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

 Jun 3, 2020
 #1
avatar+21953 
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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.

 Jun 3, 2020
 #2
avatar+1326 
+1

A triangle has side lengths 3, 4, and 5.  Find the radius of the inscribed (trianglecheeky) circle.

 

Semiperimeter is:         s = (3+4+5)/2 = 6

 

Area of a triangle is:     A = 1/2 (3*4) = 6 u²

 

Inradius is:                     r = A / s = 1  indecision

 Jun 3, 2020
edited by Dragan  Jun 3, 2020

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