Two corners of an isosceles triangle are at #(3 ,9 )# and #(2 ,5 )#. If the triangle's area is #4 #, what are the lengths of the triangle's sides?

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Answer 1 · 2017-06-10T10:18:04

The lengths of the triangle's sides are #2.83#, #2.83# and #4.12#

The length of the base is

#b=sqrt((3-2)^2+(9-5)^2)=sqrt(1^2+4^2)=sqrt17#

Let the height of the triangle be #=h#

The area is

#A=1/2*b*h#

#1/2*sqrt17*h=4#

#h=(4*2)/(sqrt17)=8/sqrt17#

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

Then,

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

#c^2=(8/sqrt17)^2+(sqrt17/2)^2#

#c^2=3.76+4.25=8.01#

#c=sqrt(8.01)=2.83#