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Name the points A(2,6), B(3,9) and C(4,8)A(2,6),B(3,9)andC(4,8).
By the Distance Formula,
AB^2=(3-2)^2+(9-6)^2=10, => AB=sqrt10AB2=(3−2)2+(9−6)2=10,⇒AB=√10
AC^2=(4-2)^2+(8-6)^2=8, => AC=sqrt8AC2=(4−2)2+(8−6)2=8,⇒AC=√8
BC^2=(4-3)^2+(8-9)^2=2, => BC=sqrt2BC2=(4−3)2+(8−9)2=2,⇒BC=√2
=> AB^2=AC^2+BC^2⇒AB2=AC2+BC2
This means that DeltaABC is a right triangle, having hypotenuse AB.
The in-radius r of the inscribed circle of a triangle is : r=(2A)/p,
where A and p are the area and perimeter of the triangle.
For a right triangle, the in-radius r takes the simple form :
r=(a*b)/(a+b+c)
where a, b are sides and c is the hypotenuse.
Therefore, the in-radius r of DeltaABC is :
r= (AC*BC)/(AC+BC+AB)
=(sqrt8*sqrt2)/(sqrt8+sqrt2+sqrt10)~~0.54
