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

Guys,
I did not understand why we can not form two trapezoids of this hexagon, and solve through it. The official solution says, that we have to form 6 equiaterial triangles in the hexagon and form through it. Forming trapezoids, I got 108 as a square of the hexagon.
Please. elaborate
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you can use trapezoids as well..

total perimeter = 36...thus side = 36/6 = 6

radii of circle = 6/2 = 3

now area of 2 trapeziods = 2* 1/2 (sum of parallel sides) height.

=(6+12)h = 18h

Now to find the height....consider any triangle and u will find the height is 6sin60 = 6*\sqrt{3}/2
= 3\sqrt{3}

Thus total area of 2 trapezoids = 18*3\sqrt{3} = 54\sqrt{3} --1

We have to subtract area of one circle + 6*(area by arc which is subtending 120* at each centre)

Thus we have area of 1+ 2 circles = 3circle = 3π r^2 = 3π * 3*3 = 27π ----2

Subtract 2 from 1

we get ans =E
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In the figure above, all the circles are congruent, all the vertices of the regular hexagon coincide with the center of a circle, and all the circles have at most 1 point in common with any other circle. If a side of the hexagon is 12, what is the area of the shaded region?

Answers:
A. \(108*(2*sqrt(3)-\pi)\)
B. \(108*(4*sqrt(3)-3\pi)\)
C. \(36*(12-3\pi)\)
D. \(36*(2*sqrt(3)-3\pi)\)
E. None of the above.



Thanks!

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}

The simplest way to calculate above problem:-
Area of shaded region= Area of hexagon - area of circumscribed circle - area of each arcs inscribed by the hexagon.

Area of hexagon= (3√3/2) (side)2
= (3√3/2) x 36
= 54√3

Area of circle=9π
Combined area of six arcs inscribed= 6x (120/360) π 32
= 18 π

Thus, Area of shaded region = 54√3 - 9π - 18 π
Area of shaded region = 54√3 - 27 π

When you say , Each triangle is 1/3 rd inside do you mean as angle of arc if 120degree and 120/360 = 1/3 that means that sector of circle is inside Hexagon ???

Am I right ?
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The Area we are looking for is equal to the area of the hexagon minus the area of the circle with center O and minus the slices of the other 6 circles.

The slices of these 6 circles have an equal angle, and the sum of these angles equals the sum of the angles of the hexagon. We know that the sum of the angles of a polygon of n=6 sides is 180*(n-2). This means the sum of these angles is \(180*(6-2)=180*4=360*2\). Therefore, the area of the 6 slices equals the area of 2 circles.

If we add the area of the circle in the middle, we have that the "white" area inside the hexagon equals the area of 3 circles.

So the area we are looking for is the area of the hexagon (A) minus the area of 3 circles (B).

(A) We split the hexagon in 6 isosceles triangles. In an isosceles triangle with side \(6=2*X\), the height must be \(\sqrt{3}*X\). Therefore, because X=3, the height must be \(\sqrt{3}*3\). And then, the area of one triangle is \(\frac{1}{2}\), times the base of 6, times the height of \(\sqrt{3}*3\). This equals \(3*\sqrt{3}*3\). Times 6 triangles to find the area of the hexagon, we have: \(54*\sqrt{3}\)

(B) The area of 1 circle of radius 3 is \(pi*3^2\). Times 3 circles to find the "white" area inside of the hexagon, is: \(27*pi\).

Then, the area we are looking for is (A) minus (B):

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

Solution E.

Hi Bunuel,

I have applied central angle theorem since two adjacent equilateral triangle forms 120 degree
then central angle of all the six circle would be 120 degree.
is it correct?

Thanks & Regards,
WarriorGmat

Is this true for any trapezoid then?

HI Bunnel,

I did it using equilateral triangle. But I got a query in following statement.

Not shaded area inside the hexagon is the size of 3 circles = \(3\pi{r^2}\).

Here I consider hexagon as a 6 equilateral triangle. so each angle of the triangle is 60. Now if I will multiply 60*6 + 1 circle inside hexagon it becomes 2 circles.

How you are considering it as 3 circles inside the hexagon.

Please clarify

Thanks.