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In general, the area of a triangle is half the product of its base and height. The formula of the area of an equilateral triangle can be obtained by substituting the altitude formula. [7] Another way to prove the area of an equilateral triangle is by using the trigonometric function. The area of a triangle is formulated as the half product of ...
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.
An example of a circular triangle with three convex edges is a Reuleaux triangle, which can be made by intersecting three circles of equal size. The construction may be performed with a compass alone without needing a straightedge, by the Mohr–Mascheroni theorem. Alternatively, it can be constructed by rounding the sides of an equilateral ...
The same area formula can also be derived from Heron's formula for the area of a triangle from its three sides. However, applying Heron's formula directly can be numerically unstable for isosceles triangles with very sharp angles, because of the near-cancellation between the semiperimeter and side length in those triangles. [19]
Let A'BC be the equilateral triangle having base BC and vertex A' on the negative side of BC and let AB'C and ABC' be similarly constructed equilateral triangles based on the other two sides of triangle ABC. Then the lines AA', BB', CC' are concurrent and the point of concurrence is the 1st isogonal center. Its trilinear coordinates are
There are many ways to prove Heron's formula, for example using trigonometry as below, or the incenter and one excircle of the triangle, [7] or as a special case of De Gua's theorem (for the particular case of acute triangles), [8] or as a special case of Brahmagupta's formula (for the case of a degenerate cyclic quadrilateral).
The ratio of the area of the incircle to the area of the triangle is less than or equal to /, with equality holding only for equilateral triangles. [ 19 ] Suppose A B C {\displaystyle \triangle ABC} has an incircle with radius r {\displaystyle r} and center I {\displaystyle I} .
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.
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