Two corners of an isosceles triangle are at #(2 ,3 )# and #(1 ,4 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

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Answer 1 · 2016-10-21T20:13:49

The 3 sides are #90.5, 90.5, and sqrt(2)#

Let b = the length of the base from #(2,3)# to #(1, 4)#

#b = sqrt((1 - 2)^2 + (4 - 3)^2)#

#b = sqrt(2)#

This cannot be one of the equal sides, because the maximum area of such a triangle would occur, when it is equilateral, and specifically:

#A = sqrt(3)/2#

This conflicts with our given area, #64 units^2#

We can use the Area to find the height of the triangle:

#Area = (1/2)bh#

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

#h = 64sqrt(2)#

The height forms a right triangle and bisects the base, therefore, we can use the Pythagorean theorem to find the hypotenuse:

#c^2 = (sqrt(2)/2)^2 + (64sqrt(2))^2#

#c^2 = 8192.25#

#c~~ 90.5#