Two corners of an isosceles triangle are at #(7 ,4 )# and #(3 ,1 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

Question

Two corners of an isosceles triangle are at #(7   ,4   )# and #(3  ,1    )#.  If the triangle's area is #64   #, what are the lengths of the triangle's sides?

Aaliyah Anderson · Accepted Answer

the lengths are #5# and #1/50sqrt(1654025)=25.7218#
and #1/50sqrt(1654025)=25.7218# Let #P_1(3, 1), P_2(7, 4), P_3(x, y)# Use the formula for the area of a polygon #Area=1/2((x_1,x_2,x_3,x_1),(y_1,y_2,y_3,y_1))# #Area=1/2(x_1y_2+x_2y_3+x_3y_1-x_2y_1-x_3y_2-x_1y_3)# #64=1/2((3,7,x,3),(1,4,y,1))# #128=12+7y+x-7-4x-3y# #3x-4y=-123" "#first equation We need a second equation which is the equation of the perpendicular bisector of the segment connecting #P_1(3, 1),and P_2(7, 4)# the slope #=(y_2-y_1)/(x_2-x_1)=(4-1)/(7-3)=3/4# for the perpendicular bisector equation, we need slope#=-4/3# and the midpoint #M(x_m, y_m)# of #P_1# and #P_2# #x_m=(x_2+x_1)/2=(7+3)/2=5# #y_m=(y_2+y_1)/2=(4+1)/2=5/2# Perpendicular bisector equation #y-y_m=-4/3(x-x_m)# #y-5/2=-4/3(x-5)# #6y-15=-8x+40# #8x+6y=55" "#second equation Simultaneous solution using first and second equations #3x-4y=-123" "# #8x+6y=55" "# #x=-259/25# and #y=1149/50# and #P_3(-259/25, 1149/50)# We can now compute for the other sides of the triangle using distance formula for #P_1# to #P_3# #d=sqrt((x_1-x_3)^2+(y_1-y_3)^2)# #d=sqrt((3--259/25)^2+(1-1149/50)^2)# #d=1/50sqrt(1654025)# #d=25.7218# We can now compute for the other sides of the triangle using distance formula for #P_2# to #P_3# #d=sqrt((x_2-x_3)^2+(y_2-y_3)^2)# #d=sqrt((7--259/25)^2+(4-1149/50)^2)# #d=1/50sqrt(1654025)# #d=25.7218# God bless...I hope the explanation is useful.

Carson Arnold · Answer

undefined The lengths of the sides of the triangle are approximately 16 units, 16 units, and 8 units.