Triangle Inscribed In Hexagons

The $6$ triangular pieces can be moved to form a second congruent hexagon:

and since $1$ hexagon has an area of $8$ , $2$ hexagons have an area of $\boxed{16}$ .

The area of a regular hexagon is given by $A_H = \dfrac{3\sqrt{3}}{2}x^2$ where $x$ is the side length. Substituting, we have

$8 = \dfrac{3\sqrt{3}}{2}x^2$

$x^2=\dfrac{16}{3\sqrt{3}}$

Applying cosine rule, we have

$a^2=x^2+x^2-2x^2 \cos 120=2x^2-2x^2\left(\dfrac{-1}{2}\right)=3x^2=3\left(\dfrac{16}{3\sqrt{3}}\right)=\dfrac{16}{\sqrt{3}}$

$a=\sqrt{\dfrac{16}{\sqrt{3}}}$

The side of the equilateral triangle is $2a$ .

$2a=2\sqrt{\dfrac{16}{\sqrt{3}}}$

Let the side length of the equilateral triangle be $y=2a=2\sqrt{\dfrac{16}{\sqrt{3}}}$ . The area of an equilateral triangle is given by $A=\dfrac{\sqrt{3}y^2}{4}$ where $y$ is the side length. Substituting, we have

$A=\dfrac{\sqrt{3}\left(2\sqrt{\dfrac{16}{\sqrt{3}}}\right)^2}{4}=\dfrac{\sqrt{3}\left[4\left(\dfrac{16}{\sqrt{3}}\right)\right]}{4}=\boxed{16}$

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Marvin

We divide the inner hexagon as well as triangle $ABC$ into congruent isosceles triangles as shown above on the right. Since the inner hexagon is made up of six of these isosceles triangles, and triangle $ABC$ is made up of twelve, the area of triangle $ABC$ must be twice the area of the inner hexagon, i.e. $\boxed{16}$

In the first of the pictures below, it is shown that the area of a regular hexagon can be split up into thirds, and the second picture then shows that the area of a hexagon can be split up into sixths, by lines such as the blue ones.

Therefore, as shown below, the area of triangle ABC is composed of one red hexagon, and six green triangles, each one of which has $\frac16$ of the area of each rectangle.

$\implies \triangle ABC = 1(8) + 6 \left(\frac16 \times 8\right) = 8+8 = \boxed{16}$

The figure above shows that each isosceles triangle (in red) is $\frac{1}{6}$ of a hexagon. Since $\triangle ABC$ consist of six of these and one hexagon, the area of $ABC$ is equal to the area of two hexagons, $\boxed{16}$

Note that two small red triangles combine to give a third of the hexagon so it takes six such triangles to give a hexagon. Note that we have six red triangles which can give another hexagon. Thus the red region is just twice the area of a hexagon - $2\times8=\boxed{16}$ .

Moderator note : This solution is a bit out-of-date after Brilliant staff made some edits to this problem. The method for re-arranging the pieces is still valid, however!

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As shown above, label points $D$ , $E$ , and $F$ where Point $D$ is the midpoint of $\overline{AC}$ , Point $E$ is the center of $\triangle DEF$ , and Point $F$ is the midpoint of $\overline{AD}$ .

Now, to solve this problem, consider $\triangle ABC$ to be equilateral because of the fact that all three of its side lengths are the sum of two congruent hypotenuses of two congruent isosceles triangles. Furthermore, consider $\triangle DEF$ being a $30° - 60° - 90°$ triangle because $\overline{EF}$ is drawn perpendicular to $\overline{AD}$ and $\angle{DEF}$ is $\frac{1}{6}$ of $360°$ . We can use this information to solve for $\overline{DF}$ as such

$\overline{DF}$ = $\frac{\overline{EF}}{2} \times {\sqrt{3}}$

$\therefore \overline{DF}$ = $\frac{1}{2} \times {\sqrt{3}}$ $\Rightarrow$ $\overline{DF}$ = $\frac{\sqrt{3}}{2}$

Now that we have solved for $\overline{DF}$ , we can solve for $\overline{AC}$ using midpoints $F$ and $D$ as such:

