Two corners of an isosceles triangle are at (9 ,6 ) and (3 ,2 ). If the triangle's area is 48 , what are the lengths of the triangle's sides?

1 Answer
Aug 20, 2016

sqrt(2473/13)

Explanation:

Let the distance between the given points be s.
then s^2 = (9-3)^2 + (6-2)^2
s^2 = 52
hence s = 2sqrt13
The perpendicular bisector of s, cuts s sqrt13 units from (9;6).
Let the altitude of the triangle given be h units.
Area of triangle = 1/22sqrt13.h
hence sqrt13h = 48
so h = 48/sqrt13
Let t be the lengths of the equal sides of the given triangle.
Then by Pythagoras' theorem,
t^2 = (48/sqrt13)^2 + sqrt13^2
= 2304/13 + 169/13
= 2473/13
hence t = sqrt(2473/13)