What is the area of a regular hexagon whose sides are each 12 inches long?
467774
The area of the regular hexagon is given by:\n\nA = 2.598 s^2\n\nSo if the sides are 12 inches each;\n\nA = 2.598 (12 in)^2\n\nA = 374 in^2
509644
Formula of area 'A' of any regular 'n'-sided polygon of each side 'a' is\nA = 0.25*n*a^2*cot(pi/n)\n\nPut a=10 and n=6 in this formula.\nA = 0.25*6*10^2*cot(pi/6)\n  = 150*sqrt(3)   [since cot(pi/6) = sqrt(3)]\n  = 259.8 sq units
797386
Consider a regular n-gon in a circle of radius r with side length l\nDraw in all radii from vertices to centre.\nThis forms n isosceles triangles with the sides of the n-gon as their bases\n\nThen angle at centre = 360/n° and each angle at base = (90 - 180/n)°\n\nBy sine rule: sin(90 - 180/n)° /r = sin(360/n)° / l \nWhence l = r * sin(360/n)°/ sin(90 - 180/n)°\n= r * sin(2 * 180/n)°/ cos(180/n)°\n= r * 2 sin(180/n)°cos(180/n)°/cos(180/n)°\n=2rsin(180/n)°\n\nSo the perimeter of the polygon is 2nrsin(180/n)°\n\nNote: As n →∞ Perimeter  → circumference of circle\nie lim  n →∞ 2nrsin(180/n)° = 2πr\nSo lim  n →∞ nsin(180/n)° = π\nOr for large n, sin(180/n)° ≈ π/n; which makes sense given that for small x (in radians) sinx ≈ x\n\nGiven area of a triangle = ½absinC\nthen area of 1 isosceles triangle = ½r²sin(360/n)°\nSo the area of the polygon is ½nr²sin(360/n)°\n\nThus for n = 9 and r = 10;\nl = 2rsin(180/n)° \n= 20 sin 20°\n≈ 6.84 in
1128959
d six sided is called hexagon, while 7 sided is called heptagon 4rm femgeorge29@yahoo.com
787617
(m)\n\n14 sides: tetradecogon
760569
small area=2in^2\nbig sides are 3x as long so area is 3^2 x as long\n2*3^2=18 in^2
644806
if the triangle is a regular triangle\n\na = (1/2)bh\n\na = (1/2)(6)(6 * sin60)\n\na = (1/2)(6)(3√3)\n\na = 9√3 = 15.588\n\n\nfor a regular polygon, \narea = (n*t^2) / (4*tan(π/n))\n\nwhere n = number of sides & t = length of side
800499
a hexangle would be 6 degrees or an offset of 6 degrees if you are speaking of a tool or a deflection.\n-------\nIf you are speaking of a hexagon:\n\nAll regular polygons have a total of 360 degrees for their center  angles, and all are equal.\n\n360/6=60 degrees.\n----------\n\nThe sum of the angles of a triangle is 180 degrees.\n\nSo each segment of the hexagon will form a triangle with the center point which is an isosceles triangle, with an apex angle of 60. The other two angles will be equal as they are opposite sides of equal length, and they will total 180-60=120. The three angles of the triangle will be 60, 60 and 60, so we know it is also an equilateral triangle.\n\nThe angle between one segment of the side of the hexagon and the next will then have to be 2x60=120 degrees because all the segments will be the same.
462376
20\n\n 12 edges and 8 vertices
956626
Figure out how many square inches on the cube.  One side is 10" x 10" = 100 inches square.\n\n6 sides makes 600 inches square.\n\nA square foot is 12 x 12 inches = 144 inches square\n\nSquare footage is 600 / 144 = 4.17
431862
There are two angles formed wherever two sides of the hexagon meet. The angle inside the hexagon is 120 degrees and the angle outside the hexagon is 360 - 120 = 240 degrees. You can see this easily by placing a point at the center of the hexagon and connecting it to the 6 points where the edges of the hexagon meet. This forms 6 equilateral triangles and every one of the angles of the 6 triangles is 60 degrees (180/3). You can see that the angle inside the hexagon where 2 sides of the hexagon meet is equal to 2 of these 60 degree angles; hence 120 degrees. Draw it and you'll see it right away.
189870
answer depends on arrangement.  \n\nSA = area of top + bottom + left side + right + front + back.\nif 1x1x24, SA = 24+24+24+24+1+1 = 98 sq cm, \nif 1x2x12, SA = 24+24+12+12+2+2 = 76 sq cm,\nif 1x3x8, SA = 24+24+8+8+3+3 = 70 sq cm,\nif 1x4x6, SA = 24+24+6+6+4+4 = 68 sq cm,\nif 2x2x6, SA = 12+12+12+12+4+4 = 56 sq cm,\nif 2x3x4, SA = 12+12+8+8+6+6 = 52 sq cm.\n\nnotice how the SA gets smaller as the arrangement becomes more compact.