$\overline{AC}$ = $\overline{AD} \times {2}$

$\overline{AD}$ = $\overline{DF} \times {2}$

$\therefore \overline{AC}$ = $\overline{DF} \times {4}$ $\Rightarrow$ $\overline{AC}$ = $\frac{\sqrt{3}}{2} \times {4}$ $\Rightarrow$ $\overline{AC}$ = $2\sqrt{3}$

Now that we have solved for $\overline{AC}$ , we can compute the area of equilateral $\triangle{ABC}$ as such:

Area of Equilateral Triangle = $\frac{s^{2}\sqrt{3}}{4}$

$\therefore A_{\triangle{ABC}}$ = $\frac{(2\sqrt{3})^{2}\sqrt{3}}{4}$ $\Rightarrow$ $\frac{12\sqrt{3}}{4}$ $\Rightarrow$ $3\sqrt{3}$ $\Rightarrow$ $A_{\triangle{ABC}}$ $\approx \boxed{5.196}$

Divide the triangle into four component triangles of equal area by joining the corners of the central hexagon which are at the midpoint of each side of the large triangle. If each 1/3 of the three small triangles has area x, the central triangle has area 3x, therefore x = 8/6. The total area = 12x = 12 x 8/6 = 16

In the triangle ABC there are 6 small triangles next to the hexagon If we look at a hexagon, we see that 2 small triangles make up half the area of a hexagon The area of a hexagon consists of 2/3 of the rectangle and one third of the two small triangles. A small triangle corresponds to 1/6 of the area of the hexagon. 6 small triangles correspond to the area of a hexagon. Therefore the area of the triangle ABC is 8 + 8 = 16

Area of a regular hexagon will be given by

(1/2)×b×h + (1/2)×b×h + b×B = 8

~ b×h + B×h = 8

~b(h+H) = 8

Now area of the triangle is given by

(1/2)×2b×(B+B+h+h)

~ (1/2)×2b×2(B+h)

~ 2b×(B+h) = 2×8 = 16

The hexagon is composed of six equilateral triangles each of area [sqrt(3)/4]s^2. Therefore, 6 (sqrt(3)/4)s^2 = 8, and s^2 = (16/9)sqrt(3). The triangular pieces each have an area equal to (1/2)(s^2)sin(120) = (1/2)(16/9)(sqrt(3) sqrt(3)/2 = 4/3. Then the red area = 6*(4/3) + 8 = 16. Ed Gray

The area if the red triangle equals 1 hexagon plus x (6 small triangles). Shift the red triangle down by the side length of the hexagon. Then you see that the area of the red triangle equals 3 hexagons minus x. Thus, x = 1. The red triangle equals 2 hexagons; hence, has area 16.

There is one hexagon and six hexagon peices in the triangle giving you a total area of 16

The area of the hexagon( $A_{Hexagon}$ ) is $\frac{3a^2\sqrt3}{2}$ ,the area of the hexagon in the picture is 8.So,we have

$\frac{3a^2\sqrt3}{2}=8$ so $a^2=\frac{16}{3\sqrt3}$

Because each three small triangle( $3 \times A_{small-\triangle}$ ) in the picture is $360^\circ$ degree so each ( $A_{small-\triangle}$ ) have one $120^\circ$ .

.Let call $b$ is $\frac{1}{2}$ the side of the big triangle.Use the cosine rule,we have

$b^2=2a^2-2a^2\cos120=2a^2-2a^2 \times \frac{-1}{2}=2a^2--a^2=2a^2+a^2=3a^2=3 \times \frac{16}{3\sqrt3}=\frac{16}{\sqrt3}$ so $b=\sqrt\frac{16}{\sqrt3}$

So the side of the big triangle( $A_{big-\triangle}$ ) is

$2b=2 \times \sqrt\frac{16}{\sqrt3}$

The area of the equilateral triangle is $\frac{a^2\sqrt3}{4}$ ,So the area of the ( $A_{big-\triangle}$ ) is

$\frac{(2a)^2\sqrt3}{4}=\frac{4a^2\sqrt3}{4}=a^2\sqrt3=(\sqrt\frac{16}{\sqrt3})^2\sqrt3=\frac{16}{\sqrt3} \times \sqrt3=\color{#3D99F6}{\boxed{\large{16}}}$