area of a regular six pointed star

Firstly, find the area of the large equilateral triangle that is inscribed in the circle. Draw lines from the centre to each of the three vertices of the triangle. The area is then $3 \cdot \frac{1}{2} (4)(4) \sin 120º = 12 \sqrt{3}$ using the formula $A = \frac{1}{2}ab \sin C$ .

Then subdivide the figure into 12 smaller equilateral triangles. The large equilateral triangle covers $5 + 3 + 1 = 9$ of these smaller triangles. Therefore, the area of the figure is $12 \sqrt{3} \cdot \frac{12}{9} = \boxed{16 \sqrt3}.$

Draw lines at the base of each of the six triangles. This breaks the figure up into six triangles and a regular hexagon. The hexagon can further be broken up into 6 identical copies of the same triangles. Therefore, the total area is 12 (=6+6) times the area of each triangle. We can draw a radius of the circle to the one of the vertices on the triangle which lies on the circle. We can see that the radius is twice the height of the triangle. Because the radius of the circle is 4, this means that the height of the triangle is 2. By the Pythagorean Theorem, this means the side of each triangle is 4 sqrt(3) / 3. Therefore, the area of each triangle is (1/2) (side length) height = (1/2) (4 sqrt(3)/3) 2 = 4 sqrt(3) / 3. Therefore, the total area, which is 12 times the area of each triangle, is 16 sqrt(3)