A circle is incribed in a right angled triangle of hpotenuse 10 units.if the radius of incribbed circle is 1 unit what is he perimenter of triangle

consider the figure

(AB)^2=(AC)^2+(BC)^2 ----------- Pythagoras

consider AD = BF = x

and CF=DC=1 because radius is equal to 1

100=(x+1)^2+(x+1)^2

which implies x^2+2x-49

solving through quad equation we get x=-1+5(2)^1/2 and -1-5(2)^1/2

perimeter = AB+BC+AC = 10+x+1+x+1 = 2x+12 = 2(x+6)

possible value of x = -1+5(2)^1/2

subtituting tha valu of x gives u perimeter = 2(5+5(2)^1/2) = 10+10(2)^1/2

(2)^1/2 means root 2

10(2)^1/2 means 10 root 2

a²+b²=c² (c=10)-------- (1)

from Δ^le ICF ; IC=√2

→ CE= 1+√2

In Δ^leCEB ; cos(C/2) = CE/a where C=90^o

→ a= 2+√2

→ put a= 2+√2 and c=10 in (1) and b = 2√ (26-√2)

hence perimeter = a+b+c = 12+√2+2√(26-√2)