Question

Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?A. \(108-18\pi\)B. \(54\sqrt{3}-9\pi\)C. \(54\sqrt{3}-18\pi\)D. \(108-27\pi\)E. \(54\sqrt{3}-27\pi\)

Answer 1

Answer:

E. (54\sqrt{3}-27\pi\)

Step-by-step explanation:

I draw the figure described by the statement down there so you can see easily how to solve it.

Notice 3 things:

- The length of one side of the hexagon is simply 36 divided by 6. = 6 This is beacause, by nature, every single side of the hexagon has the same length.

- The radius of each circle is half the length of one side of the hexagon. This would mean that the radius of each circle is 3.

- The outer circles have 120°/360° of its area enclose. We get this value because by definition, the hexagon can be divided into 6 regular triangles, each one with every angle equal to 60°. The size of the internal angles of the hexagon will be twice this value.

So, in summary, The area enclosed by the circles is expressed like this:

Acircles = 6 * $\frac{120}{360}$ * π * (r)² + π*r² = 3 * π * r² = 27 π

Now all we need is the area of the regular hexagon, which is simply:

Ahexagon = $\frac{1}{2}$* p * a

Where p is the perimeter, given by the problem, and a is its apothem, the distance between the center of the hexagon and the middle of one of its sides, that is found by multiplying the length of a side of the hexagon times $\frac{\sqrt{3}}{2}$.

Ahexagon = $\frac{1}{2}$ * 36 * 6 * $\frac{\sqrt{3}}{2}$ = 54$\sqrt{3}$.

Then, the area of the shaded area is equal to:

Ashaded = Ahexagon - Acircles = 54$\sqrt{3}$ - 27π