Hexagon Area

Let $O$ be the center of the hexagon. $ABO$ is an equilateral triangle, and specifically $\angle AOB$ is 60 degrees. Since $A^\prime$ is the midpoint of $AB$ . The triangle $A^\prime BO$ is a right-triangle with $A^\prime B$ has length 2, and the measure of $\angle A^\prime OB$ is 30.

We then see that $A^\prime O$ has length $2\sqrt{3}$ . So triangle $A^\prime B^\prime O$ is an equilateral triangle with side length $2\sqrt{3}$ . Let $P$ be the midpoint of $A^\prime B^\prime$ . Then $A^\prime P O$ is another 30-60-90 right triangle. $\angle A^\prime O P$ has measurement 30, and $A^\prime P$ has length $\sqrt{3}$ . So $OP$ , which is the height of triangle $A^\prime B^\prime O$ , has length 3.

Therefore, triangle $A^\prime B^\prime O$ has area $3\sqrt{3}$ . Hexagon $A^\prime B^\prime C^\prime D^\prime E^\prime F^\prime$ is composed of six such equailateral triangles, and hence $X$ is $18\sqrt{3}$ . $X^2$ is thus 972.

As angle in a regular hexagon is 120 degree So angle ABC =120 degree But also 2 AB 0.5 BC 0.5 cos(120)=AB 0.5 AB 0.5+BC 0.5 BC 0.5-A'B' A'B' 2 2 2 (-1/2)=4+4-A'B'^2 A'B'=2 SQRT(3) NOW AREA OF HEXAGON IS 6 AREA OF EQUILATERAL TRIANGLE WITH SIDE LENGTH 2sqrt(3) Area=6 sqrt(3) (1/4) 4 3 Area area=18 18 3=972 answer

To calculate the area of the A'B'C'D'E'F' hexagon all we need to do is calculate its side length,because it is a regular hexagon. Since A' is the midpoint of AB, it follows that: $A'B=BB'=\frac{AB}{2}=2$ . We also know that $\angle{A'BB'}=(180-\frac{360}{6})^\circ=120^\circ$ . Now we are ready to apply the law of cosines to A'BB' in order to compute A'B'.

${A'B'}^2={A'B}^2+BB'^2-2A'B\cdot{BB'}cos(120^\circ)$

$A'B'^2=12$

$X=6\cdot{(A'B'O)}$ where O the center of mass of both hexagons and (A'B'O) the area of the A'B'O triangle. $(A'B'O)=\frac {A'B'^2\cdot\sqrt{3}}{4}=3\sqrt{3}$

Thus it follows that $X=18\sqrt{3}$

and $X^2=972$

We know how to find the area of hexagon $ABCDEF$ , as well as triangles $A'F'A$ . So one way to find the area of $A'B'C'D'E'F'$ is to find the area of $ABCDEF$ and then subtract the area of the $6$ triangles each with an area congruent to that of $A'F'A$ .

The area of $ABCDEF$ is the area of $6$ equilateral triangles with side length $4$ . We can split each into $2$ equal 30-60-90 (degree) triangles, with side lengths $2$ , $2\sqrt{3}$ , and $4$ . So the area of each equilateral triangle is $2(\frac{1}{2}bh) = 2(\frac{1}{2} \times 2 \times 2\sqrt{3} )= 4\sqrt{3}$ , and the area of $ABCDEF = 6 \times 4\sqrt{3} = 24\sqrt{3}$ .

The area of $A'F'A$ can be found by splitting it into $2$ equal 30-60-90 degree triangles as well, with side lengths $1$ , $\sqrt{3}$ , and $2$ . So the area of $A'F'A$ is $2(\frac{1}{2}bh) = 2(\frac{1}{2} \times 1 \times \sqrt{3}) = \sqrt{3}$ .

Then $X = ($ area of $ABCDEF)- 6($ area of $A'F'A)$ $= 24\sqrt{3} - 6(\sqrt{3})$ $= 18\sqrt{3}$ .

Thus, $X^2 = 972$ .

