#### In triangle ABC in ratio of a:b:c=3:4:5 then find the ratio of raduis of circumcicle to that of inner circle ?Option: 1 1.5Option: 2 2Option: 3 2.5Option: 4 3.5

In-Circle and In-Centre -

In-Circle and In-Centre

The point of intersection of the internal angle bisectors of a triangle is called the in-centre of the triangle. Also,  a circle that can be inscribed within a triangle such that it touches each side of the triangle internally is called in-circle of a triangle. The in-centre of a triangle is denoted by I.

The radius of the inscribed circle of a triangle is called the in-radius and it is denoted by ‘r’.

$\\\mathrm{1.\quad In-radius,\;\mathit{r}\;is\;given\;by\;=\frac{\mathit{\Delta}}{\mathit{s}}}\\\mathrm{2.\quad \mathit{r}\;=\mathit{(s-a)}\tan\frac{A}{2}=\mathit{(s-b)}\tan\frac{B}{2}=\mathit{(s-c)}\tan\frac{C}{2}}\\\mathrm{3.\quad \mathit{r}\;=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}$

Proof :

1.

Consider the triangle, ABC

We know that, area of ABC = area of IBC + area of IBA + area of ICA

$\\\mathrm{\quad\quad\Delta=\frac{1}{2}\mathit{ar}+\frac{1}{2}\mathit{br}+\frac{1}{2}\mathit{cr}}\\\\\mathrm{\quad\quad\quad=\frac{1}{2}\mathit{r}(\mathit{a+b+c})}\\\\\mathrm{\quad\quad\quad=\frac{1}{2}\mathit{r}(\mathit{2s})\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[\because \mathit{2s=a+b+c}]}\\\\\mathrm{\quad\quad\quad=\mathit{rs}}$

2.

From the half-angle formula of tangent

$\\\mathrm{\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}}$

Multiply both sides with (s - a)

$\\\mathrm{\mathit{(s-a)}\tan\frac{A}{2}=\mathit{(s-a)}\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}}\\\\\mathrm{\quad\quad\quad\quad\quad\quad= \mathit{\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}}}\\\\\mathrm{\quad\quad\quad\quad\quad\quad= \mathit{{\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}}}}\\\\\mathrm{\quad\quad\quad\quad\quad\quad= \mathit{{\frac{\Delta}{s}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; [\because\;\Delta=\sqrt{s(s-a)(s-b)(s-c)}]}$

In a similar fashion, other formula can be proved

3.

From the half angle formula of sine

$\\\mathrm{\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}}\\\\\mathrm{\sin\frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{ac}}}\\\\\mathrm{\sin\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}}$

$\\\mathrm{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=4R\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-c)}{ac}}\sqrt{\frac{(s-a)(s-b)}{ab}}}\\\\\mathrm{\quad\quad\quad\quad\quad\quad\quad\quad\quad=4R\frac{\mathit{(s-a)(s-b)(s-c)}}{\mathit{abc}}}\\\\\mathrm{\quad\quad\quad\quad\quad\quad\quad\quad\quad=\frac{4R}{abc}\frac{\mathit{s(s-a)(s-b)(s-c)}}{\mathit{s}}}\\\\\mathrm{\quad\quad\quad\quad\quad\quad\quad\quad\quad=\frac{1}{\Delta}\frac{\Delta^2}{s}=\frac{\Delta}{s}=r}$

$\mathrm{\left [ We\;are\;using\;the\;fact\;that\;\;\Delta=\frac{abc}{4R}\;\;and\;\;\Delta=\sqrt{\mathit{s(s-a)(s-b)(s-c)}} \right ]}$

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Diagram

$AB\perp BC\\ so\ AC\ is\ the\ radius\ of\ circumcircle \\ 2R=AC=2R=5\\ EC=FC=4-r\\ AD=AF=3-r\\ AC=AF+FC\\ 5=3-r+4-r\\ r=1\\ \frac{R}{r}=\frac{5}{2}$