Question

$\sf{s_1 , s_2}$ are inscribed and circumscribed circles of a Triangle with sides 3,4,5 then $\sf\dfrac{area\:of\:s_1}{area\:of\:s_2}$ is ?

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Answer 1

✯ Your Question :-

• $\sf{s_1 , s_2}$ are inscribed and circumscribed circles of a Triangle with sides 3,4,5 then $\sf\dfrac{area\:of\:s_1}{area\:of\:s_2}$ is ?

✮ Given :-

• $\sf{s_1 , s_2}$ are inscribed and circumscribed circles of a Triangle with sides 3,4,5.

✮ To Find :-

• $\sf\dfrac{area\:of\:s_1}{area\:of\:s_2}$

✯ Explanation :-

☘ Sides of triangle ➣ 3,4,5

✪ 3² + 4² = 5² [right angled]

.°. Given triangle is a right angled triangle.

$\\ ✫\:\color{darkviolet}\tt \dfrac{a}{sin \: A} = 2R \: \implies \: a \: = \: 2R$

$✫\:\color{purple}\bf R \: : \implies \: \frac{a}{2} = \frac{5}{2}$

$✪\:\color{red}\tt r\: = \: \frac{ \triangle}{5} \: = \: \frac{\frac{1}{2} \times 4 \times 3}{ \frac{5 + 4 + 3}{2} } = 1$

$❀\:\bf{\underline{Thus}}, \: \: \color{darkgreen}\sf \dfrac{The \: area \: of \: inscribed \: circle \: s_{1}}{\: area \: of \: circumscribed \: circle \: s_{2 }}$

$➲ \: \color{blue} \tt \frac{\pi r {}^{2} }{\pi R {}^{2} }\:\:\:\: ➲ \: \orange{\tt \dfrac{1 {}^{2} }{ (\frac{5}{2}) {}^{2} } } \:\:\:\:➲ \: \tt \color{darkviolet} \: \frac{1 \times 1}{ \frac{5 \times 5}{2 \times 2} } ➲ \: \tt \color{darkblue} \: \frac{1}{ \frac{25}{4} } \:\:\:\:➲ \: \: \color{darkgreen} \tt 1\times \left(\frac{4}{25}\right)$

$➥ \: \: \underline{\boxed{\bf \frac{4}{25}}}$

$✰ \: \: \boldsymbol{\therefore{ \underline{\:\:Hence}, } \: \: \color{purple}\sf \dfrac{The \: area \: of \: inscribed \: circle \: s_{1}}{\: area \: of \: circumscribed \: circle \: s_{2 }} \: \: \color{black}➣ \: \: \color{red}\underline{\boxed{ \bf\frac{4}{25} } }}$

Note :- Please slide left or right to see the full ans.

✰ Hope it helps u ✰

Answer 2

$\large{ \mathbb{ \colorbox{blac} { \boxed{ \boxed{ \colorbox{blue} {-:Answer:-}}}}}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{red}{Given::}}}}}}$

$\pink{➠}{ \sf{s_1 , s_2}are \: inscribed \: and \:circumscribed\:circles \: of \: a \: triangle\: with \: sides\: 3,4 \: and \: 5}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{b}{To \: find::}}}}}}$

$\pink{➠}{ \sf{\dfrac{Area\:of\:inscribed \: circle(s_1)}{Area\:of\:circumscribed \: circle(s_2)}}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{pink}{Formula \: used::}}}}}}$

$\pink{➠}{ \sf{\dfrac{a}{sin \: A} = 2R \: \: ...(i) \{Area \: circumradius \: formula \}}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{gold}{Concept \: used::}}}}}}$

$\pink{➠}{ \pmb{ \bf{In \: right-angled \: triangle,}}}$

${:: \implies \sf {(Hypotenuse)}^{2} = {(Base)}^{2} + {(Perpendicular)}^{2} \: \: ...(ii) \{ By \: pythagoras \: theorem \}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{plum}{According \: to \: Question::}}}}}}$

$\pmb{ \bf{Let's \: start \: now \: from \: concepts!!!}}$

$\pink{➠}{ \sf {(Hypotenuse)}^{2} = {(Base)}^{2} + {(Perpendicular)}^{2} \: \: \{ By \:equation \: (ii) \}}$

${:: \implies{ \sf {(5)}^{2} = {(3)}^{2} + {(4)}^{2} \: \:}}$

${:: \implies{ \sf {25} = 9 + 16 \: \:}}$

${:: \implies{ \sf {25} = 25\:}}$

$\bf{So, \triangle ABC \: is \: right-angled \: triangle.}$

$\pmb{ \bf{Then,by \: using \: the \: formula;}}$

$\pink{➠}{ \sf{\dfrac{a}{sin \: A} =Diameter \: \: \{From \: equation \: (i) \}}}$

$\pink{➠}{ \sf{\dfrac{a}{sin \: A} = 2R \: \: }}$

$: : \implies{ \sf{{a} = 2R \: \: }}$

${: : \implies{ \sf{Radius \: of \: circumscribed \: circle(R) = \frac{a}{2} = \frac{5}{2} } }}$

${\pink{➠}{ \sf{Radius \: of \: inscribed \: circle (r)= \frac{Area \: of \: triangle}{semi - perimeter} }}}$

${: : \implies{ \sf{ Radius \: of \: inscribed \: circle(r)= \frac{ \frac{1}{2} \times 4 \times 3 }{ \frac{3 + 4 + 5}{2} } }}}$

${: : \implies{ \sf{Radius \: of \: inscribed \: circle (r)= \frac{ \frac{1}{ \cancel 2} \times \cancel 4 \times 3 }{ \frac{12}{2} } }}}$

${: : \implies{ \sf{ Radius \: of \: inscribed \: circle(r)= \frac{2\times 3 }{ 6 } }}}$

${: : \implies{ \sf{Radius \: of \: inscribed \: circle( r)= \frac{6 }{ 6 } }}}$

${: : \implies{ \sf{Radius \: of \: inscribed \: circle (r)= 1 }}}$

$\pmb{ \bf{After \: that,}}$

$\pink{➠}{ \sf{\dfrac{Area\:of\:inscribed \: circle(s_1)}{Area\:of\:circumscribed \: circle(s_2)}}}$

$\pink{➠}{ \sf{\dfrac{\pi{r}^{2}}{\pi{R}^{2} } = \frac{ {(1)}^{2} }{ ({ \frac{5}{2}) }^{2} } }}$

$\pink{➠}{ \sf{\dfrac{\pi{r}^{2}}{\pi{R}^{2} } = \frac{4}{25} }}$

$\bf{Hence,}$

${\pink{➠}{ \sf{ \frac{4}{25} \: is \: the \: correct \: answer.}}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{green}{Diagram::}}}}}}$

${\pink{➠}{ \sf{Kindly \:refer \: the \: attachments.. }}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{orange}{Know \: more \: about \: circle::}}}}}}$

${\pink{➠}{ \bf{ Inscribed \: circle:{ \sf{A \: triangle \: is \: a \: circle \: contained \: in \: the \:circle \: such \: that \: just }}}}}$

${\sf{touches\: the \: side \: of \: triangle. }}$

${\pink{➠}{ \bf{Circumscribed \: circle:{ \sf{A \: triangle \: is \: a \: circle \: containing \: the \: triangle\:such \: that }}}}}$

$\sf{\:vertices \:of \: the \: triangle \: touches \:the \: circumference \: of \: circle.\: \: \: \: \: }$

$\bf{Note:}$

$\pink{➠}\sf{Kindly\:swipe\:left-right\: to \:see\: full\: answer.}$