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