[[Image:Triangle.Equilateral.svg|Equilateral Triangle]] | [[Image:Triangle.Isosceles.svg|Isosceles triangle]] | [[Image:Triangle.Scalene.svg|Scalene triangle]] |

Equilateral | Isosceles | Scalene |

[[Image:Triangle.Right.svg|Right triangle]] | [[Image:Triangle.Obtuse.svg|Obtuse triangle]] | [[Image:Triangle.Acute.svg|Acute triangle]] |

Right | Obtuse | Acute |

[[Image:Triangle.Centroid.svg|frame|left|The [[centroid]] is the center of gravity.]] A [[median (geometry)|median]] of a triangle is a straight line through a vertex and the midpoint of the opposite side, and divides the triangle into two equal areas. The three medians intersect in a single point, the triangle's [[centroid]]. This is also the triangle's [[center of gravity]]: if the triangle were made out of wood, say, you could balance it on its centroid, or on any line through the centroid. The centroid cuts every median in the ratio 2:1, i.e. the distance between a vertex and the centroid is twice as large as the distance between the centroid and the midpoint of the opposite side. [[Image:Triangle.NinePointCircle.svg|frame|right|[[Nine-point circle]] demonstrates a symmetry where six points lie on the same circle.]] The midpoints of the three sides and the feet of the three altitudes all lie on a single circle, the triangle's [[nine-point circle]]. The remaining three points for which it is named are the midpoints of the portion of altitude between the vertices and the [[orthocenter]]. The radius of the nine-point circle is half that of the circumcircle. It touches the incircle (at the [[Feuerbach point]]) and the three [[excircle]]s.

[[Image:Triangle.EulerLine.svg|frame|left|[[Euler's line]] is a straight line through the centroid (orange), orthocenter (blue), circumcenter (green) and center of the nine-point circle (red).]] The centroid (yellow), orthocenter (blue), circumcenter (green) and center of the nine-point circle (red point) all lie on a single line, known as [[Euler's line]] (red line). The center of the nine-point circle lies at the midpoint between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half that between the centroid and the orthocenter. The center of the incircle is not in general located on Euler's line. If one reflects a median at the angle bisector that passes through the same vertex, one obtains a [[symmedian]]. The three symmedians intersect in a single point, the [[symmedian point]] of the triangle.

== Computing the area of a triangle == Calculating the area of a triangle is an elementary problem encountered often in many different situations. Various approaches exist, depending on what is known about the triangle. What follows is a selection of frequently used formulae for the area of a triangle.http://mathworld.wolfram.com/TriangleArea.html ===Using geometry=== The [[surface area|area]] ''S'' of a triangle is ''S'' = ½''bh'', where ''b'' is the length of any side of the triangle (the ''base'') and ''h'' (the ''altitude'') is the perpendicular distance between the base and the vertex not on the base. This can be shown with the following geometric construction. [[image:Triangle.GeometryArea.svg|frame|center|The triangle is first transformed into a [[parallelogram]] with twice the area of the triangle, then into a rectangle.]] To find the area of a given triangle (green), first make an exact copy of the triangle (blue), rotate it 180°, and join it to the given triangle along one side to obtain a [[parallelogram]]. Cut off a part and join it at the other side of the parallelogram to form a rectangle. Because the area of the rectangle is ''bh'', the area of the given triangle must be ½''bh''. [[image:Triangle.VectorArea.svg|frame|right|The area of the parallelogram is the magnitude of the cross product of the two vectors.]] The product of the [[inradius]] and the [[semiperimeter]] of a triangle also gives its area. ===Using vectors=== The area of a parallelogram can also be calculated by the use of [[Vector (spatial)|vectors]]. If ''AB'' and ''AC'' are vectors pointing from A to B and from A to C, respectively, the area of parallelogram ABDC is |''AB'' × ''AC''|, the magnitude of the [[cross product]] of vectors ''AB'' and ''AC''. |''AB'' × ''AC''| is also equal to |''h'' × ''AC''|, where ''h'' represents the altitude ''h'' as a vector. The area of triangle ABC is half of this, or ''S'' = ½|''AB'' × ''AC''|. The area of triangle ABC can also be expressed in term of [[dot product]]s as follows: :$\backslash frac\{1\}\{2\}\; \backslash sqrt\{(\backslash mathbf\{AB\}\; \backslash cdot\; \backslash mathbf\{AB\})(\backslash mathbf\{AC\}\; \backslash cdot\; \backslash mathbf\{AC\})\; -(\backslash mathbf\{AB\}\; \backslash cdot\; \backslash mathbf\{AC\})^2\}\; =\backslash frac\{1\}\{2\}\; \backslash sqrt\{\; |\backslash mathbf\{AB\}|^2\; |\backslash mathbf\{AC\}|^2\; -(\backslash mathbf\{AB\}\; \backslash cdot\; \backslash mathbf\{AC\})^2\}$ [[Image:Triangle.TrigArea.svg|frame|left|Applying trigonometry to find the altitude ''h''.]] ===Using trigonometry=== The altitude of a triangle can be found through an application of [[trigonometry]]. Using the labelling as in the image on the left, the altitude is ''h'' = ''a'' sin γ. Substituting this in the formula ''S'' = ½''bh'' derived above, the area of the triangle can be expressed as ''S'' = ½''ab'' sin γ. It is of course no coincidence that the area of a parallelogram is ''ab'' sin γ. If one uses :$\backslash cos\; C=\backslash frac\{a^2+b^2-c^2\}\{2ab\}$ and :$\backslash sin\; C\; =\; \backslash sqrt\{1-\; \backslash cos^2\; C\}$ and also the formula shown above, then one arrives at the following formula for area :$\backslash frac\{1\}\{4\}\; \backslash sqrt\{2(a^2\; b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)\}$ [Note that, this is a multiplied out form of Heron's formula] ===Using coordinates=== If vertex A is located at the origin (0, 0) of a [[Cartesian coordinate system]] and the coordinates of the other two vertices are given by B = (''x''