A triangle has side length 3, 4, and 5. Find the radius of the inscribed triangle.

Guest Jun 3, 2020

#1**0 **

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,

m_{1} = slope of one of the sides, m_{2} = slope of the other side):

m = [ m_{1}·m_{2} + sqrt( 1 + m_{1}^{2} )·sqrt(1 + m_{2}^{2} ) - 1 ] / [ m_{1} + m_{2} ]

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**.

geno3141 Jun 3, 2020