Question

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?

Answer 1

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`