Two corners of an isosceles triangle are at #(9 ,6 )# and #(3 ,2 )#. If the triangle's area is #48 #, what are the lengths of the triangle's sides?

1 Answer
Rory K.
Aug 20, 2016

#sqrt(2473/13)#

Explanation:

Let the distance between the given points be s.
then #s^2# = #(9-3)^2 + (6-2)^2#
#s^2# = 52
hence s = 2#sqrt13#
The perpendicular bisector of s, cuts s #sqrt13# units from (9;6).
Let the altitude of the triangle given be h units.
Area of triangle = #1/2##2sqrt13.h#
hence #sqrt13#h = 48
so h = #48/sqrt13#
Let t be the lengths of the equal sides of the given triangle.
Then by Pythagoras' theorem,
#t^2# = #(48/sqrt13)^2# + #sqrt13^2#
= #2304/13# + #169/13#
= #2473/13#
hence t = #sqrt(2473/13)#

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