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This proof depends on the readily-proved proposition that the area of a triangle is half its base times its height—that is, half the product of one side with the altitude from that side. [2] Let ABC be an equilateral triangle whose height is h and whose side is a. Let P be any point inside the triangle, and s, t, u the perpendicular distances ...
An equilateral triangle with a side of 2 has a height of √ 3, as the sine of 60° is √ 3 /2. The legs of either right triangle formed by an altitude of the equilateral triangle are half of the base a {\displaystyle a} , and the hypotenuse is the side a {\displaystyle a} of the equilateral triangle.
In geometry, calculating the area of a triangle is an elementary problem encountered often in many different situations. The best known and simplest formula is where b is the length of the base of the triangle, and h is the height or altitude of the triangle. The term "base" denotes any side, and "height" denotes the length of a perpendicular ...
Ptolemy's Theorem yields as a corollary a pretty theorem [2] regarding an equilateral triangle inscribed in a circle. Given An equilateral triangle inscribed on a circle and a point on the circle. The distance from the point to the most distant vertex of the triangle is the sum of the distances from the point to the two nearer vertices.
The canonical Sierpiński triangle uses an equilateral triangle with a base parallel to the horizontal axis (first image). Shrink the triangle to 1 / 2 height and 1 / 2 width, make three copies, and position the three shrunken triangles so that each triangle touches the two other triangles at a corner (image 2).
An equilateral triangle base and three equal isosceles triangle sides It gives 6 isometries, corresponding to the 6 isometries of the base. As permutations of the vertices, these 6 isometries are the identity 1, (123), (132), (12), (13) and (23), forming the symmetry group C 3v , isomorphic to the symmetric group , S 3 .
draw an equilateral triangle that has the middle segment from step 1 as its base and points outward. remove the line segment that is the base of the triangle from step 2. The first iteration of this process produces the outline of a hexagram. The Koch snowflake is the limit approached as the above steps are followed indefinitely.
A regular hexagon can also be created as a truncated equilateral triangle, with Schläfli symbol t{3}. Seen with two types (colors) of edges, this form only has D 3 symmetry. A truncated hexagon, t{6}, is a dodecagon, {12}, alternating two types (colors) of edges. An alternated hexagon, h{6}, is an equilateral triangle, {3}.