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The area of a given hexagon is equal to the area of an equilateral triangle whose perimeter is 36 inches. can you please help me Find the length of a side of the regular hexagon.?
2*6^1/2
Expert answered|latefisher|Points 1675|
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Asked 5/9/2012 1:41:59 AM
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When chords intersect in a circle, the vertical angles formed intercept congruent arcs.?
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can you please help me Find the area of an equilateral triangle (regular 3-gon) with the given measurement.? 6-inch side. A = sq. in.?
Weegy: Here are the shortcut formulas, perimeter is P=3s, area is A=(1/2)sh, h=[(sqrt 3)/2]s where h is height and s is side, all sides are equal, if the given is radius it means it is in the center, so from any point to center is equal length. r=(2/3)h. [ I also have my own equation if r is given to get the side instantly, s^2=(2x)^2-x^2 where x is radius. i'll answer #1 to give idea. So given r=3in, apply s^2=(2x)^2-x^2, so s^2=(2*3)^2-3^2 => s=sqrt (36-9) => s=sqrt 27 => s=sqrt (9*3) => s=3(sqrt 3). Then h=(3/2)r => h=3*3/2 => h=9/2 then A=(1/2)*3(sqrt 3)*(9/2) => A=27(sqrt 3)/4 in^2 answer. If you want to use h=[(sqrt 3)/2]s to get s, you should get h 1st from r, so since h=9/2 then (9/2)*[2/sqrt 3)]=s => 9/sqrt 3=s to cancel fraction multiply by sqrt 3/sqrt 3, so s=9(sqrt 3)/3 => s=3(sqrt 3). So in hexagon you just break it in 6 parts, the radius given in #2 is already the side, after getting 1 area just multiply by 6 for total area. the rest are repetitions. Hope this helps ] (More)
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Asked 5/7/2012 1:19:07 AM
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can you please help me Find the area of an equilateral triangle (regular 3-gon) with the given measurement.? 6-inch radius A = sq. in.?
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Asked 5/7/2012 1:24:21 AM
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9-inch perimeter how many will it be A = sq. in.?
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Asked 5/7/2012 2:47:14 AM
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(regular 3-gon)
Weegy: The regular 3-gon, known as the equilateral triangle, was constructed in I.1, while the regular 4-gon, known as the square, was constructed in I.46. In book IV, regular 5-gons and regular 6-gons have been constructed. [ An application of III.30 (which was used in this proposition) can double the number of sides of a regular polygon, and therefore regular polygons with 8, 10, 12, 16, 20, 24, etc., sides can be constructed. This proposition shows how to use a regular m-gon and a regular n-gon to produce a regular mn-gon, provided that m and n are relatively prime numbers. That produced a 15-gon, and from that we can produce regular polygons with 30, 60, 120, etc., sides. Thus, a regular n-gon can be constructed if the only prime numbers that divide n are 2, 3, and 5, where 2 can be a repeated factor, but 3 and 5 are not repeated. ] (More)
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