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

1 Answer
Nov 28, 2016

If a,ba,b and cc are the lengths of the sides and SS is the area of the triangle the radius of the inscribed circle is given by the formula:

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

Explanation:

The area SS can be computed with Eron's formula:

S=sqrt(p(p-a)(p-b)*(p-c))S=p(pa)(pb)(pc)

where p=(a+b+c)/2p=a+b+c2.

Substituting in the formula:

r=(sqrt(p(p-a)(p-b)*(p-c)))/(p)r=p(pa)(pb)(pc)p

Start calculating the lengths of the sides:

a=sqrt((x_2-x_1)^2+(y_2-y_1)^2)a=(x2x1)2+(y2y1)2
b=sqrt((x_3-x_2)^2+(y_3-y_2)^2)b=(x3x2)2+(y3y2)2
c=sqrt((x_1-x_3)^2+(y_1-y_3)^2)c=(x1x3)2+(y1y3)2

a = sqrt((9-4)^2+(2-7)^2) = sqrt(25+25) = 5sqrt(2)a=(94)2+(27)2=25+25=52
b= sqrt((5-4)^2+(8-7)^2) =sqrt(1+1) =sqrt(2)b=(54)2+(87)2=1+1=2
c= sqrt((9-5)^2+(8-2)^2) =sqrt(16+36) =2sqrt(13)c=(95)2+(82)2=16+36=213

p=(5sqrt(2)+sqrt(2)+2sqrt(13))/2 = 3sqrt(2)+sqrt(13)p=52+2+2132=32+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