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Another proof that uses triangles considers the area enclosed by a circle to be made up of an infinite number of triangles (i.e. the triangles each have an angle of dπ at the center of the circle), each with an area of β 1 / 2 β · r 2 · dπ (derived from the expression for the area of a triangle: β 1 / 2 β · a · b · sinπ ...
The nine-point circle is tangent to the incircle and excircles. In geometry, the nine-point circle is a circle that can be constructed for any given triangle. It is so named because it passes through nine significant concyclic points defined from the triangle. These nine points are: [28] [29] The midpoint of each side of the triangle; The foot ...
In geometry, the circumscribed circle or circumcircle of a triangle is a circle that passes through all three vertices. The center of this circle is called the circumcenter of the triangle, and its radius is called the circumradius .
Regular polygons; Description Figure Second moment of area Comment A filled regular (equiliteral) triangle with a side length of a = = [6] The result is valid for both a horizontal and a vertical axis through the centroid, and therefore is also valid for an axis with arbitrary direction that passes through the origin.
The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
The largest possible ratio of the area of the inscribed square to the area of the triangle is 1/2, which occurs when =, = /, and the altitude of the triangle from the base of length is equal to . The smallest possible ratio of the side of one inscribed square to the side of another in the same non-obtuse triangle is 2 2 / 3 {\displaystyle 2 ...
The circle and the triangle are equal in area. Proposition one states: The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference of the circle.
The area A of any triangle is the product of its inradius (the radius of its inscribed circle) and its semiperimeter: A = r s . {\displaystyle A=rs.} The area of a triangle can also be calculated from its semiperimeter and side lengths a, b, c using Heron's formula :