Two corners of an isosceles triangle are at #(4 ,8 )# and #(1 ,3 )#. If the triangle's area is #2 #, what are the lengths of the triangle's sides?

1 Answer
May 24, 2016

Lengths of the triangle's sides are # AC = BC =3.0, AB=5.83#

Explanation:

Let ABC be the isocelles triangle of which AB is base and AC=BC and the corners are A#(4,8)# and B #(1,3)#. Base #AB=sqrt((3-8)^2+(1-4)^2) = sqrt 34 # Let CD be the altitude(h) drawn from corner C on AB at point D , which is the mid point of AB. We know #area = 1/2*AB * h # or #2 = sqrt34 * h/2 or h = 4/sqrt34# Hence side #AC^2= (sqrt34/2)^2 + (4/sqrt34)^2 or AC =3.0 = BC # since #AC^2=AD^2+CD^2# # :.AC = BC =3.0, AB=sqrt 34=5.83# [Ans]