Aregular 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%

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.

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

Letter a i would say