Can I square the triangle?

Question

I know I can't construct a square with the same area as a given circle (because $\pi$ is transcendental).

Can I construct (ruler and compass) a square with the same area as a given triangle?

I think I can because Heron's formula only includes a radical (order 2) so it should be constructable. However, I'm not sure because some of the terms under the radical (when expanded out) are of degree 4. Does the fact that they are under the radical and hence 'really' of degree 2 importance help?

Heron's formula:

$$A = \sqrt{s(s-a)(s-b)(s-c)} = \frac14\sqrt{(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4)}$$

Answer 1

Yes.

Two steps:

(1) Since the area of the triangle is half the area of a rectangle with the same base and altitude, construct a rectangle with the same area as the triangle.

(2) Euclid knew how to construct a square with the same area as a given rectangle. Look it up.

Answer 2

See construction number 12 on the page linked here.

A summary of the construction is:

Drop a perpendicular from one vertex of the triangle to the opposite side. You now have a base $b$ and altitude $h$ of the triangle, so the area of the triangle is $\frac12 bh$.

Find the geometric mean of $b$ and $h$, that is, construct a segment of length $\sqrt{bh}$.

Construct a square of side $\sqrt{bh}$. Its area is $bh$.

Join the midpoints of all four sides of the square with segments, forming a smaller square. This square's area is $\frac12 bh$, as desired.