Question

If a circle is inscribed in a right angled triangle having right angle at $$B$$, then find the diameter of the circle.

Answer 1

Hint: First we have to define what the terms we need to solve the problem are,
The right angled triangle $ABC$having right angle at$B$,
The area of right angled triangle = ${\displaystyle \frac{1}{2}}$ (base$\times$perpendicular)

Complete step-by-step solution:

As we know, $△ABC$ is right angle triangle,
Since, area of right angle triangle = ${\displaystyle \frac{1}{2}}$ (base$\times$perpendicular)
Base = $AB$ and perpendicular = $BC$that is area of $△ABC$=${\displaystyle \frac{1}{2}}AB\times BC$
Also we can see in the figure that
Area of $△ABC$= area of $△AOB$$+$ area of $△BOC$$+$ area of $△AOC$ $(2)$
For area of$△AOB$, base = $r$ and perpendicular = $AB$
That is area of $△AOB$= ${\displaystyle \frac{1}{2}}r\times AB$(base $\times$perpendicular of$△AOB$)
For area of$△BOC$, base = $r$ and perpendicular = $BC$
Similarly, area of $△BOC$= ${\displaystyle \frac{1}{2}}r\times BC$
And for area of$△AOC$, base = $r$ and perpendicular = $CA$
That is area of $△AOC$= ${\displaystyle \frac{1}{2}}r\times CA$
Now substitute all equations in $(2)$we get
Area of $△ABC$= ${\displaystyle \frac{1}{2}}r\times AB$$+$ $+$${\displaystyle \frac{1}{2}}r\times CA$
Therefore area of $△ABC$= ${\displaystyle \frac{1}{2}}r\times (AB+BC+CA)$[taking common terms] $(3)$
Equating equation $(1)$ and $(3)$we get
${\displaystyle \frac{1}{2}}AB\times BC$= ${\displaystyle \frac{1}{2}}r\times (AB+BC+CA)$
Cancelling ${\displaystyle \frac{1}{2}}$ on both sides
$AB\times BC$= $r\times (AB+BC+CA)$[cross multiplying on both sides]
$2r={\displaystyle \frac{2AB\times BC}{AB+BC+CA}}$[Multiply 2 on both sides]
Now $(AB+BC{)}^2-(A{B}^2+B{C}^2)=2AB\times BC$ substitute in above equation
$2r={\displaystyle \frac{{(AB+BC)}^2-(A{B}^2+B{C}^2)}{AB+BC+CA}}$
Now by Pythagoras theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the square of other two sides which is,$A{C}^2=A{B}^2+B{C}^2$ substitute this theorem in above equation we get
$2r={\displaystyle \frac{{(AB+BC)}^2-A{C}^2}{AB+BC+CA}}$
Thus $2r={\displaystyle \frac{(AB+BC-AC)\times (AB+BC+CA)}{(AB+BC+CA)}}$[taking common terms out]
Cancelling ${\displaystyle \frac{AB+BC+CA}{AB+BC+CA}}$
Hence $2r=AB+BC-CA$

Note: The diameter of the circle is and having right triangle at $△ABC$
Pythagoras theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the square of other two sides which is (perpendicular)${}^2$= (base)${}^2$$+$(hypotenuse)${}^2$