A triangle has corners at #(3 , 2 )#, #(6 ,7 )#, and #(2 ,4 )#. What is the radius of the triangle's inscribed circle?

1 Answer

#0.842#

Explanation:

The area #\Delta# of triangle with vertices #(x_1, y_1)\equiv(3, 2)#, #(x_2, y_2)\equiv(6, 7)# & #(x_3, y_3)\equiv(2, 4)# is given by following formula

#\Delta=1/2|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|#

#=1/2|3(7-4)+6(4-2)+2(2-7)|#

#=5.5#

Now, the lengths of all three sides say #a, b# & #c# of given triangle are computed by using distance formula as follows

#a=\sqrt{(3-6)^2+(2-7)^2}=\sqrt34#

#b=\sqrt{(6-2)^2+(7-4)^2}=5#

#c=\sqrt{(3-2)^2+(2-4)^2}=\sqrt5#

hence, the semi-perimeter #s# of given triangle is computed as follows

#s=\frac{a+b+c}{2}#

#=\frac{\sqrt34+5+\sqrt5}{2}=6.533#

hence, the radius of inscribed circle is given as

#\frac{\Delta}{s}#

#=\frac{5.5}{6.533}#

#=0.842#