Two corners of an isosceles triangle are at #(6 ,4 )# and #(9 ,7 )#. If the triangle's area is #36 #, what are the lengths of the triangle's sides?

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Answer 1 · 2017-06-09T18:37:30

The lengths of the sides are #=4.24#, #17.1# and #17.1#

The length of the base is

#b=sqrt((9-6)^2+(7-4)^2)=sqrt(3^2+3^2)=3sqrt2#

Let the height of the triangle be #=h#

The area is

#A=1/2*b*h#

#1/2*3sqrt2*h=36#

#h=(36*2)/(3sqrt2)=24/sqrt2=12sqrt2#

Let the lengths of the second and third sides of the triangle be #=c#

Then,

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

#c^2=(12sqrt2)^2+(3sqrt2/2)^2#

#c^2=288+9/2=587/2#

#c=sqrt(585/2)=17.1#