Two corners of an isosceles triangle are at #(3 ,2 )# and #(9 ,1 )#. If the triangle's area is #12 #, what are the lengths of the triangles sides?

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Answer 1 · 2017-12-08T10:25:23

Measure of the three sides are (6.0828, 3.6252, 3.6252)

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Length #a = sqrt((9-3)^2 + (1-2)^2) = sqrt 37 = 6.0828#

Area of #Delta = 12#
#:. h = (Area) / (a/2) = 12 / (6.0828/2) = 6 / 3.0414 = 1.9728#

#side b = sqrt((a/2)^2 + h^2) = sqrt((3.0414)^2 + (1.9728)^2)#
#b = 3.6252#

Since the triangle is isosceles, third side is also #= b = 3.6252#

Measure of the three sides are (6.0828, 3.6252, 3.6252)