## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$$h=2\sqrt{3}$$
We draw a line straight through the middle of the triangle to form two right triangles. We see that the base of this right triangle will be 2 and its hypotenuse will be 4. Thus: $$h=\sqrt{4^2-2^2}\\\ h=\sqrt{12}\\ h=2\sqrt{3}$$