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

1 Answer
Jan 1, 2018

Lengths of sides: #{1,128.0,128.0}#

Explanation:

The vertices at #(1,3)# and #(1,4)# are #1# unit apart.
So one side of the triangle has a length of #1#.

Note that the equal length sides of the isosceles triangle can not be both equal to #1# since such a triangle could not have an area of #64# sq. units.

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If we use the side with length #1# as the base then the height of the triangle relative to this base must be #128#
(Since #A=1/2 * b * h# with the given values: #64=1/2 * 1 * hrarr h=128#)

Bisecting the base to form two right triangles and applying the Pythagorean Theorem, the lengths of the unknown sides must be
#sqrt(128^2+(1/2)^2)=sqrt(16385)~~128.0009766#
(Notice that the height to base ratio is so great, there is no significant difference between the height and the length of the other side).