The hexagon is composed of 6 equilateral triangles of side 4 and interior angle \frac {\pi}{3}

Area of a triangle = \frac {vertical height {\times} base}{2}

Vertical height = 4 \times \sin \frac {\pi}{3} = 2\sqrt {3}

Therefore the area of one triangle is \frac {4 \times 2 \sqrt {3}}{2} = 4 \sqrt {3}

Hence total area of hexagon = 6 \times 4\ sqrt{3] = 24 \sqrt {3}

When smaller hexagon is inserted 6 smaller triangles are also formed between the two hexagons. These triangles are isosceles with side length 2 cm and included angle \frac {2{\pi}}{3}

Each of these triangles can be split into two right angled triangles with hypotenuse 2 cm and interior angles of \frac {\pi}{3} and \frac {\pi}{6}

The height of the smaller triangle is 2 \times \sin \frac {\pi}{6} = 2 \times \frac {1}{2} = 1

The length of the base of each right angled triangle is 2 \cos \frac {\pi}{6} = \sqrt {3}

Each right angled triangle has area of \frac {1 \times \sqrt {3}}{2} = \frac {\sqrt {3}}{2}

There are 12 of these small right-angled triangles and so their total area is 12 \frac {\sqrt {3}}{2} = 6 \sqrt {3}

Subtracting the areas for the small triangles from the area of the large hexagon gives 24 \sqrt {3} - 6 \sqrt {3} = 18 \sqrt {3} = X

So X^2 = {18 \sqrt {3}}^2 = 972

The interior angle of a hexagon is equal to 120 degrees. By connecting the midpoints A'B'C'D'E'F a new hexagon was formed.

To get the length of the new hexagon, connect A' to F' and a triangle AF'A' was formed. Knowing that AF' is 2(because of midpoint) and angle F'AA' is 120 degrees, Create a new point perpendicular from A to A'F' (midpoint)' and name it T. Angle ATF' is equal to 90 degrees, angle F'AT is 60 degrees (bisected 120 degrees) and AF'T is 30 degrees. By the use of 30-60-90 angle theorem, we can say that F'T is equal to \sqrt{3} and A'F' is equal to twice F'T which is 2 \sqrt{3}

Now, the formula for the area of a hexagon is A=1/2(apothem)(perimeter). To get the apothem, create a point from the center of the new hexagon and name it X and connect it to T. XT is perpendicular to A'F'. Connect A' and F' to X also. This creates a triangle A'XF'. Angle A'XF' is equal to 60 degrees but XT bisected it and turned angle A'XT to 30 degrees, which then makes TA'X equal to 60 degrees. By using the 30-60-90 theorem and A'T equal to \sqrt{3} we can conclude that the apothem XT is equal to 3.

The formula for the Area is A=1/2(apothem)(Perimeter) A=1/2(3)(6 * 2 \sqrt{3} ) A= 18\sqrt{3}

This is the area, but we are looking for square of the Area which is equal to (18\sqrt{3})^2 = 972.

First off, since the midpoints of the original hexagon are connected, and it is obvious that the each internal angle of the hexagon is $120^\circ$ , each side of the new hexagon can be easily found by the cosine rule: Letting the new side = $s$ , we have $s^2 = 2^2 + 2^2 - 2(2)(2) \cos 120^ \circ = 4 + 4 - 8(-\frac{1}{2}) = 12$ . Hence $s = \sqrt{12}$ .

To find the area of a hexagon with side length $s$ , we simply divide it into 6 equilateral triangles by joining opposite vertices. It is clear that the area of each equilateral triangle = $\frac{1}{2} (s)(\frac{\sqrt{3}}{2} s) = \frac{\sqrt{3}}{4} s^2$ , and area of the hexagon = $\frac{3 \sqrt{3}}{2} s^2$ . In this case, $s = \sqrt{12}$ , hence area = $\frac{3 \sqrt{3}}{2} (12) = 18 \sqrt{3} = X$ . Thus the required answer $= X^2 = 972$ .

Let [ABCDEF] refer to the area of ABCDEF. Note that the area of a regular hexagon is $\frac{3\sqrt{3}}{2}$ times the square of the side length.

Note that the angles of a regular hexagon are $120^\circ$ each. So F'AA', A'BB', B'CC', ..., E'FF' are similar triangles (AAS). Also $\angle AF'A' = 30 ^ \circ$ . So by the sine rule, $\frac{F'A'}{\sin 120^\circ} = \frac{F'A}{\sin 30^\circ} \Rightarrow F'A' = 2\sqrt{3}$ .

It is obvious that A'B'C'D'E'F' is a regular hexagon. So $[A'B'C'D'E'F']^2 = (F'A'^2 \cdot \frac{3\sqrt{3}}{2})^2 = 972$

Given: ABCDEF is a regular hexagon of side 4. The hexagon A'B'C'D'E'F' constructed from the midpoints of the sides of ABCDEF is also regular: the six triangles A'AB', B'BC', C'CD', D'DE', E'EF' and F'FA' are congruent isosceles triangles with the equal sides (A'A, AB' etc.) of length 2 each and the angles (A'AB etc.) equal to 120 \textdegree . Now, in the isosceles triangle A'AB, A'A = AB' = 2, and A'AB = 120 \textdegree . Then, by the cosine rule for the triangle A'AB, we get A'B' = 2 \sqrt{3} . Similarly B'C' = C'D' = D'E' = E'F' = F'A' = 2 \sqrt{3} . Therefore A'B'C'D'E'F' is a regular hexagon with side length 2 \sqrt{3} , and hence , its area is 6 \times $1/2 \times 2 \sqrt{3} \times 3$ = 18 \sqrt{3} . We are asked the square of this area, which is 18^2 \times 3 = 324 \times 3 = 972.

