How many different shapes can you make out of 5 equalateral triangle?
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It depends what you mean by "different" shapes.  For example, is a long chain, 3 lengths on the bottom, 2 lengths on the top different from the same shape upside-down?  Assuming that the usual set of transformations ensuring congruence (translation, reflection and rotation) are allowed as not making different shapes, and also assuming you're looking for closed shapes (rather than, say, 5 triangles arranged so that no sides are touching, only meeting at vertices), then there are only 4 different arrangements.\n\nI'll do my best to try to describe them:\n\n- All 5 triangles in a straight line, 3 facing up, 2 facing down.\n- A hexagon with 1/6 missing (like a pie with 1 piece eaten).\n- A larger equilateral triangle (4 pieces) with one extra triangle on one of the sides (doesn't matter where, due to reflections and rotations)\n- A chain of 4 triangles, making a parallelogram, with one extra triangle in one of the locations that will make a different shape than the one above.
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trianglular pyramid has 8 edges and 5 verticles
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What type of triangle is it?\nIf it is an equilateral triangle, all sides will be the same length, so p=(1.25 X 3)m\n\nHope this helps
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The shape with 120 sides is called Hexakis Icosahedron.\n\nThe Hexakis Icosahedron has 62 vertices, 120 faces and 180 edges. (The Hexakis Icosahedron is the only polyhedron with 120 faces.) \n\nThe Hexakis Icosahedron contains 120 triangles. \n\nThe dual of the Hexakis Icosahedron is the great rhombicosidodecahedron. \n\nI hope I answered your question. All the best.
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An isosceles triangle is a triangle with (at least) two equal sides and two equal angles.\n\nhttp://mathworld.wolfram.com/IsoscelesTriangle.html
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Cut from the tip of vertical line to the tip of the horizontal,all the way around, so it looks like a kite. Then cut straight through the 2 lines, you now have 4 triangles. Rotate each one 180 degrees and push them together. YOUR SQUARE.
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Triangle, quadrilateral, pentagon, hexagon, heptagon, octagon, nonagon, decagon
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The first five squares are easy:\n1, 4, 9, 16, 25.\n\nRectangular:\n2, 6, 12, 20, 30\n\nTriangular:\n1, 3, 6, 10, 15\n\nPentagonals are:\n1, 5, 12, 22, 35
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Geometry Session 2, Part B: Linkage-Strip Constructions  \n... (or the hands-on version) to try to build quadrilaterals with the given lengths. ... Can a set of four lengths make two different quadrilaterals? ...learner.org/channel/courses/learningmath/.../part_b/constructing.html \nEasy Construction of Bicentric Quadrilateral II  \nEasy Construction of Bicentric Quadrilateral ... By construction, four quadrilaterals MPND, MPLC, LPKB, and KPNA are also cyclic. ...www.cut-the-knot.org/Curriculum/Geometry/BicentricQuadri2.shtml \nBrahmagupta Quadrilaterals  \nIn this paper we give a construction of Brahmagupta. quadrilaterals from rational Heron triangles. ... Construction of Brahmagupta quadrilaterals ...forumgeom.fau.edu/FG2002volume2/FG200221.pdf \nEasy Construction of Bicentric Quadrilateral  \nEasy Construction of Bicentric Quadrilateral ... appears as a consequence of another construction of bicentric quadrilaterals. ...www.cut-the-knot.org/Curriculum/Geometry/BicentricQuadri.shtml \nOnline Lesson On Geometrical Construction Using iMovies  \n... On Geometrical Construction Using iMovies. LESSON ... Module 3 - Construction of Quadrilaterals. After this module, the students should be able to. 1. ...ali.apple.com/ali_sites/sgli/exhibits/1001179/Lesson_Objectives.html
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I'm afraid that with the given information, this would be impossible to determine. The only thing one could conclude from this information is that this polygon has an even number of sides. \n\nIt could be a square with sides of 3cm each\nIt could be a hexagon with sides of 2cm each\nIt could be an octagon with sides of 1.5cm each\nIt could be a decagon with sides of 1.2cm each\nIt could be a 12-gon with sides of 1cm each\n\nThe list is infinite unless you are given more specific information, such as the length of each side or the angles within the shape.
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Square and regular hexagon together \n\nEach decagon is surrounded on each alternate side by a square and a hexagon\n\nAlternately instead of the regular hexagon you could put 6 equilateral triangles tesselated into a regular hexagon or a combination of hexagons and equilateral triangles.
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Draw a triangle with the letter "M" INSIDE and the triangle should cut the top 2 vertices and the 3 bottom vertices.\n\nyou should get 8 triangles and the last 1 is the big triangle with "M" inside.
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A pentagon has 5 sides of equal length if that is what you are asking.
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A circle has 360 degrees. Now imagine that from the icosagon you can make 20 equally-sized triangles. The angle made by each triangle is 360/20=18 degrees. The other two angles of each triangle are both 81 degrees.\n\nThe basic formula for the area of a triangle is 1/2 base x height. How do we find these? Well, imagine that from each of the 20 triangles we make two additional right triangles. In doing this we know several things:\n\n-The hypotenuse is equal to 6.5 cm (the radius).\n-The angles of the smaller triangle are now 9 degrees, 81 degrees, and 90 degrees.\n-The sine of 9 degrees is equal to the opposite side divided by the hypotenuse. If we multiply this value by 6.6 we get the length of the opposite side ("1/2 base").\n-The sine of 81 degrees multiplied by 6.5 gives us the "height".\n-Once you compute each value (1/2 base and height) you multiply them by each other and you get the area of the triangle.\n-Multiply the area of that triangle by 20, since you have 20 of them. Voila.
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Actually the only way it would be possible is if you use Elliptic or Riemannian geometry, and would require a manifold.  A representation of such a construct would be similar to drawing a 4 dimensional shape in 3 or even 2 dimensions.\n\nEuclidian geometry doesn't allow triangles with a sum of all angles to be greater than 180 degrees.
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An equilateral triangle is a regular polygon with three equilateral sides (sides measuring the same length) and three equiangular angles (angles that are the same measure) these being of 60 degrees.
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For any polygon, you take the number of sides, subtract 2, and then multiply by 180.\n\nEx.  Triangle  (3-2)180 = 180\n     Quadrilateral (4-2)180 = 360\n     Pentagon (5-2)180 = 540\n\nAny polygon with n sides  (n-2)*180\n\n\nthen, if it is a regular polygon, (all sides and angles congruent), then divide that number by the number of sides.\n\n( (n-2)*180 ) /n