Two corners of an isosceles triangle are at #(2 ,4 )# and #(1 ,4 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

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Answer 1 · 2016-06-09T14:30:19

#{1,124.001,124.001}#

Let #A = {1,4}#, #B = {2,4}# and #C = {(1+2)/2,h}#
We know that #(2-1) xx h/2 = 64# solving for #h# we have

#h = 128#.

The side lengths are:

#a = norm(A-B) = sqrt((1-2)^2+(4-4)^2) = 1#
#b = norm(B-C) =sqrt((2-3/2)^2+(4-128)^2) =124.001#
#a = norm(C-A) =sqrt((3/2-1)^2+(128-4)^2) =124.001 #