As it is a regular polygon , due to symmetry , let us consider one part of the two hexagonal system. we have triangle A’BB’ with < A’BB’=120(coz it is a regular hexagon its interior angle is 120) A’B=B’B=2 therefore <A’B’B=<BB’A
<A’BB’ + <A’B’B + <BB’A=180 \Rightarrow <A’B’B=<BB’A=30 using sine rule, A’B/sin30=A’B’/sin 120 \Rightarrow A’B’=2 \sqrt{3} regular hexagonal A’B’C’D’E’F’ is of side 2\sqrt{3} it consists of six equilateral triangles of side 2\sqrt{3} so the area of the hexagonal is 6*3\sqrt{3}=18\sqrt{3} and its square is 972

Call G the center of ABCDEF and A'G the apothem. By drawing AG, BG, CG..., we can divide the hexagon into 6 congruent triangles, all of which must be equilateral (each interior angle must be 120 degrees - $\frac{180(6-2)}{6}=120^\circ$ , and since all triangles are congruent, AG, BG, etc. are all angle bisectors, creating $60^\circ$ angles).

The area of each triangle can be found by drawing a perpendicular line to the midpoint of any side. Let's use AB. A'G will be $2\sqrt{3}$ since it is the side opposite the $60^\circ$ angle and the side opposite the $30^\circ$ is 2 (half of the 4 as given), giving us an area of $4\sqrt{3}$ for each small triangle. Since the hexagon is made of 6 such triangles, its area is $24\sqrt{3}$ .

We can now look at the areas of A'BB', B'CC', C'DD'... Angle A'BB' is $120^\circ$ as shown above, so we can divide triangle A'BB' into two congruent 30-60-90 triangles by dropping a perpendicular from B to the midpoint of A'B'. The hypotenuse of each of these triangles is 2, making the height of the triangle 1 (opposite the 30) and the base $\sqrt{3}$ (opposite the 60). This makes the area of A'BB' $\sqrt{3}$ .

There are 6 of these cutout triangles making a combined area of $6\sqrt{3}$ . We subtract that amount from the area of the hexagon to get $18\sqrt{3}$ . This is the area of A'B'C'D'E'F'. The square of $18\sqrt{3}$ is 972.

You could also relatively easily show the length of the apothem of A'B'C'D'E'F' is 3 and its side lengths are $2\sqrt{3}$ given its 30-60-90 triangles, making the area of A'B'G $3\sqrt{3}$ . Multiply that by 6, and you will also get $18\sqrt{3}$ , the square of which is 972.

First, we find the side length of $A'B'C'D'E'F'$ . The distance from one vertex of $ABCDEF$ to its midpoint is two, and the two sides of a vertex make an angle of 120 degrees. Using law of cosines or 30-60-90 triangles, the side length is found to be $2\sqrt{3}$ .

From here, we can use the formula for finding the area of a regular hexagon from a side length $s$ : $A=\frac{3}{2}s^2\sqrt{3}$ . The area of $A'B'C'D'E'F'$ turns out to be $18\sqrt{3}$ , which, squared, is $972$ .

The problem gives regular hexagon ABCDEF with side lengths 4, and embedded hexagon A'B'C'D'E'F' with side lengths x.

The problem also gives A' is the midpoint of AB, so A'A = 2 and A'B = 2; similarly, B'B = 2 and B'C = 2, C'C and C'D = 2 ... F'F and F'A = 2.

First we find the value of each angle of a regular hexagon. The formula for the total degrees in a convex polygon is 180(n-2) where n is the number of sides of the polygon.(This can be proved by noting n-2 non overlapping triangles can fill the n-sided polygon.) Applying this to a hexagon, we find that a hexagon has 720 ^ \circ total in it. Because this is a regular hexagon each angle will have the same value, and there are 6 angles in the hexagon, so each angle will have \frac {720}{6} = 120 ^\circ.

