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

1 answer:

6 0

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.

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Solve the following equation simultaneously 2a-3b+10=0, 10a+6b=8

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