Two corners of an isosceles triangle are at #(4 ,2 )# and #(1 ,5 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

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Answer 1 · 2018-05-22T14:47:16

#color(blue)(a=b=sqrt(32930)/6 and c=3sqrt(2)#

Let #A=(4,2)# and #B=(1,5)#

If #AB# is the base of an isosceles triangle then #C=(x,y)# is the vertex at the altitude.

Let The sides be # a,b,c#, #a=b#

Let h be the height, bisecting AB and passing through point C:

Length #AB = sqrt((4-1)^2+(2-5)^2)=sqrt(18)=3sqrt(2)#

To find #h#. We are given area equals 64:

#1/2AB*h=64#

#1/2(3sqrt(2))h=64=>h=(64sqrt(2))/3#

By Pythagoras' theorem:

#a=b=sqrt(((3sqrt(2))/2)^2+((64sqrt(2))/3)^2)=sqrt(32930)/6#

So the lengths of the sides are:

#color(blue)(a=b=sqrt(32930)/6 and c=3sqrt(2)#