Question

Two corners of an isosceles triangle are at `(1 ,2 )` and `(3 ,1 )`. If the triangle's area is `2`, what are the lengths of the triangle's sides?

Answer 1

Find the triangle's height and use Pythagoras.

Explanation:

Start by recalling the formula for the height of a triangle `H=(2A)/B`. We know that A=2, so the beginning of the question can be answered by finding the base.
The given corners can produce one side, which we will call the base. The distance between two coordinates on the XY plane is given by the formula `sqrt((X1-X2)^2+(Y1-Y2)^2)`. Plug`X1=1, X2=3, Y1=2,` and `Y2=1` to get `sqrt((-2)^2+1^2)` or `sqrt(5)`. Since you don't have to simplify radicals in work, the height turns out to be `4/sqrt(5)`.

Now we need to find the side. Noting that drawing the height inside an isosceles triangle makes a right triangle consisting of half of the base, the height and the leg of the full triangle, we find that we can use Pythagoras to calculate the hypotenuse of the right triangle or the leg of the isosceles triangle. The base of the right triangle is `4/sqrt(5)/2` or `2/sqrt(5)` and the height is `4/sqrt(5)`, meaning that the base and height are in a `1:2` ratio, making the leg `2/sqrt(5)*sqrt(5)` or `2`.