Regular hexagon ABCDEF has a perimeter of 36. O is the cente

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

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}

Sure we can: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}\), as side=2*raius, \(Area_{equilateral}=4r^2*\frac{\sqrt{3}}{4}=r^2*\sqrt{3}\) --> \(Area_{hexagon}=6*Area_{equilateral}=6*\sqrt{3}*r^2\);

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

Area of the shaded region: \(6*\sqrt{3}*r^2-3\pi{r^2}\).

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

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.

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 \(\sqrt{72}\)?

Please check the explanation in the attached file

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 !

While this Q can be solved in multiple ways one of which includes utilizing knowledge of trigonometry, the most intuitive way is visualizing the diagram and solving for it.

(1) Area of a sector = Pi *R^2 (Arc angle /360)
= Pi *3^3 (120/360)
= 3*Pi

Area of 6 sectors = 6 * ( 3*Pi) = 18*Pi

(2) Area of innermost circle = Pi * R^2 = Pi * 3^2 = 9* Pi

(3) Area of hexagon = 6 * (Area of equilateral triangles)
= 6 * (Sq Rt 3/4 * (side)^2 )
= 6 * (Sq Rt 3/4 * 36 )
= 6 * (9 Sq Rt 3)
= 54 Sq Rt 3

Therefore , area of shaded part = Area of hexagon - (area of 6 sectors + Area of innermost circle)
= 54 Sq Rt 3 - (18*Pi +9* Pi )
= 54 Sq Rt 3 - 27* Pi

Option E

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

given hexagaon of perimeter 36 has side =6
and a hexagon has 6 equilateral ∆'s so the area of the hexagon ; 6*36*√3/4 ; 54√3
to find the area of shaded region we see that for the 6 circles each has radius of 3 and is at 120* so each circle has area coverage under hexagon of 120/360 * pi * 3^2 = 3pi
since 6 circles are present ; 6*3Pi = 18 pi
and area of central circle of radius 3 (O) ; 9pi
we get \(54\sqrt{3}-27\pi\) shaded region area.
IMO E

Rule 1: you can Divide a Regular Hexagon into 6 Equilateral Triangles with Side = Side of the Regular Hexagon by connecting a Straight Line from Each Vertex to the Center of the Regular Hexagon

Also, I'm assuming that All of the Circles are Congruent and Equal.


Since the Perimeter of the Regular Hexagon = 36 and Every Side is Congruent in a Regular Hexagon:

6 * (side) = 36

Side of Regular Hexagon = 1 Side of the 6 Equilateral Triangles = 6


Further, the way the Diagram is drawn, from Point F to Point C (through Center O ---- the Diagonal FC = 2 * (Side Length of 1 Equilateral Triangle) = 2 * 6 = 12


2nd)Since there are 4 Equal Radii on a Co-linear Diagonal from Point F to Point C, we can find the Radius of Each of the Circles.

12/4 = 3 = Radius of 1 Circle


3rd) The Degree measure of Each Interior Angle of a Regular Hexagon = 120 degrees.

Therefore, Each Circle that has 1 of the Hexagon's Vertices as its Center, has a Central Angle = 120 Degrees that is contained within the Hexagon

This means that: 120 / 360 = 1/3 of Each of these Circle's AREA is located within the Regular Hexagon.


Solution) to Find the AREA that is Shaded inside the Regular Hexagon, but outside the Circles:

(Area of Regular Hexagon w/ Side 6) - (Entire Area of Circle in the Center) - (the 1/3rd Area of Each of the 6 Circles around the Perimeter of the Regular Hexagon) = SOLUTION

[6 * (6)^2 * sqrt (3) / 4 ] - [ (pi) * (3)^2 ] - 6 * [ 1/3 * (pi) * (3)^2 ] = SOLUTION

[54 * sqrt (3)] - [9 (pi)] - [6 * 9 (pi) / 3] =

54 * sqrt (3) - 27 (pi)

Answer -E-

Area of regular polygon = s^2 x n/(4 x tan (180/4)
Plug values = 54√3
Area of circular regions
Total 120 degree segments = 6+3 = 9 . Therefore area = 9*9pi /3 = 27pi
Hence 54✓3 -27pi

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Okay so a regular hexagon can be divided into 6 equilateral triangles.

How do we know that the circles mentioned in the question above are equal ?

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Don't forget the inner circle!!!

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. 108−18π
B. 543‾√−9π
C. 543‾√−18π
D. 108−27π
E. 543‾√−27π

1. Find the area of the hexagon. If you ever forget the formula, just remember that a regular hexagon is 6x the area of the equilateral triangles embedded inside.

Area(hexagon) = ( 3√4 x 6^2 ) x 6 = 54√3

2. Find the area of one circle

Area(circle) = 9 x pi

3. Find the area of the circle that is inside the hexagon.

9 pi /3 = 3 pi

We get 3 pie because we know the 2 angles in the centre sum to 120 (since equilateral triangle). So 360/120 = 3. Since we get a parallelogram forming, the opposite angle must also be 120.

4. Multiply the area of 1/3 of a circle by 6.

3 pi x 6 = 18 pi

5. Add the area of the inner circle (9 pi)

18 pi + 9 pi = 27 pi

6. Subtract the pieces of the circle in step 5 from the area of the hexagon to get the shaded portion.

54√3 - 27 pi

Given: 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.

Asked: 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)?

Area of the hexagon = \(6*\sqrt{3}*6^2/4 = 54\sqrt{3}\)

Radius of each circle = 6/2 = 3

Area of the 6 parts of the circles = \(6* \pi * 3^2 * 120/360 = 18 \pi\)
Area of the full circle =\( 3^2 \pi= 9 \pi\)

The area of the shaded region (inside the hexagon but outside the circles) = \(54\sqrt{3} - 18\pi - 9\pi = 54\sqrt{3} - 27\pi\)

IMO E

Hi Bunuel please advise if my approach is correct since each interior angle in a hexagon is 120, and there are six circle surrounding internal circle. Then area of shaded region= area of 6 equilateral triangle - area of 3 circle(3* π(3^2)

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