This is a good approach to the problem. We can also approach like this: If we remember our special triangles, we'll remember that in a 30-60-90 triangle, the sides are $$x-x\sqrt{3}-2x$$ Here, our shortest leg is half the length of the hypotenuse, just like in the special triangle. This means that we have a 30-60-90 triangle, and the other leg of the triangle must be the length of the shortest leg times the square root of 3:
$$\frac{y}{2}-\frac{y\sqrt{3}}{2}-y$$
So now we have the length of our other leg without using the pythagorean theorem. We can then solve for area:
$$\frac{1}{2}\cdot\frac{y}{2}\cdot\frac{y\sqrt{3}}{2}$$ $$\frac{y^2\sqrt{3}}{8}$$
Now we can plug 2 in for y^2: $$\frac{2\sqrt{3}}{8}$$ $$\frac{\sqrt{3}}{4}$$
This approach isn't necessarily any easier on this particular problem, but special triangles can speed things up on other problems.

LUANDATO wrote:

A flat triangular cornfield has the dimensions shown in the figure above. If y^2=2, what is the area of the field in square miles?

$$A.\ \frac{1}{4}$$
$$B.\ \frac{\sqrt{3}}{4}$$
$$C.\ \frac{1}{2}$$
$$D.\ \frac{\sqrt{3}}{2}$$
$$E.\ 1$$

Since y^2 = 2, y = √2.

Let's now determine the base of the triangle using the Pythagorean theorem:

b^2 + (√2/2)^2 = (√2)^2

b^2 + 1/2 = 2

b^2 = 3/2

b = √(3/2)

So the area is:

1/2 x √2/2 x √(3/2)

√2/4 x √(3/2) = √3/4

Answer: B