Two corners of an isosceles triangle are at #(2 ,1 )# and #(7 ,5 )#. If the triangle's area is #4 #, what are the lengths of the triangle's sides?

1 Answer
Mar 21, 2016

There are three possibilities:
#color(white)("XXX"){6.40,3.44,3.44}#
#color(white)("XXX"){6.40, 6.40, 12.74}#
#color(white)("XXX"){6.40, 6.40, 1.26}#

Explanation:

Note the distance between #(2,1)# and #(7,5)# is #sqrt(41)~~6.40#
(using the Pythagorean Theorem)

Case 1
If the side with length #sqrt(41)# is not one of the equal length sides

then using this side as a base the height #h# of the triangle can be calculated from the area as
#color(white)("XXX")((hsqrt(41))/2=4) rArr (h=8/sqrt(41))#

and the two equal length sides (using Pythagorean Theorem) have lengths
#color(white)("XXX")sqrt((sqrt(41)/2)^2+(8/sqrt(41))^2) ~~3.44#

Case 2
If the side with length #sqrt(41)# is one of the sides of equal length

then if the other side has a length of #a#, using Heron's Formula
#color(white)("XXX")#the semiperimeter, #s# equals #a/2+sqrt(41)#
and
#color(white)("XXX")"Area" = 4 = sqrt((a/2+sqrt(41))(a/2)(a/2)(sqrt(41)-a/2))#
#color(white)("XXXXXXXXX")=a/2sqrt(41-a^2)#
which can be simplified as
#color(white)("XXX")a^4-164a^2+256=0#
then substituting #x=a^2# and using the quadratic formula
we get:
#color(white)("XXX")a=12.74 or a=1.26#