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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 Δle ICF ; IC=√2
→ CE= 1+√2
In ΔleCEB ; cos(C/2) = CE/a where C=90o
→ 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)
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