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The inradius of the incircle in a triangle with sides of length , , is given by [7] = () (), where s = 1 2 ( a + b + c ) {\displaystyle s={\tfrac {1}{2}}(a+b+c)} is the semiperimeter. The tangency points of the incircle divide the sides into segments of lengths s − a {\displaystyle s-a} from A {\displaystyle A} , s − b {\displaystyle s-b ...
Inradius r of a regular convex polygon with n sides and side length s: = ... Einstein's field equation of general relativity: ...
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 :
The incircle is the circle that lies inside the triangle and touches all three sides. Its radius is called the inradius. There are three other important circles, the excircles; they lie outside the triangle and touch one side, as well as the extensions of the other two. The centers of the incircles and excircles form an orthocentric system. [26]
Euler's theorem: = | | = In geometry, Euler's theorem states that the distance d between the circumcenter and incenter of a triangle is given by [1] [2] = or equivalently + + =, where and denote the circumradius and inradius respectively (the radii of the circumscribed circle and inscribed circle respectively).
If r and R are the inradius and the circumradius respectively, then the area K satisfies the inequalities [14] 4 r 2 ≤ K ≤ 2 R 2 . {\displaystyle \displaystyle 4r^{2}\leq K\leq 2R^{2}.} There is equality on either side only if the quadrilateral is a square .
where r is the inradius and R is the circumradius of the triangle. Here the sign of the distances is taken to be negative if and only if the open line segment DX (X = F, G, H) lies completely outside the triangle. In the diagram, DF is negative and both DG and DH are positive. The theorem is named after Lazare Carnot (1753–1823).
where r is the incircle radius and R is the circumcircle radius; hence the circumradius is at least twice the inradius (Euler's triangle inequality), with equality only in the equilateral case. [7] [8] The distance between O and the orthocenter H is [9] [10]