A base angle of an isosceles triangle measures 30°, and the length  of one of the legs is 12. What is the length of the altitude drawn from the vertex to the base of the triangle?

To solve this problem, we need to use some trigonometry.

1. First, we know that an isosceles triangle has two equal sides and two equal angles. The base angles are the ones at the bottom of the triangle, and they are equal. In this case, they are each 30°.

2. The altitude of the triangle splits the isosceles triangle into two 30-60-90 right triangles. In a 30-60-90 triangle, the sides are in the ratio of 1:√3:2.

3. The longest side (the hypotenuse) is across from the 90° angle, the shortest side (which we'll call "a") is across from the 30° angle, and the remaining side (which we'll call "b") is across from the 60° angle.

4. In this case, the hypotenuse of the 30-60-90 triangle is the leg of the isosceles triangle, which is 12. So, we can say that 2a = 12, which means a = 6.

5. The altitude of the triangle is the side "b" of the 30-60-90 triangle, which is across from the 60° angle. We can find it by using the formula b = a√3.

6. Substituting a = 6 into the formula, we get b = 6√3.

So, the length of the altitude drawn from the vertex to the base of the triangle is 6√3 units.