Question

Show that the radius of incircle of right angle triangle is equal to the difference of half of the perimeter and twice the length of its hypotenuse of that triangle.

Answer 1

Let r be the radius of radius of incircle of right angled triangle ABC, as shown.
It can be seen ODBE on a square with side = r.
CE = CF & AD = AF (fom C & A respectively).
Perimeter of $\Delta ABC$ = AB + BC + CA
= AD + r + r + CE + CF + AF.
= 2r + 2AD + 2CF
= 2(r + AD + CF)
= 2(r + AF + CF)
= 2(r + hypoteneses) = 2r + 2 $\times$ hypotenese
$\therefore r=\frac{\frac{Perimeter}{2}-Hypotense\times 2}{Proved}$