In triangle ABC, AB=AC and ∠BAC = 40°. The bisector of ∠ABC meets side AC at D. What is ∠BDC?

The sum of angles in a triangle is $180^\circ$. Since $AB = AC$, $\angle ABC = \angle ACB$.

$$\angle ABC + \angle ACB + \angle BAC = 180^\circ$$ $$2 \cdot \angle ABC + 40^\circ = 180^\circ$$ $$2 \cdot \angle ABC = 180^\circ - 40^\circ$$ $$2 \cdot \angle ABC = 140^\circ$$ $$\angle ABC = 70^\circ$$

So, $\angle ACB = 70^\circ$.

BD bisects $\angle ABC$.

$$\angle DBC = \frac{1}{2} \cdot \angle ABC$$ $$\angle DBC = \frac{1}{2} \cdot 70^\circ$$ $$\angle DBC = 35^\circ$$

The sum of angles in triangle BDC is $180^\circ$.

$$\angle BDC + \angle DBC + \angle BCD = 180^\circ$$ $$\angle BDC + 35^\circ + 70^\circ = 180^\circ$$ $$\angle BDC + 105^\circ = 180^\circ$$ $$\angle BDC = 180^\circ - 105^\circ$$ $$\angle BDC = 75^\circ$$

The correct option is C.

$$\boxed{\text{C) } 75^\circ}$$

Let $s$ be the side length of the equilateral triangle. The formula for the altitude $h$ of an equilateral triangle is:

$$h = \frac{\sqrt{3}}{2}s$$

Given $h = 6\sqrt{3}$.

$$6\sqrt{3} = \frac{\sqrt{3}}{2}s$$

Divide both sides by $\sqrt{3}$:

$$6 = \frac{1}{2}s$$

Multiply both sides by 2:

$$s = 12$$

The formula for the area $A$ of an equilateral triangle is:

$$A = \frac{\sqrt{3}}{4}s^2$$

Substitute $s = 12$:

$$A = \frac{\sqrt{3}}{4}(12)^2$$ $$A = \frac{\sqrt{3}}{4}(144)$$ $$A = 36\sqrt{3}$$

The correct option is D.

$$\boxed{\text{D) } 36\sqrt{3}}$$

The formula for the circumference $C$ of a circle is $C = 2\pi r$.
Given $C = 24\pi$.

$$24\pi = 2\pi r$$

Divide both sides by $2\pi$:

$$r = 12$$

The formula for the area $A_{circle}$ of a circle is $A_{circle} = \pi r^2$.

$$A_{circle} = \pi (12)^2$$ $$A_{circle} = 144\pi$$

The area of a sector is a fraction of the total area, determined by the central angle $\theta$.

$$A_{sector} = A_{circle} \times \frac{\theta}{360^\circ}$$

Given $\theta = 30^\circ$.

$$A_{sector} = 144\pi \times \frac{30^\circ}{360^\circ}$$ $$A_{sector} = 144\pi \times \frac{1}{12}$$ $$A_{sector} = 12\pi$$

The correct option is B.

$$\boxed{\text{B) } 12\pi}$$