Looking at isosceles triangle F'AA' we know \angle F'AA' = 120 ^\circ from the previous discussion, and we know that \angle AF'A' = \angle AA'F because it is an isosceles triangle. If we let this value be \alpha, then 120 + \alpha + \alpha = 180 because the angles of a triangle sum up to 180^\circ.

120 + \alpha + \alpha = 180

2\alpha = 60

\alpha = 30.

An altitude from A to A'F' to point M makes a 30-60-90 triangle where AM is opposite the 30, AF' is opposite the 90, and MF' is opposite the 60.

AF' = 2 from the beginning of the discussion, so AM must = 1 and MF' must be \sqrt{3}.

MF' is also half of F'A', so 2MF' or 2\sqrt{3} is the side length of the hexagon.

Now we solve for the area by splitting the hexagon into 6 identical equilateral triangles from the centroid of the hexagon to each vertex. Each side length of the equilateral triangle is 2\sqrt{3} by the previous discussion.

The area of a triangle is \frac {a b sin(c)}{2} so the area of this triangle is \frac {2\sqrt{3}(2\sqrt{3})(\frac{\sqrt{3}}{2})}{2} which simplifies to 3\sqrt{3}. (2)

Since there are six of these triangles, we multiply (2) by 6 to obtain X = 18\sqrt{3}.

X^2 is therefore (18\sqrt{3})^2 = 972.

Let O be the center of both hexagons. Now, we observe triangle AA'O. L is a point on AO such that A'L is the altitude of the triangle AA'O. We know that <A'AO = <A'LO = 60degrees and correspondingly <AOA' = <AA'L = 30degrees, and AA' = 2. So solving these triangles we get OA = 4, LA = 1 and hence OL = 3; A'L = sqrt(3). In hexagon A'B'C'D'E'F' there are 12 congruent triangles AA'O. So X = [3 sqrt(3)]/2 *12 = 18 sqrt(3). X^2 = 324*3 = 972

A'B'C'D'E'F' is clearly also a regular hexagon. So, we can find the side length of this hexagon and find the area.

We see that AE = A'D' and the side length of A'B'C'D'E'F' is one half of A'D'. By some trigonometry, we find that AE = 4sqrt(3). So, the side length is 2sqrt(3). Thus, the area of the hexagon is ((2sqrt(3))^2 * sqrt(3) / 4) * 6 = 18sqrt(3). So, X^2 is 972.

since it is a regular hexagon,each angle is equal to 120 degrees.the area of the hexagon will be 24 root3 . the total areas of the smaller triangles leaving the required area will be 6 root3.therefore required area will be 18 root3.its square is 972.

Let us find the area of ABCDEF first. Area of ABCDEF is $\frac{1}{2} ap=\frac{1}{2} \cdot 2\sqrt{3} \cdot 4 \cdot 6=24\sqrt{3}$ , a is length of apothem and p is perimeter. (Length of apothem is $\frac{sin 60^\circ}{4}=2\sqrt{3}$ ). The ratio of ABCDEF to A'B'C'D'E'F' is $(\frac{AB}{A'B'})^2$ . Let us see the triangle A'BB', by using sine rule, we have $\frac{A'B'}{\sin 120^\circ}=\frac{A'B}{\sin 30^\circ}\Rightarrow A'B'={\sin 120^\circ}\cdot \frac{2}{\sin 30^\circ}=2\sqrt{3}$ . So, The ratio of ABCDEF to A'B'C'D'E'F' is $(\frac{AB}{A'B'})^2=(\frac{4}{2\sqrt{3}})^2=\frac{4}{3}$ . $X=24\sqrt{3} \cdot \frac{3}{4}=18\sqrt{3}\Rightarrow$ $X^2=(18\sqrt{3})^2=972$

Let $[PQRS]$ denote the area of the figure $PQRS$ . Let $O$ be the center of the hexagon. Triangle $OAB$ is an equilateral triangle with side length equal to $4$ and $[OAB] = \frac{1}{2}\cdot 4 \cdot 4 \cdot \sin60^\circ = 4 \sqrt{3}$ , so $[ABCDEF] = 6\times (4 \sqrt{3}) = 24 \sqrt{3}$ .

$A’B’C’D’E’F’$ is also a regular hexagon, and is similar to the larger hexagon by a factor of $\frac {OA’}{OA}$ . Consider right triangle $O A A'$ , since $OA=4, AA'=2$ , thus $OA' = 2 \sqrt{3}$ . Hence $X= \left(\frac {OA’}{OA} \right) ^2 \times 24 \sqrt{3} = \frac {3}{4} \times 24 \sqrt {3} = 18 \sqrt {3}$ $\Rightarrow X^2 = 972$ .