Two corners of an isosceles triangle are at #(5 ,8 )# and #(9 ,2 )#. If the triangle's area is #36 #, what are the lengths of the triangle's sides?

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Answer 1 · 2017-07-16T12:49:46

The lengths of the sides are #=10.6#, #10.6# and #=7.2#

The length of the base is

#b=sqrt((9-5)^2+(2-8)^2)=sqrt(16+36)=sqrt52=2sqrt13=7.2#

Let the altitude of the triangle be #=h#

Then

The area of the triangle is #A=1/2*b*h#

#h=2A/b=2*36/(2sqrt13)=36/sqrt13#

The sides of the triangle are

#=sqrt(h^2+(b/2)^2)#

#=sqrt(36^2/13+13)#

#=10.6#