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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 , and the hypotenuse is the side of the equilateral triangle.
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 ...
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 ...
Triangles have many types based on the length of the sides and on the angles. A triangle whose sides are all the same length is an equilateral triangle, [3] a triangle with two sides having the same length is an isosceles triangle, [4] [a] and a triangle with three different-length sides is a scalene triangle. [7]
Heron's formula. A triangle with sides a, b, and c. In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of the three side lengths Letting be the semiperimeter of the triangle, the area is [1] It is named after first-century engineer Heron of Alexandria (or Hero) who ...
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 .
The extended sides of the orthic triangle meet the opposite extended sides of its reference triangle at three collinear points. [22] [23] [21] In any acute triangle, the inscribed triangle with the smallest perimeter is the orthic triangle. [24] This is the solution to Fagnano's problem, posed in 1775. [25]
These include the Calabi triangle (a triangle with three congruent inscribed squares), [10] the golden triangle and golden gnomon (two isosceles triangles whose sides and base are in the golden ratio), [11] the 80-80-20 triangle appearing in the Langley's Adventitious Angles puzzle, [12] and the 30-30-120 triangle of the triakis triangular tiling.