Question

Two corners of an isosceles triangle are at `(5 ,2 )` and `(2 ,1 )`. If the triangle's area is `6`, what are the lengths of the triangle's sides?

Answer 1

`sqrt{10}`, `13/sqrt{10}` and `13/sqrt{10}`, respectively.

Explanation:

One side of the triangle is immediate - its length is the distance between the points `(5,2)` and `(2,1))`, which is
`sqrt{(5-2)^2+(2-1)^2}=sqrt{10}`

Now, there are two possibilities - either this is the base of the isosceles triangle, or it is one of the two equal sides. Let us tackle these cases one by one.

First, if this side is the base of our triangle, then its perpendicular bisector passes through the opposite vertex. The length of this segment, which is the height, can be easily found from the area to be `2 xx 6 /sqrt{10}`

We can easily use Pythagoras' theorem to find out that the length of the other two sides is given by

`sqrt{(12/sqrt{10})^2+(sqrt{10}/2)^2}=sqrt{144/10+10/4}=sqrt{169/10}=13/sqrt{10}`

As for the second case, if you assume that the length of the base is `a`, the height must be `12/a`, and thus `a` must satisfy

`(a/2)^2+(12/a)^2 = 10`

It can be easily seen that this will give rise to a quadratic equation in `a^2` which has no real roots.