Question 971679: A regular hexagon ABCDEF has vertices at A(4,4√3), B(8,4√3), C(10,2√3), D(8,0), E(4,0), and F(2,2√3). Suppose the sides of the hexagon are reduced by 40% to produce a similar regular hexagon. What are the perimeter and area of the smaller hexagon rounded to the nearest tenth?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A regular hexagon ABCDEF has vertices at A(4,4√3), B(8,4√3), C(10,2√3), D(8,0), E(4,0), and F(2,2√3). Suppose the sides of the hexagon are reduced by 40% to produce a similar regular hexagon. What are the perimeter and area of the smaller hexagon rounded to the nearest tenth?
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Find the lengths of the sides.
eg, AB = 4
 = 4
It's a regular hexagon, all sides = 4
Perimeter = 6*4 = 24 units
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 where n = 6 & s = 4
Area =~ 41.5692 sq units
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Reducing by 40% --> 0.6*Perimeter
P = 24*0.6 = 14.4 units
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Area is a function of the square of the side length --> 0.36 x original
Area = 41.5692*0.6^2
Area =~ 14.965 sq units
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