990570
You could use the law of cosines to find the lengths of the diagonals.\n\nThe measure of interior angle B = 128 4/7 degrees.\nAC^2 = AB^2 + BC^2 - 2(AB)(BC)(cos B).\nSince the heptagon is regular, BC = AB, so you can substitute:\nAC^2 = 2AB^2 + - 2(AB)^2(cos B)\nAC^2 = 2AB^2(1 - cos B)\nAC^2 = 2AB^2(1 - cos (128 4/7))\nAC = sqrt [2AB^2(1 - cos (128 4/7))]\n\nThe measure of angle BCD = 128 4/7 degrees.\nThe measure of angle BCA = 25 5/7 degrees because that triangle is isosceles.\nTherefore, the measure of angle DCA is 102 6/7 degrees.\n\nAD^2 = AC^2 + CD^2 - 2(AC)(CD)(cos DCA)\n\nAnd so on...\n\nCheating with a calculator, it's this:\n\nAC = sqrt [2AB^2(1 - (-0.6234898))]\nAC = sqrt [2AB^2(1.6234898)]\nAC = sqrt [AB^2(3.2469796)]\nAC = sqrt [3.2469796](AB)\nAC = 1.8019377358(AB)\n\nThe measure of angle BCD = 128 4/7 degrees.\nThe measure of angle BCA = 25 5/7 degrees because that triangle is isosceles.\nTherefore, the measure of angle DCA is 102 6/7 degrees.\n\nAD^2 = AC^2 + CD^2 - 2(AC)(CD)(cos DCA)\nAgain, CD = AB and you've already found AC, so substitute:\nAD^2 = [1.8019377358(AB)]^2 + AB^2 - 2[1.8019377358(AB)](AB)(cos DCA)\nAD^2 = 3.2469796(AB^2) + AB^2 - 3.60838754716(AB)^2(cos DCA)\nAD^2 = 4.2469796(AB^2) - 3.60838754716(AB)^2(cos DCA)\nAD^2 = 4.2469796(AB^2) - 3.60838754716(AB)^2(cos 102 6/7)\nAD^2 = 4.2469796(AB^2) - 3.60838754716(AB)^2(-0.22252)\nAD^2 = 4.2469796(AB^2) + 0.8019377358(AB)^2\nAD^2 = 4.2469796(AB^2) + 0.8019377358(AB)^2\nAD^2 = 5.04891734(AB)^2\nAD = sqrt [5.04891734(AB)^2]\nAD = 2.2469796(AB)\n\n1/AC + 1/AD = 1/1.8019377358(AB) + 1/2.2469796(AB)\n1/AC + 1/AD = 2.2469796/4.04891734(AB) + 1.8019377358/4.04891734(AB)\n1/AC + 1/AD = 4.04891734/4.04891734(AB)\n1/AC + 1/AD = 1/AB\n\nI like this problem, and I'll give it a shot without a calculator later this afternoon.
999307
(a) You can flip a hexagon horizontally or vertically across a line, and you'll end up with the same hexagon.\n(b) Assuming that point is in the middle, every 60° you'll have the same hexagon.\n(c) Rotate it 30° from the center and the hexagon will appear rotated.\n(d) 120° is a multiple of 60°.\n\nThe answer is C.
584072
I would use the trapezoid rule myself. For seven vertices:\nV0 = (x(0),y(0))\nV1 = (x(1),y(1))\nV2 = (x(2),y(2))\n...\nV6 = (x(6),y(6))\n\nFor notational convenience, define V7 = V0.\nThat is, x(7) = x(0) and y(7) = y(0).\n\nThen the area is:\n\n 6\nSUM ( x(i+1)*y(1) - x(i)*(y(i+1) ) / 2\ni=0\n\n= ( x(1)*y(0) - x(0)*y(1)\n + x(2)*y(1) - x(1)*y(2)\n + x(3)*y(2) - x(2)*y(3)\n + x(4)*y(3) - x(3)*y(4)\n + x(5)*y(4) - x(4)*y(5)\n + x(6)*y(5) - x(5)*y(6)\n + x(0)*y(6) - x(6)*y(0) ) / 2\n\n\nIf the heptagon is regular, you don't need to know the coordinates of the vertices, just the length of the regular side(s):\n\nA = (7/4) * s^2 * cot(π/7)
23271
The area of a triangle can be found by multiplying the base times the one-half the height.\n\nIf a triangle has a base of length 6 inches and a height of 4 inches, its area is 6*2=12 square inches
935014
If the square has sides of 12, the perimeter of the square is 4x12 = 48 inches. This is the length of the wire.\n\nTo find out the sides of the equilateral triangle, divide 48 by 3, since the triangle has 3 equal sides.