draw the figure and solve?

sanju09 wrote: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) 118 – 18 π
(B) 54 √3 – 9 π
(C) 54 √3 – 18 π
(D) 108 – 27 π
(E) 54 √3 – 27 π
[spoiler]Is the stem sufficient to draw the figure and solve?[/spoiler]

See the figure.



Since ABCDEF is a regular hexagon with perimeter 36, so each side of the hexagon = 6
So, the radius of circles O, A, B, C, D, E, and F is 6/2 = 3
Hence, area of circle with center at O = (pi)(3)² = 9(pi)
Now, each internal angle of hexagon = (6 - 2) * 180º/6 = 120º {The Interior Angle (or Internal Angle) of a regular polygon with "n" sides can be calculated using: (n - 2) * 180° / n}
Next, Area of Sector = 1/2 * (θ * pi/180) * r² (when θ is in degrees)
So, Area of the circles that are inside the hexagon = 6 * {1/2 * (120 * pi/180) × 3²} = 18(pi)

A regular hexagon has 6 equilateral triangles. Now, we need to find the area of hexagon.
Look at the following figure.



First we'll find the area of 1 equilateral triangle.
By Pythagoras Theorem, OD = √{s² - (s/2)²} = √(3s²/4) = s√3/2
So, area of 1 equilateral triangle = (1/2)* s * s√3/2 = s²√3/4
Therefore, area of hexagon ABCDEF = 6s²√3/4 = 6(36)√3/4 = 54√3

So, Area of shaded region = 54√3 - {(9(pi) + 18(pi)} = 54√3 - 27(pi)

The correct answer is E.