Find the area of the shaded region. round answers to the nearest tenth. assume all inscribed polygons are regular.

We can calculate for the area of the shaded region by 1) calculating the area of circle, 2) calculating for area of triangle, and 3) calculating for area of the segment below the triangle. Then subtract 2 and 3 from 1.

1) Area of circle, Ac = π r^2

Ac = π (6)^2

Ac = 113.10

2) Area of triangle, At

The triangle is an equilateral triangle with angles on each corner equal to 60 degrees. Meanwhile, the 3 angles at the center is 120 degrees each since a circle is 360 degree. We know that the radius (line from centerpoint to corner) is equivalent to 6. Using the cosine law, we can calculate for the length of one side.

s^2 = 6^ + 6^2 – 2 (6) (6) cos 120

s^2 = 108

s = 10.4

Since this is an equilateral triangle therefore all sides are equal. The area for this is:

At = (sqrt3 / 4) * s^2

At = 46.77

3) Area of segment, As is given as:

As = [ (r^2) / 2 ] ( θ rad – sin θ )

As = [(6^2) / 2] (120 * π / 180 – sin 120)

As = 22.11

Therefore the area of shaded region is:

A = 113.10 - 46.77 - 22.11

A = 44.22

answer: 1722

step-by-step explanation:

given: the mass of saturn=

the mass of mercury=

the number of times the mass of saturn is greater than the mass of mercury

hence, the mass of saturn is 1722 times greater than the mass of mercury.