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

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Answer 1 · 2017-12-08T10:53:02

Measure of the three sides are (6.0828, 4.2435, 4.2435)

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

Area of #Delta = 9#
#:. h = (Area) / (a/2) = 9 / (6.0828/2) = 9 / 3.0414 = 2.9592##

#side b = sqrt((a/2)^2 + h^2) = sqrt((3.0414)^2 + (2.9592)^2)#
#b = 4.2435#

Since the triangle is isosceles, third side is also #= b = 4.2435#

Measure of the three sides are (6.0828, 4.2435, 4.2435)