Question

In a $\mathrm{\Delta }ABC$, right angled at B, the in –radius is:
A. ${\displaystyle \frac{AB+BC-AC}{2}}$
B. ${\displaystyle \frac{AB+AC-BC}{2}}$
C. ${\displaystyle \frac{AB+BC+AC}{2}}$
D. None

Answer 1

Hint: The radius of a triangle is equal to ${\displaystyle \frac{\mathrm{\Delta }}{s}}$. Where ${}^{\prime }{\mathrm{\Delta }}^{\prime }$ 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 ${}^{\prime }{\mathrm{\Delta }}^{\prime }$ 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.

$$\begin{array}{rl}& Areaofatriangleis={\displaystyle \frac{1}{2}}\times \left(base\right)\times \left(height\right)\\ & \Rightarrow \mathrm{\Delta }={\displaystyle \frac{1}{2}}\times \left(AB\right)\times \left(BC\right)\\ & Semiperimeterof\mathrm{\Delta }ABC={\displaystyle \frac{Perimeterof\mathrm{\Delta }ABC}{2}}\\ & \Rightarrow s={\displaystyle \frac{AB+BC+AC}{2}}\\ & \Rightarrow s={\displaystyle \frac{a+b+c}{2}}.......\left(1\right)\end{array}$$
Now, let us put calculated values of ${}^{\prime }{\mathrm{\Delta }}^{\prime }$ and ‘s’ in the formula of inradius;
$\begin{array}{rl}& r={\displaystyle \frac{\mathrm{\Delta }}{s}}\\ & \Rightarrow r={\displaystyle \frac{({\displaystyle \frac{1}{2}}\times AB\times BC)}{\left({\displaystyle \frac{AB+BC+CA}{2}}\right)}}\end{array}$
Multiplying both numerator and denominator by 2, we will get,
$\Rightarrow r={\displaystyle \frac{AB\times BC}{AB+BC+AC}}$
Putting AB = c, BC = a and AC = b, we will get;
$\begin{array}{rl}& \Rightarrow r={\displaystyle \frac{c\times a}{c+a+b}}\\ & \Rightarrow r={\displaystyle \frac{ac}{a+b+c}}...........\left(2\right)\end{array}$
Subtracting equation (1) from equation (2), we will get,
$\Rightarrow r-s={\displaystyle \frac{ac}{a+b+c}}-{\displaystyle \frac{a+b+c}{2}}$
Taking LCM and subtracting, we will get,
$\Rightarrow r-s={\displaystyle \frac{2ac-{(a+b+c)}^2}{2(a+b+c)}}$
 We know ${(a+b+c)}^2=a^2+b^2+c^2+2ab+2bc+2ca$.
Using these identity, we will get,
$\begin{array}{rl}& \Rightarrow r-s={\displaystyle \frac{2ac-(a^2+b^2+c^2+2ab+2bc+2ca)}{2(a+b+c)}}\\ & \Rightarrow r-s={\displaystyle \frac{2ac-a^2-b^2-c^2-2ab-2bc-2ca}{2(a+b+c)}}\end{array}$
Taking “-1” common, we will get,
$\Rightarrow r-s={\displaystyle \frac{-(a^2+b^2+c^2+2ab+2bc)}{2(a+b+c)}}$
As $\mathrm{\Delta }ABC$ is right angled at B,
 $\begin{array}{rl}& {\left(hypotenuse\right)}^2={\left(side1\right)}^2+{\left(side2\right)}^2\\ & \Rightarrow b^2=a^2+c^2\end{array}$
By replacing $a^2+c^2$with $b^2$, we will get,
$\Rightarrow r-s={\displaystyle \frac{-(b^2+b^2+2ab+2bc)}{2(a+b+c)}}$
Multiplying both sides of equation by “-1”, we will get,
$\begin{array}{rl}& \Rightarrow -1\times (r-s)=(-1)\times \left[{\displaystyle \frac{-(2b^2+2ab+2bc)}{2(a+b+c)}}\right]\\ & \Rightarrow s-r={\displaystyle \frac{2b^2+2ab+2bc}{2(a+b+c)}}\end{array}$
Taking “2b” common from the numerator in RHS, we will get,
$$\Rightarrow s-r={\displaystyle \frac{2b(b+a+c)}{2(a+b+c)}}$$
Dividing both numerator and denominator by 2(a + b + c), we will get,
$\begin{array}{rl}& \Rightarrow s-r=b\\ & \Rightarrow r=s-b\end{array}$
Putting $s={\displaystyle \frac{a+b+c}{2}}$, we will get,
$r=\left({\displaystyle \frac{a+b+c}{2}}\right)-{\displaystyle \frac{b}{1}}$
Taking LCM and subtracting, we will get,
$\begin{array}{rl}& r={\displaystyle \frac{a+b+c-2b}{2}}\\ & \Rightarrow r={\displaystyle \frac{a-b+c}{2}}\end{array}$
Now, putting a = BC, b = AC, c = AB, we will get,
$\begin{array}{rl}& \Rightarrow r={\displaystyle \frac{BC-AC+AB}{2}}\\ & \Rightarrow r={\displaystyle \frac{AB+BC-AC}{2}}\end{array}$
Hence, the inradius of $\mathrm{\Delta }ABC$ is equal to ${\displaystyle \frac{AB+BC-AC}{2}}$ 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.