A triangle has side length 3, 4, and 5. Find the radius of the inscribed triangle.
I'm assuming that you want the radius of the inscribed circle.
The only way that I can see to find the radius of the inscribed circle is to use coordinate geometry and a very nasty formula; so, if anyone else has a nicer solution, please post it!
A 3-4-5 triangle is a right triangle and the center of the inscribed circle is the point of intersection of the angle bisectors.
Let A = (0,3), B = (4,0), and C = (0,0).
The angle bisector of angle(C) is the line whose equation is y = x.
To find the angle bisector of angle(B), we can use this formula (where m = slope of the bisector,
m1 = slope of one of the sides, m2 = slope of the other side):
m = [ m1·m2 + sqrt( 1 + m12 )·sqrt(1 + m22 ) - 1 ] / [ m1 + m2 ]
The slope of BC = 0; the slope of AB = (0 - 3) / (4 - 0) = -3/4
m = [ (0)·(-3/4) + sqrt( 1 + (0)2 )·sqrt(1 + (-3/4)2 ) - 1 ] / [ (0) + (-3/4) ]
= [ 0 + sqrt( 1 )·sqrt( 25/16 ) - 1 ] / [ -3/4 ]
= -1/3 <--- this is the slope of the angle bisector at vertex B
The equation of the angle bisector at B is: y - 0 = (-1/3)(x - 4) ---> y = (-1/3)x + 4/3
The two lines: y = x and y = (-1/3)x + 4/3 intersect at th point (1,1)
because: x = (-1/3)x + 4/3
(4/3)x = 4/3
x = 1
The distance from the point (1,1) to the x-axis is 1; as is the distance to the y-axis and also to line AB.
Thus, the radius of the incircle is 1.