Two corners of an isosceles triangle are at #(2 ,4 )# and #(8 ,5 )#. If the triangle's area is #4 #, what are the lengths of the triangle's sides?

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Answer 1 · 2017-12-05T13:01:04

Three sides of the triangle measure #color(red)(6.0828, 3.3136, 3.3136#

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Length #a = sqrt((8-2)^2 + (5-4)^2) = sqrt 37 = 6.0828#

Area of #Delta = 4#
#:. h = (Area) / (a/2) = 4 / (6.0828/2) = 4 / 3.0414 = 1.3152#

#side b = sqrt((a/2)^2 + h^2) = sqrt((3.0414)^2 + (1.3152)^2)#
#b = 3.3136#

Since the triangle is isosceles, third side is also #= b = 3.3136#