Rectangle in an hexagon

Assuming that $b = c = 1$ , we first note that since each of the interior angles of a regular hexagon have a measure of $120$ degrees, and since $a || d$ , we have that

$d = a + 2b*\sin(30) = 2 + 2*(\frac{1}{2}) = 3$ ,

and that the second dimension of the rectangle is

$2c*\cos(30) = 2*(\frac{\sqrt{3}}{2}) = \sqrt{3}$ .

Thus the area of the rectangle is $\boxed{3\sqrt{3}}$ .

Solution: Let e be the horizontal line joining the midpoints of the adjacent sides of the hexagon. You can get the size of e by using a the cosine rule: - $e^{2} = 1^{2}+ 1^{2} - 2 \times 1 \times 1 \times cos(120°) = 3$ - $e = \sqrt{3}$ - Now we need to find d (of the image). Observe that the diagonal of the rectangle is equal to the height of the hexagon. It is equal to twice the apothema, which is equal to the height of a equlateral triangle of side 2. The height is: $2^{2} = 1^{2} + x^{2} therofore: x= \sqrt{3}$ - So the diagonal of the rectangle is : $2\sqrt{3}$ - Using pithaghoras we now can get the size of d (of the image): - $(2x)^{2} = e^{2} + d^{2}$ - $(2 \sqrt{3})^{2} = \sqrt{3}^{2} + d^{2}$ solving we get: $d=\sqrt{9} =3.$ -The area of the rectangle is $d \times e = 3\sqrt{3}$