Two corners of an isosceles triangle are at #(1 ,6 )# and #(2 ,9 )#. If the triangle's area is #24 #, what are the lengths of the triangle's sides?

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Answer 1 · 2018-05-09T02:02:39

base #sqrt{10},# common side #sqrt{2329/10} #

Archimedes' Theorem says the area #a# is related to the squared sides #A, B# and #C# by

# 16a^2 = 4AB-(C-A-B)^2#

# C=(2-1)^2+(9-6)^2 = 10#

For an isosceles triangle either #A=B# or #B=C#. Let's work out both. #A=B# first.

#16 (24^2) = 4A^2 - (10-2A)^2#

#16(24^2) = -100 +40A #

# A = B = 1/40 ( 100+ 16(24^2)) = 2329/10 #

#B=C# next.

#16 (24)^2 = 4 A (10) - A^2 #

# (A - 20)^2 = - 8816 quad # has no real solutions

So we found the isosceles triangle with sides

base #sqrt{10},# common side #sqrt{2329/10} #