3. Find the area of a regular hexagon whose sides are all 2 cm.

The key to this problem is to note that a regular hexagon

can be broken up into 6 equilateral triangles.

Since a hexagon has 4x180o = 720o, each angle in a regular hexagon is 720o/6 = 120o. When we bisect them we get 60o angles. Since each triangle has two 60o angles, they are all equilateral triangles. So all we have to do is to find the area of an equilateral triangle where all of the sides are 2cm long.

The area of a triangle is one half the base times the height. The base is obviously 2 cm, so we just need to find the height. To find the height, drop a perpendicular from the top down to the base

When you drop a perpendicular from the vertex of an isosceles triangle to its base, it meets the base at its midpoint. The height is part of the following right triangle.

We can now use the Pythagorean Theorem to find the height.

22 = 12 + h2

4 = 1 + h2

3 = h2

So the area of the equilateral triangle is

So the area of the hexagon is 6 times that