Three regular hexagons are drawn such that their diagonals cut each ot...
Solution:
Let the smallest hexagon have side length 'a', and the largest hexagon have side length 'c'. Then the side length of the middle hexagon is 'b'.
Area Ratio:
The area of a regular hexagon of side length 'x' is A = (3√3/2) x^2. Therefore, the area ratio is given by:
A':A":A''' = (3√3/2) a^2 : (3√3/2) b^2 : (3√3/2) c^2
= a^2 : b^2 : c^2
= 1 : 2 : 3
Therefore, b^2 = 2a^2 and c^2 = 3a^2.
Diagonal Ratio:
The diagonals of a regular hexagon of side length 'x' intersect at a point that is equidistant from all vertices. Therefore, the point of intersection of the diagonals of the three hexagons is the same. Let this point be O.
The diagonal of a regular hexagon of side length 'x' is 2x. Therefore, the ratio of the diagonals of the three hexagons is:
OA' : OA" : OA''' = 2a : 2b : 2c
= a : b : c
Therefore, b/a = A''/A' = 2/1 and c/a = A'''/A' = 3/1.
Final Ratio:
We have b^2 = 2a^2 and c^2 = 3a^2. Therefore, c^2/b^2 = 3/2. Substituting the value of c/a = 3/1, we get:
(3a)^2/b^2 = 3/2
b^2/a^2 = 4/3
Substituting the value of b/a = 2/1, we get:
(a/2)^2/a^2 = 4/3
a = √(3/4)
Therefore, the ratio of the side lengths of the three hexagons, from smallest to largest, is:
a : b : c
= √(3/4) : 2√(3/4) : 3√(3/4)
= √3 : √3(2) : √3(3)
= 1 : √3 : √3(3)
Hence, the required ratio of the length of the sides of the regular hexagons (from smallest to the largest) is 1 : √3 : √3(3).
Three regular hexagons are drawn such that their diagonals cut each ot...
Side is proportional area^2
thus 1:root 2 :root3
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