[tex]{ \dashrightarrow{ \boxed{ \green{ \sf{area \: of \: circle = \pi {r}^{2} }}}}}\\ { \dashrightarrow{ \boxed { \green{ \sf{circumference \: of \: circle = 2\pi \times r}}}}}[/tex]
[tex]{ \rm{ \leadsto{let \: the \: area \: of \: circle \: 1 \: be \: 9x \: {unit}^{2} }}} \\ { \rm{ \: \: \: \: \: \: and \: that \: of \: circle \: 2 \: be \: 4x {unit}^{2} }}[/tex]
[tex]{ \boxed{ \underline{ \pink{ \tt{ circle \: 1}}}}}[/tex]
[tex]{ \underline{ \underline{ \huge{ \red{ \rm{Question}}}}}}[/tex]
The areas of two circles are in ratio of 9:4 , then what is the ratio of their circumferences ?
[tex]{ \underline{ \underline{ \huge{ \red{ \rm{ Solution}}}}}}[/tex]
[tex]{ \underline{ \underline{ \tt{ \orange{given} \: : – }}}}[/tex]
[tex]{ \dag{ \sf { \pink{ \: \: \: ratio \: of \: area \: of \: two \: circles \: = 9 : 4}}}}[/tex]
[tex]{ \underline{ \underline{ \tt{ \orange{ \: to \: find \: } : – }}}}[/tex]
[tex]{ \dag{ \sf{ \: \: \: ratio \: of \: their \: circumferences \: }}}[/tex]
[tex]{ \underline{ \underline{ \tt{ \orange{ formula \: used} \: : – }}}}[/tex]
[tex]{ \dashrightarrow{ \boxed{ \green{ \sf{area \: of \: circle = \pi {r}^{2} }}}}}\\ { \dashrightarrow{ \boxed { \green{ \sf{circumference \: of \: circle = 2\pi \times r}}}}}[/tex]
[tex]{ \rm{ \leadsto{let \: the \: area \: of \: circle \: 1 \: be \: 9x \: {unit}^{2} }}} \\ { \rm{ \: \: \: \: \: \: and \: that \: of \: circle \: 2 \: be \: 4x {unit}^{2} }}[/tex]
[tex]{ \boxed{ \underline{ \pink{ \tt{ circle \: 1}}}}}[/tex]
[tex]{ \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}}} } [/tex]
[tex]{ \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}}}}[/tex]
[tex]{ \boxed{ \underline{ \tt{ \pink{circle \: 2}}}}}[/tex]
[tex]{ :{ \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}}}}[/tex]
[tex]{ :{ \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}}} } [/tex]
[tex]{ \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}}}}}}[/tex]
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
[tex]\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}[/tex]
Circumference of a circle is given by,
C = 2πr
The ratio of their circumferences :
= 2πr₁ : 2πr₂
= r₁ : r₂
= 3 : 2