Math, asked by MichWorldCutiestGirl, 3 days ago


\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} \  \textless \ br /\  \textgreater \
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Answers

Answered by user0888
27

\Huge\text{$\dfrac{S_{1}}{S_{2}}=\dfrac{4}{25}$}

\huge\textbf{Question}

\textrm{A triangle has side lengths 3, 4, 5.}

\textrm{$S_{1}$ is the area of the inscribed circle.}

\textrm{$S_{2}$ is the area of the circumscribed circle.}

\textrm{Find $\dfrac{S_{1}}{S_{2}}$.}

\huge\textbf{Explanation}

\textrm{We know that (3,4,5) is a Pythagorean triple.}

\textrm{$\therefore$Given triangle is a right triangle.}

\Large\textrm{Inradius and the Area}

\cdots\longrightarrow S=\dfrac{1}{2}\times r\times(a+b+c)

\textrm{We know that the area of the triangle is 6.}

\large\textrm{(By formula)}

6=\dfrac{1}{2}\times r\times(3+4+5)

6=\dfrac{1}{2}\times r\times12

6=6r

r=1

\Large\textrm{Properties of Circumcircle}

\textrm{For a right triangle, -}

\textrm{it is located at the midpoint of the hypotenuse.}

\textrm{Hence, the radius is half the hypotenuse, $\dfrac{5}{2}$.}

\textrm{Lastly, -}

\dfrac{S_{1}}{S_{2}}=\dfrac{1^{2}}{(5/2)^{2}}

\dfrac{S_{1}}{S_{2}}=\dfrac{4}{25}

\huge\textbf{Answer}

\textrm{Find $\dfrac{S_{1}}{S_{2}}$.}

\textrm{The required answer is $\dfrac{S_{1}}{S_{2}}=\dfrac{4}{25}$.}

Answered by brainpower463
13

QUESTION :-

A triangle has side lengths 3, 4, 5.S₁ is the area of the inscribed circle. S₂ is the area of the circumscribed circle.Find S1 /S2

GIVEN :-

A triangle has side lengths 3, 4, 5

S₁ is the area of the inscribed circle

S₂ is the area of the circumscribed circle

TO FIND :-

Find S1 /S2 = ?

SOLUTION :-

first we have formula :-

r = area of traingle / semi - perimeter of traingle

R = Length of hypotenuse / 2 × sine of central angle

ACCORDING TO THE QUESTION :-

= 1/2 × 3 × 4 / 3 + 4 + 5 / 2 6/6 = 1

then, we have :-

R = Length of hypotenuse / 2 × sine of central angle

= 5 / 2 × sin 90° = 5/2

ratio of area is

πr² / πR² = 1² / ( 5/2) = 1/25 /4 = 4/25

so, S1/S2 = 4/25

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