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

1 Answer
Oct 12, 2016

Sides:
#color(white)("XXX"){3.162, 2.025, 2.025}#
or
#color(white)("XXX"){3.162,3.162,1.292}#

Explanation:

There are two cases that need to be considered (see below).

For both cases I will refer to the line segment between the given point coordinates as #b#.
The length of #b# is
#color(white)("XXX")abs(b)=sqrt((4-1)^2+(2-3)^2)=sqrt(10)~~3.162#

If #h# is the altitude of the triangle relative to base #b#
and given that the area is 2 (sq.units)
#color(white)("XXX")abs(h)=(2xx"Area")/abs(b)=4/sqrt(10)~~1.265#

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Case A: #b# is not one of the equal sides of the isosceles triangle.
enter image source here
Notice that the altitude #h# divides the triangle into two right triangles.
If the equal sides of the triangle are denoted as #s#
then
#color(white)("XXX")abs(s)=sqrt(abs(h)^2+(abs(b)/2)^2~~2.025#
(using the previously determined values for #abs(h)# and #abs(b)#)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Case B: #b# is one of the equal sides of the isosceles triangle.
enter image source here

Note that the altitude, #h#, divides #b# into two sub-line segments which I have labelled #x# and #y# (see diagram above).

Since #abs(x+y)=abs(b)~~3.162#
and #abs(h)~~1.265#
(see prologue)

#color(white)("XXX")abs(y)~~sqrt(3.162^2-1.265^2)~~2.898#

#color(white)("XXX")abs(x)=abs(x+y)-abs(y)#
#color(white)("XXXX")=abs(b)-abs(y)#
#color(white)("XXXX") ~~3.162-2.898~~0.264#

and
#color(white)("XXX")abs(s)=sqrt(abs(h)^2+abs(x)^2)=sqrt(1.265^2+0.264^2)~~1.292#