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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The area of a regular hexagon with side lengths of 8cm is about 166.3cm2
Side length is about 58cm and the perimeter is about 348cm
The area of a given hexagon is equal to the area of an equilateral triangle whose perimeter is 36 inches. Find the length of a side of the regular hexagon.Click once to select an item at the bottom of the problem.
To find the area of a regular hexagon with side length of 40cm, consider that since it is regular, then it consists of 6 equilateral triangles of side 40cm. Half of each of those triangles is a right triangle. By the pythagorean theorem, we know that if the hypotenuse is 40cm, and one side is 20cm, then the other side is the square root of (40cm squared - 20cm squared) or about 34.64cm. That makes the area of each of those 12 right triangle to be about 692.8cm, so the total area of the hexagon is about 8313.8cm.
Length of one side squared x 1.5 x square root of 3, for a REGULAR hexagon.