The sides of triangle abc are 3, 4, and 5 inches long. how far is the incenter of the triangle from the circumcenter of the triangle?

(√5)/2

Step-by-step explanation:

In the attached figure, we have labeled the circumcenter point D and the incenter point E. The points of tangency of the incircle with sides AB, BC and CA are labeled G, H, and F, respectively.

The distances from any vertex to the two points of tangency from that vertex are the same. So, AG = FA, BG = BH, and CF = CH. If we call the radius of the incircle "r", then we have ...

AG = FA = r, BG = BH = 3-r, CF = CH = 4-r

so the side length BC is ...

BC = BH +CH = (3-r) +(4-r) = 7-2r

We already know that side length BC is 5, so ...

5 = 7 -2r

r = (7 -5)/2 = 1

Of course, the circumcenter of a right triangle is the midpoint of the hypotenuse, so the circumradius "R" is 5/2 = 2.5.

The formula for the distance between the two centers is ...

d = √(R(R -2r)) = √(2.5(2.5 -2)) = √1.25 = (√5)/2

_____

Comment on this answer

We have used a formula for the center-to-center distance found using a web search. The attached diagram shows the coordinates of the two centers, so the distance can be found from those. It is the same.

answer: the angel abc is 30°,and the abc is a triangle where the angel bac is 90°

so ac is the half of 24=12ft,and from the pythagorean theorem we find ba

ba=√bc^2-ac^2=√576-144=√432=12√3ft

so bac is 12+12√3=12(1+√3)ft