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**Hint:**The radius of a triangle is equal to $\dfrac{\Delta }{s}$. Where $'\Delta '$ is the area of the triangle and ‘s’ is the semi- perimeter of the triangle. We can find the area by taking the product of base and height and then dividing it by 2. We can find the semi-perimeter by adding up all the side lengths and then dividing it by 2. Find $'\Delta '$ and ‘s’ of triangles and put in the formula to get the inradius.

**Complete step-by-step answer:**

Given triangle ABC is right angled at B.

\[\begin{align}

& Area\ of\ a\ triangle\ is\ =\dfrac{1}{2}\times \left( base \right)\times \left( height \right) \\

& \Rightarrow \Delta =\dfrac{1}{2}\times \left( AB \right)\times \left( BC \right) \\

& Semiperimeter\ of\ \Delta ABC=\dfrac{Perimeter\ of\ \Delta ABC}{2} \\

& \Rightarrow s=\dfrac{AB+BC+AC}{2} \\

& \Rightarrow s=\dfrac{a+b+c}{2}.......\left( 1 \right) \\

\end{align}\]

Now, let us put calculated values of $'\Delta '$ and ‘s’ in the formula of inradius;

$\begin{align}

& r=\dfrac{\Delta }{s} \\

& \Rightarrow r=\dfrac{\left( \dfrac{1}{2}\times AB\times BC \right)}{\left( \dfrac{AB+BC+CA}{2} \right)} \\

\end{align}$

Multiplying both numerator and denominator by 2, we will get,

$\Rightarrow r=\dfrac{AB\times BC}{AB+BC+AC}$

Putting AB = c, BC = a and AC = b, we will get;

$\begin{align}

& \Rightarrow r=\dfrac{c\times a}{c+a+b} \\

& \Rightarrow r=\dfrac{ac}{a+b+c}...........\left( 2 \right) \\

\end{align}$

Subtracting equation (1) from equation (2), we will get,

$\Rightarrow r-s=\dfrac{ac}{a+b+c}-\dfrac{a+b+c}{2}$

Taking LCM and subtracting, we will get,

$\Rightarrow r-s=\dfrac{2ac-{{\left( a+b+c \right)}^{2}}}{2\left( a+b+c \right)}$

We know ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$.

Using these identity, we will get,

$\begin{align}

& \Rightarrow r-s=\dfrac{2ac-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca \right)}{2\left( a+b+c \right)} \\

& \Rightarrow r-s=\dfrac{{2ac}-{{a}^{2}}-{{b}^{2}}-{{c}^{2}}-2ab-2bc-{2ca}}{2\left( a+b+c \right)} \\

\end{align}$

Taking “-1” common, we will get,

$\Rightarrow r-s=\dfrac{-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc \right)}{2\left( a+b+c \right)}$

As $\Delta ABC$ is right angled at B,

$\begin{align}

& {{\left( hypotenuse \right)}^{2}}={{\left( side\ 1 \right)}^{2}}+{{\left( side\ 2 \right)}^{2}} \\

& \Rightarrow {{b}^{2}}={{a}^{2}}+{{c}^{2}} \\

\end{align}$

By replacing ${{a}^{2}}+{{c}^{2}}$with ${{b}^{2}}$, we will get,

$\Rightarrow r-s=\dfrac{-\left( {{b}^{2}}+{{b}^{2}}+2ab+2bc \right)}{2\left( a+b+c \right)}$

Multiplying both sides of equation by “-1”, we will get,

$\begin{align}

& \Rightarrow -1\times \left( r-s \right)=\left( -1 \right)\times \left[ \dfrac{-\left( 2{{b}^{2}}+2ab+2bc \right)}{2\left( a+b+c \right)} \right] \\

& \Rightarrow s-r=\dfrac{2{{b}^{2}}+2ab+2bc}{2\left( a+b+c \right)} \\

\end{align}$

Taking “2b” common from the numerator in RHS, we will get,

\[\Rightarrow s-r=\dfrac{2b\left( b+a+c \right)}{2\left( a+b+c \right)}\]

Dividing both numerator and denominator by 2(a + b + c), we will get,

$\begin{align}

& \Rightarrow s-r=b \\

& \Rightarrow r=s-b \\

\end{align}$

Putting $s=\dfrac{a+b+c}{2}$, we will get,

$r=\left( \dfrac{a+b+c}{2} \right)-\dfrac{b}{1}$

Taking LCM and subtracting, we will get,

$\begin{align}

& r=\dfrac{a+b+c-2b}{2} \\

& \Rightarrow r=\dfrac{a-b+c}{2} \\

\end{align}$

Now, putting a = BC, b = AC, c = AB, we will get,

$\begin{align}

& \Rightarrow r=\dfrac{BC-AC+AB}{2} \\

& \Rightarrow r=\dfrac{AB+BC-AC}{2} \\

\end{align}$

Hence, the inradius of $\Delta ABC$ is equal to $\dfrac{AB+BC-AC}{2}$ and option (A) is the correct answer.

**Note:**inradius of a triangle is the radius of the circle inscribed in the triangle. If you don’t remember the formula for inradius. Calculate inradius using geometry. But this will be a very lengthy method, so try to memorize the formula.

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