The last paragraph at the Mathworld piece on equilateral triangles gives the answer, and cites Madachy, J. S. Madachy's Mathematical Recreations. New York: Dover, pp. 115 and 129-131, 1979.
EDIT (in response to request from Taha Akbari for more detail): Let the square have horizontal and vertical sides. Consider an equilateral triangle with one vertex at the lower left corner, $A$, of the square, and one vertex at the upper left corner, $B$, of the square, and the third vertex, $Z$, inside the square. Now consider moving the triangle vertex at $B$ to the right, toward the upper right corner, $C$, of the square, while moving $Z$ so as to keep the triangle equilateral. This increases the area of the triangle, since it increases the length of the side of the triangle, since the second vertex, $X$, of the triangle is moving away from the first vertex of the triangle.
Eventually, the triangle vertex $Z$ lies on the right side of the square, and you can't move $X$ any farther right without pushing $Z$ outside the square, so you've made the triangle as large as possible. Now the question is, why are the angles $BAX$ and $ZAD$ 15 degrees (where $D$ is the lower right corner of the square)?
The triangles $BAX$ and $ZAD$ are congruent, since $BA=AD$, $AX=AZ$, and the angles at $B$ and $D$ are equal. So the angles $BAX$ and $ZAD$ are equal. But they, together with the 60 degree angle $XAZ$, add up to the 90 degree angle $BAD$. So, they measure 15 degrees.
The fewer steps the better; I found I had to use the geometric mean construction and I wonder whether this step might be avoidable.
Something like a geometric mean is unavoidable since the problem is a quadratic equation for the side length of the equilateral triangle.
Here is a relatively efficient construction using a geometric mean. The savings is in re-using one side of ABC as the base of the equilateral triangle, and in using a non-perpendicular line to measure altitude, and allowing the semicircle construction of geometric mean to be applied.
On one of the sides, say AB, build an equilateral triangle ABD. Extend line CD to intersect AB at P. Find a length $g$ equal to the geometric mean of PC and PD, and take a point H on PD with PH = $g$. The equilateral triangle whose "height" measured along line PCD (from P) is $g$ has the same area as the given triangle.
The construction of the geometric mean is made easier by extending line CD to E so that P is the midpoint of EC, then using ED as diameter of a circle and taking $g=|PG|$ for PG a perpendicular to CD with G on the circle. Then draw the parallel to AB through H and intersect it with AD (at point K) to cut off the correct side length (AK) of equilateral triangle.
Best Answer
I assume Hartshorne is talking about a Euclidean triangle, so assuming the formula for the area holds, you then want to show that $\sqrt[4]{3}$ is Hilbert constructible.
However, I think the issue here is that the field of Hilbert constructible numbers, $\Omega$, is a proper subset of $K$, the field of constructible numbers, as the square root operation is limited to $a\mapsto\sqrt{1+a^2}$, not $a\mapsto\sqrt{a}$.
Of course $\sqrt[4]{3}\in K$, since it is obtainable from rationals with the $\sqrt{}$ operation. However, take a look at Exercise 28.9 before this, which states a number $\alpha$ is constructible with Hilbert's tools if and only if $\alpha$ is constructible and totally real. Already, we know $\sqrt[4]{3}$ is constructible, but it is not totally real. We see this since the minimal polynomial of $\sqrt[4]{3}$ is $X^4-3$ which has roots $\pm\sqrt[4]{3}$ and $\pm i\sqrt[4]{3}$, so not all the conjugate elements of $\sqrt[4]{3}$ are real. Hence $\sqrt[4]{3}$ is not totally real, and thus not constructible by Hilbert's tools, although it is constructible with ruler and compass.