Two corners of an isosceles triangle are at (9 ,2 ) and (4 ,7 ). If the triangle's area is 64 , what are the lengths of the triangle's sides?

1 Answer
Nov 2, 2016

Solution. root2{34018}/10~~18.44

Explanation:

Let's take the points A(9;2) and B(4;7) as the base vertices.
AB=root2{(9-4)^2+(2-7)^2}=5root2{2}, the height h can be taken out from formula of the area 5root2{2}*h/2=64. In such a way h=64*root2{2}/5.

The third vertex C must be on the axis of AB that is the line perpendicular to AB passing through its medium point M(13/2;9/2).
This line is y=x-2 and C(x;x-2).

CM^2=(x-13/2)^2+(x-2-9/2)^2=h^2=2^12*2/5^2.
It gets x^2-13x+169/4-2^12/25=0 that solved yelds to values possible for the third vertex, C=(193/10,173/10) or C=(-63/10,-83/10).

The length of the equal sides is AC=root2{(9-193/10)^2+(2-173/10)^2}=root2{(103/10)^2+(-153/10)^2}=root2{34018}/10~~18.44