Search results
Results from the WOW.Com Content Network
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 ...
In geometry, Fagnano's problem is an optimization problem that was first stated by Giovanni Fagnano in 1775: For a given acute triangle determine the inscribed triangle of minimal perimeter . The solution is the orthic triangle , with vertices at the base points of the altitudes of the given triangle.
In this example, the triangle's side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. As long as they obey the strict triangle inequality, they define a triangle in the Euclidean plane whose area is a positive real number.
The solution to the quadrilateral isoperimetric problem is the square, and the solution to the triangle problem is the equilateral triangle. In general, the polygon with n sides having the largest area and a given perimeter is the regular polygon, which is closer to being a circle than is any irregular polygon with the same number of sides.
The parameters most commonly appearing in triangle inequalities are: the side lengths a, b, and c;; the semiperimeter s = (a + b + c) / 2 (half the perimeter p);; the angle measures A, B, and C of the angles of the vertices opposite the respective sides a, b, and c (with the vertices denoted with the same symbols as their angle measures);
A parabolic segment is the region bounded by a parabola and line. To find the area of a parabolic segment, Archimedes considers a certain inscribed triangle. The base of this triangle is the given chord of the parabola, and the third vertex is the point on the parabola such that the tangent to the parabola at that point is parallel to the chord.
Malfatti's assumption that the two problems are equivalent is incorrect. Lob and Richmond (), who went back to the original Italian text, observed that for some triangles a larger area can be achieved by a greedy algorithm that inscribes a single circle of maximal radius within the triangle, inscribes a second circle within one of the three remaining corners of the triangle, the one with the ...
The three splitters concur at the Nagel point of the triangle. A cleaver of a triangle is a line segment that bisects the perimeter of the triangle and has one endpoint at the midpoint of one of the three sides. So any cleaver, like any splitter, divides the triangle into two paths each of whose length equals the semiperimeter.