Question

Two corners of an isosceles triangle are at `(1 ,3 )` and `(1 ,4 )`. If the triangle's area is `64`, what are the lengths of the triangle's sides?

Answer 1

Lengths of sides: `{1,128.0,128.0}`

Explanation:

The vertices at `(1,3)` and `(1,4)` are `1` unit apart.
So one side of the triangle has a length of `1`.

Note that the equal length sides of the isosceles triangle can not be both equal to `1` since such a triangle could not have an area of `64` sq. units.

If we use the side with length `1` as the base then the height of the triangle relative to this base must be `128`
(Since `A=1/2 * b * h` with the given values: `64=1/2 * 1 * hrarr h=128`)

Bisecting the base to form two right triangles and applying the Pythagorean Theorem, the lengths of the unknown sides must be
`sqrt(128^2+(1/2)^2)=sqrt(16385)~~128.0009766`
(Notice that the height to base ratio is so great, there is no significant difference between the height and the length of the other side).