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Imagine drawing a straight line from each vertex of the hexagon to the centre of the circle.
The hexagon is now split into six identical triangles.
The triangles are equilaterals. How do we know this?
The angle at the centre is ${360^o \over 6}=60^o$ and the two edges next to this vertex are radii so have length $1$. Because these two edges have the same length the angles at the remaining two vertices must be equal. Angles in a triangle add up to $180^o$ so all the angles must be $60^o$.
The area of a single triangle is
$$\mbox{Area of triangle} = {1\over 2} \times \mbox{height} \times \mbox{base}.$$
Consider one of the triangles making up the hexagon. Dropping a perpendicular from one of the points to the opposite edge and using Pythagoras' Theorem gives that the height of the triangle is ${\sqrt 3\over 2}$.
Therefore the area of one of the triangles is
$$\mbox{Area of triangle} = {1\over 2} \times {\sqrt 3\over 2} \times 1={\sqrt 3\over 4}.$$
And the area of the hexagon is $6$ times this, which is ${3 \sqrt 3\over 2}$.
To find the area of the equilateral triangle, draw it on the same diagram as the first part. Notice that each edge of the large triangle cuts two of the smaller triangles and divides their area in half. Therefore the area of the equilateral triangle in the unit circle is half that of the hexagon, which is ${3\sqrt 3\over 4}$.
Inscribed in a Circle
Age 14 to 16
Challenge Level 
Imagine drawing a straight line from each vertex of the hexagon to the centre of the circle.
The hexagon is now split into six identical triangles.
The triangles are equilaterals. How do we know this?
The angle at the centre is ${360^o \over 6}=60^o$ and the two edges next to this vertex are radii so have length $1$. Because these two edges have the same length the angles at the remaining two vertices must be equal. Angles in a triangle add up to $180^o$ so all the angles must be $60^o$.
The area of a single triangle is
$$\mbox{Area of triangle} = {1\over 2} \times \mbox{height} \times \mbox{base}.$$
Consider one of the triangles making up the hexagon. Dropping a perpendicular from one of the points to the opposite edge and using Pythagoras' Theorem gives that the height of the triangle is ${\sqrt 3\over 2}$.
Therefore the area of one of the triangles is
$$\mbox{Area of triangle} = {1\over 2} \times {\sqrt 3\over 2} \times 1={\sqrt 3\over 4}.$$
And the area of the hexagon is $6$ times this, which is ${3 \sqrt 3\over 2}$.
To find the area of the equilateral triangle, draw it on the same diagram as the first part. Notice that each edge of the large triangle cuts two of the smaller triangles and divides their area in half. Therefore the area of the equilateral triangle in the unit circle is half that of the hexagon, which is ${3\sqrt 3\over 4}$.
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