SOLUTION: What is the area of a regular hexagon whose sides are each 12 inches long? (round to the nearest square inch..) (Clue:Draw the figure.) Show all work.
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-> SOLUTION: What is the area of a regular hexagon whose sides are each 12 inches long? (round to the nearest square inch..) (Clue:Draw the figure.) Show all work.
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Question 161610: What is the area of a regular hexagon whose sides are each 12 inches long? (round to the nearest square inch..) (Clue:Draw the figure.) Show all work.
Thanxs Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website! found the answer on the web.
take the hexagon and draw diagonals through each of the opposite corners.
this breaks the hexagon up into 6 equilateral triangles.
each of the sides of these equilateral triangles will be 12 inches long.
area of a triangle is 1/2 base * height.
we have the base.
we need the height.
take one of the equilateral triangles and drop a perpendicular from opposite side.
this becomes the height of the triangle.
the right triangles become 30-60-90 triangles.
in a 30-60-90 triangle, the sides opposite the angles are in the following ratio:
side opposite 30 degrees = .5
side opposite 60 degrees =
side opposite 90 degrees = 1
height is side opposite 60 degrees.
base is side opposite 30 degrees.
since the base is (1/2)*L and the hypotenuse equals L, the height is equal to *L
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we now have enough to compute the area of the hexagon.
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area of the hexagon is 6 * the area of each equilateral triangle.
area of each equilateral triangle is 1/2 * base * height.
base = 12
height = *12
formula becomes
6 * (1/2) * 12 * *12
answer is:
374.1229744.
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formula for area of the hexagon had become
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