\dfrac{3^2}{2^2}} \\ \tt \dfrac{r_1}{r_2} = \dfrac{3}{2}

Circumference of a circle is given by,

C = 2πr

The ratio of their circumferences :

= 2πr₁ : 2πr₂

= r₁ : r₂

= 3 : 2

Answered by Ridvisha
35

{ \underline{ \underline{ \huge{ \red{ \rm{Question}}}}}}

The areas of two circles are in ratio of 9:4 , then what is the ratio of their circumferences ?

{ \underline{ \underline{ \huge{ \red{ \rm{ Solution}}}}}}

{ \underline{ \underline{ \tt{ \orange{given} \:  : -  }}}}

{ \dag{   \sf { \pink{ \:  \:  \: ratio \: of \: area \: of \: two \: circles \:  = 9 : 4}}}}

{ \underline{ \underline{ \tt{ \orange{ \: to \: find \: } : - }}}}

{ \dag{ \sf{   \:  \:  \:  ratio \: of \: their \: circumferences \: }}}

{ \underline{ \underline{ \tt{ \orange{ formula \: used} \:   :  - }}}}

{ \dashrightarrow{ \boxed{ \green{ \sf{area \: of \: circle = \pi {r}^{2} }}}}}\\ { \dashrightarrow{ \boxed { \green{ \sf{circumference \: of \: circle = 2\pi \times r}}}}}

{ \rm{ \leadsto{let \: the \: area \: of \: circle \: 1 \: be \: 9x \: {unit}^{2} }}} \\ { \rm{ \:  \:  \:  \:  \:  \:  and \: that \: of \: circle \: 2 \: be \: 4x {unit}^{2} }}

{ \boxed{ \underline{ \pink{ \tt{ circle \: 1}}}}}

{ \sf{ :{\implies { \blue{ \: radius 1 (r1)\: }}}}} \\ \\  \:  \:  \:  \:  \:\: { \rightarrow { \blue{ \sf{  \: \: area1 = 9x = \pi {r1}^{2} }}}} \\  \:  \:  \:  \:  \:  \:  { \rightarrow{ \sf{ \blue{ \:  \:  {r1}^{2}  =  \frac{9x}{\pi} }}}} \\  \:  \:  \:  \:  \:  \: { \rightarrow{ \sf{ \blue{ \:  \: r1 =  \sqrt{ \frac{9x}{\pi}}}}} }  \\  \:  \:  \:  \:  \:  \: { \rightarrow{ \sf{ \blue{ \:  \:  r1 = 3 \sqrt{ \frac{x}{\pi}} \: unit}}} }

{ \sf{ :{ \implies{ \orange{ \: circumference 1(c1) }}}}} \\ \\   \:  \:  \:  \:  \:  \:  { \rightarrow{ \blue{ \sf{ \:  \:  \:  c1 = 2\pi \times r1}}}} \\  \:  \:  \:  \:  \:  \: { \rightarrow{ \blue{ \sf{ \:  \:  \: c1 = 2\pi \times 3 \ \sqrt{ \frac{x}{\pi} } \:  \:  units}}}}

{ \boxed{ \underline{ \tt{ \pink{circle \: 2}}}}}

{ :{ \implies{ \blue{ \sf{radius2(r2)}}}}} \\  \\  \:  \:  \:  \:  \:  \: { \rightarrow{ \blue{ \sf{ \:  \: area \: 2 =4x =  \pi {(r2)}^{2} }}}} \\  \:  \:  \:  \:  \:  \: { \rightarrow{ \blue{ \sf{ \:  \:  {(r2)}^{2}  =  \frac{4x}{\pi}}}}}  \\  \:  \:  \:  \:  \:  \: { \rightarrow{ \blue{ \sf{ \:  \: {r2 =  \sqrt{ \frac{4x}{\pi} } }}}}} \\  \:  \:  \:  \:  \:  \: { \rightarrow{ \blue{ \sf{ \:  \: r2 = 2 \sqrt{ \frac{x}{\pi} } unit}}}}

{ :{ \implies{ \orange{ \sf{circumference \: 2(c2)}}}}} \\  \\  \:  \:  \:  \:  \:  \: { \rightarrow{ \blue{ \sf{ \:  \: c2 = 2\pi \times r2}}}} \\   \:  \:  \:  \:  \:  \: { \rightarrow{ \blue{ \sf{ \:  \: c2 = 2\pi \times 2 \sqrt{ \frac{x}{\pi}} \: units}}} }

{ \underline{ \bold{ \green{ \: ratio \: of \: c1 \: and \: c2}}}} \\  \\  \\{ : { \implies{\green{ \sf  \frac{c1}{c2} =  \frac{2\pi \times 3 \sqrt{ \frac{x}{\pi} } }{2\pi \times 2 \sqrt{ \frac{x}{\pi} } }} } }} \\  \\  \\ {  : { \implies{\green{ \sf{ \frac{c1}{c2}  =  \frac{3}{2} }}}}} \\  \\   { \leadsto{ \boxed{ \boxed{ \red{ \sf{c1 : c2 = 3 : 2}}}}}}

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