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Let the vertices of the hexagon be A, B, C, D, E and F. Join AD, BE and CF. These lines intersect at point M.
The external angle of a hexagon =360/6 =60° . The corresponding internal angle is 120°.
The lines drawn bisect the internal angle at each vertex. The angles of ∆ABM are 60°, 60° and thus the angle at M is also 60° - the triangle is therefore equilateral with a side length of 6 cm. There are 6 congruent equilateral triangles forming the hexagon.
Draw a line MN where N is the mid point of AB then ∆MNA is a right angled triangle with a hypotenuse (MA) length of 6 cm and a base (AN) length of 3 cm.
By Pythagoras, the length of the altitude (MN) is √(62 - 32) = √27 cm
The area of ∆ABM = 2 x area of ∆MNA = 3√27
The area of the hexagon = 6 x area of ∆ABM = 6 x 3√27 = 18√27 = 93.53 sq cm (2dp)
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