A regular hexagon is inscribed (vertices lie on the circles) in a circle.

a. what percent of the area of the circle is overlapped by the area of the inscribed regular hexagon?

b. if the radius of the circle is tripled and a regular hexagon is inscribed in the new circle, what percent of the area of the circle is overlapped by the area of the inscribed regular hexagon?

Radius of the circle: r

Area of the circle: A1=pi r^2

A1=3.141592654 r^2

Area of the regular hexagon: A2 = (6) (1/2) (r) (r) sin 60°

A2 = (6/2) r^2 sqrt (3) / 2

A2=3 sqrt (3) r^2 / 2

A2=3 (1.732050808) r^2/2

A2=2.598076212 r^2

a. What percent of the area of the circle is overlapped by the area of the inscribed regular hexagon?

Percentage: P = (A2/A1) * 100%

P=[2.598076212 r^2 / (3.141592654 r^2) ]*100%

P = (0.826993343) * 100%

P=82.6993343%

P=82.7%

Answer: 82.7 percent of the area of the circle is overlapped by the area of the inscribed regular hexagon.

b. if the radius of the circle is tripled and a regular hexagon is inscribed in the new circle, what percent of the area of the circle is overlapped by the area of the inscribed regular hexagon?

The same percentage.

Answer: 82.7 percent of the area of the circle is overlapped by the area of the inscribed regular hexagon.