Based on the image, what is the area of parallelogram ABCD?

To find the area of parallelogram ABCD, we can use the given information to calculate the lengths of its sides and height. Let's analyze the triangle ABK. Since angle BAK is 30 degrees and BK is perpendicular to AK, triangle ABK is a right-angled triangle. We have BK = 3. In a right-angled triangle with an angle of 30 degrees, the side opposite the 30-degree angle is half the length of the hypotenuse. Thus, $$AB = 2 cdot BK = 2 cdot 3 = 6$$ Using the Pythagorean theorem for triangle ABK, we can find AK: $$AK^2 + BK^2 = AB^2$$ $$AK^2 + 3^2 = 6^2$$ $$AK^2 + 9 = 36$$ $$AK^2 = 27$$ $$AK = \sqrt{27} = 3\sqrt{3}$$ Now let's analyze the triangle CDH. We are given that CH is perpendicular to DH and CH = $$2\sqrt{3}$$. Since ABCD is a parallelogram, AB is parallel to CD and BC is parallel to AD. Therefore, the distance between AB and CD (which is BK) is equal to the distance between BC and AD (which is CH). However, this is incorrect. Instead, let's find the area of parallelogram ABCD using the height BK and the base AD. Since ABCD is a parallelogram, AD = BC and AB = CD. We already know that AB = 6 and BK = 3. We need to find AD. Let's look at triangle CDH, where CH = $$2\sqrt{3}$$. Since the area can also be expressed as AD * BK, we have: Area = $$AD cdot BK = CD cdot CH$$ Now, let's find DH using the Pythagorean theorem for triangle CDH: $$CD^2 = CH^2 + DH^2$$ However, we don't know CD yet. We know that CD = AB = 6. $$6^2 = (2\sqrt{3})^2 + DH^2$$ $$36 = 12 + DH^2$$ $$DH^2 = 24$$ $$DH = \sqrt{24} = 2\sqrt{6}$$ Now, we know that AD = AK + KD. Also, we know that AK = $$3\sqrt{3}$$. Since ABCD is a parallelogram, AD || BC and AB || CD. Hence AD = KD = AK and CD = AB = 6. Thus, $$AD = AK + KD$$ is not what we are looking for since KD is part of the base, it is not equal to $$DH$$. However, we know that area = base * height. The height BK = 3 and CH = $$2\sqrt{3}$$. The area of parallelogram ABCD can be given as $$AD cdot BK$$ or $$AB cdot CH$$, hence $$AD cdot 3 = 6 cdot 2\sqrt{3}$$ thus: $$AD = \frac{6 cdot 2\sqrt{3}}{3} = 4\sqrt{3}$$ Therefore, the area is $$AD cdot BK = 4\sqrt{3} cdot 3 = 12\sqrt{3}$$. Answer: $$12\sqrt{3}$$