In-Circle and In-Centre - Practice Questions & MCQ

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 an in-circle of a triangle. Its centre is the in-centre of the given triangle. The in-centre of a triangle is denoted by I.

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

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

Proof:
1.Consider the triangle, $A B C$
We know that, the area of ABC = area of IBC + area of IBA + area of ICA

$
\begin{aligned}
\Delta & =\frac{1}{2} a r+\frac{1}{2} b r+\frac{1}{2} c r \\
& =\frac{1}{2} r(a+b+c) \\
& =\frac{1}{2} r(2 s) \quad[\because 2 s=a+b+c] \\
& =r s
\end{aligned}
$

2. From the half-angle formula of tangent

$
\tan \frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{s}(\mathrm{~s}-\mathrm{a})}}
$
Multiply both sides with (s - a)

$
\begin{aligned}
(s-a) \tan \frac{\mathrm{A}}{2} & =(s-a) \sqrt{\frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{s}(\mathrm{~s}-\mathrm{a})}} \\
& =\sqrt{\frac{(s-a)(s-b)(s-c)}{s}} \\
& =\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s} \\
& =\frac{\Delta}{s}=r \quad[\because \Delta=\sqrt{\mathrm{s}(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}]
\end{aligned}
$
Similarly, other formulae can be proved.

3. From the half-angle formula of sine

$
\begin{aligned}
& \sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}} \\
& \sin \frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{a c}} \\
& \sin \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{a b}}
\end{aligned}
$
So,

$
\begin{aligned}
& 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=4 R \sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{(s-a)(s-c)}{a c}} \sqrt{\frac{(s-a)(s-b)}{a b}} \\
& =4 \mathrm{R} \frac{(s-a)(s-b)(s-c)}{a b c} \\
& =\frac{4 \mathrm{R}}{\mathrm{abc}} \frac{s(s-a)(s-b)(s-c)}{s} \\
& =\frac{1}{\Delta} \frac{\Delta^2}{\mathrm{~s}}=\frac{\Delta}{\mathrm{s}}=\mathrm{r}
\end{aligned}
$

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