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

My approach:
Divide the hexagon into 6 equilateral triangles:

1. Area of hexagon. = 6 * area of a triangle (As all triangles are equilateral with same side).
= \(6 * \sqrt{3} /4 * 6^2\)
= \(54\sqrt{3}\)

2. Area of orange part in the diagram:
\(pie*r^2* 120/360\)
\(pie*9* 1/3\)
\(3* pie\)

Total of 6 areas like this + the central circle = \(18 * pie + 9 * pie = 27 * pie\)

Now area of black shaded region = area point 1 - area point 2.
= \(54\sqrt{3} - 27 * pie\)

Answer E.

Can we calculate the area of the hexagon by calculating the area of 1 equilateral triangle andmultiply by 6 to get the area of the hexagon?

Area of Equilateral Triangle = (s^2/4)*\sqrt{3}

Area of a regular hexagon (6 equilateral triangles) = \(6a^2\)\(\frac{\sqrt{3}}{4}\) = \(6*36*\)\(\frac{\sqrt{3}}{4}\)

= \(54\sqrt{3}\)

Internal angles of regular hexagon = 120 degrees. So each circle has a third of its area inside the hexagon. So 6 circles = \(6*\frac{1}{3}*\pi*r^2\) = \(18\pi\)

Area of the circle entirely inside = \(9\pi\)

So required area = \(54\sqrt{3}\) - \(18\pi\) - \(9\pi\)

= \(54\sqrt{3}\) - \(27\pi\)

Answer is E

1. Check the diagram below:
2. Cannot follow you. How did you get that the hypotenuse, which is the side of the hexagon, equals to \(\sqrt{72}\)?
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Hello ,

Just adding this as an alternative approach which i think would be pretty easy to get through :

Consider :

Regular Hexagon : ABCDEF

thus Equilateral triangle : ABO (See attached image )

we need to calculate the shaded portion so what we can do is Calculate the area of equilateral triangle ABO and deduct the area of smaller quadrants within that triangle (that are formed by extending the radius of all the triangle's) : so here we will deduct quadrant made from Circle A , Circle B and Circle O from the bigger equlilateral triangle and hence we can conclude the remaining area !

Now : as regular hexagon (ab=6) thus (a+b / 2) =3 the radius of all the circle

Area of ABO (bigger equilateral triangle ) = [[square_root]3 /4 ]s*s so ([square_root]3 / 4) *6 *6 = 9 [square_root]3

Area of 1 smaller quadrant : pie * r^2 / 6 ( because if we extend all the radius of adjacent circles to O six equal quadrants would be formed because ABCDEF is a regular hexagon )
thus : pie *3^2 /6 = 3pie /2

now as 3 quadrants the total area would be (3pie /2) * 3 = 9 pie /2

no subtract 3 quadrant area from the total area of equilateral triangle to get the area of shaded portion :

subtract : 9 [square_root]3 - 9pie /2 = [18 [square_root]3 - 9 pie /2 ] multiply this into 6 to get the area of all the shaded portions thus

54 [square_root]3 - 27 pie .... Option d

Sorry for pie !

Work hard , play hard !

Bunuel
I have a doubt
here ab=6cm
so why we are taking both circles or rather all to be 3cm radius
why cannot it be like circle a has radius 2 and b has 4
also what is use of the information given in qsn " If each circle is tangent to the two circles adjacent to it and to circle O," plz reply