A triangle has corners at #(9 , 2 )#, #(4 ,7 )#, and #(5 ,8 )#. What is the radius of the triangle's inscribed circle?

1 Answer
Andrea S.
Nov 28, 2016

If #a,b# and #c# are the lengths of the sides and #S# is the area of the triangle the radius of the inscribed circle is given by the formula:

#r=(2S)/(a+b+c)#

Explanation:

The area #S# can be computed with Eron's formula:

#S=sqrt(p(p-a)(p-b)*(p-c))#

where #p=(a+b+c)/2#.

Substituting in the formula:

#r=(sqrt(p(p-a)(p-b)*(p-c)))/(p)#

Start calculating the lengths of the sides:

#a=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#
#b=sqrt((x_3-x_2)^2+(y_3-y_2)^2)#
#c=sqrt((x_1-x_3)^2+(y_1-y_3)^2)#

#a = sqrt((9-4)^2+(2-7)^2) = sqrt(25+25) = 5sqrt(2)#
#b= sqrt((5-4)^2+(8-7)^2) =sqrt(1+1) =sqrt(2)#
#c= sqrt((9-5)^2+(8-2)^2) =sqrt(16+36) =2sqrt(13)#

#p=(5sqrt(2)+sqrt(2)+2sqrt(13))/2 = 3sqrt(2)+sqrt(13)#

#r=sqrt((3sqrt(2)+sqrt(13))(3sqrt(2)+sqrt(13)-5sqrt(2))(3sqrt(2)+sqrt(13)-sqrt(2))(3sqrt(2)+sqrt(13)-2sqrt(13)))/(3sqrt(2)+sqrt(13))=sqrt((3sqrt(2)+sqrt(13))(sqrt(13)-2sqrt(2))(2sqrt(2)+sqrt(13))(3sqrt(2)-sqrt(13)))/(3sqrt(2)+sqrt(13))#

and I leave the rest of the calculation :D

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