Two corners of an isosceles triangle are at #(6 ,4 )# and #(4 ,1 )#. If the triangle's area is #8 #, what are the lengths of the triangle's sides?

1 Answer

the lengths are #a=sqrt(15509)/26# and #b=sqrt(15509)/26# and #c=sqrt13#

Also #a=4.7898129# and #b=4.7898129# and #c=3.60555127#

Explanation:

First we let #C(x, y)# be the unknown 3rd corner of the triangle.

Also Let corners #A(4, 1)# and #B(6, 4)#

We set the equation using sides by distance formula
#a=b#

#sqrt((x_c-6)^2+(y_c-4)^2)=sqrt((x_c-4)^2+(y_c-1)^2)#

simplify to obtain
#4x_c+6y_c=35" " "#first equation

Use now the matrix formula for Area:

#Area=1/2((x_a,x_b,x_c,x_a),(y_a,y_b,y_c,y_a))=#

#=1/2(x_ay_b+x_by_c+x_cy_a-x_by_a-x_cy_b-x_ay_c)#

#Area=1/2((6,4,x_c,6),(4,1,y_c,4))=#

#Area=1/2*(6+4y_c+4x_c-16-x_c-6y_c)#

#Area=8# this is given

We now have the equation

#8=1/2*(6+4y_c+4x_c-16-x_c-6y_c)#

#16=3x_c-2y_c-10#

#3x_c-2y_c=26" " " #second equation

Solving simultaneously the system
#4x_c+6y_c=35#
#3x_c-2y_c=26#
#x_c=113/13# and #y_c=1/26#

We can now solve for the lengths of sides #a# and #b#
#a=b=sqrt((x_b-x_c)^2+(y_b-y_c)^2)#

#a=b=sqrt((6-113/13)^2+(4-1/26)^2)#

#a=b=sqrt(15509)/26=4.7898129" " "#units