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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 ...
A square can be divided into an even number of triangles of equal area (left), but into an odd number of only approximately equal area triangles (right). Monsky's proof combines combinatorial and algebraic techniques and in outline is as follows: Take the square to be the unit square with vertices at (0, 0), (0, 1), (1, 0) and (1, 1).
Given a square, construct equilateral triangles on two adjacent edges, either both inside or both outside the square. Then the triangle formed by joining the vertex of the square distant from both triangles and the vertices of the triangles distant from the square is equilateral. [2]
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.
A square has a larger area than any other quadrilateral with the same perimeter. [7] A square tiling is one of three regular tilings of the plane (the others are the equilateral triangle and the regular hexagon). The square is in two families of polytopes in two dimensions: hypercube and the cross-polytope. The Schläfli symbol for the square ...
There are many ways to prove Heron's formula, for example using trigonometry as below, or the incenter and one excircle of the triangle, [8] or as a special case of De Gua's theorem (for the particular case of acute triangles), [9] or as a special case of Brahmagupta's formula (for the case of a degenerate cyclic quadrilateral).
Given an equilateral triangle ABC in the plane, and a point P in the plane of the triangle ABC, the lengths PA, PB, and PC form the sides of a (maybe, degenerate) triangle. [1] [2] Proof of Pompeiu's theorem with Pompeiu triangle ′ The proof is quick. Consider a rotation of 60° about the point B.
We need to prove that there exists a real number h consistent with the values a, b, and c, in which case this triangle exists. Triangle with altitude h cutting base c into d + (c − d). By the Pythagorean theorem we have b 2 = h 2 + d 2 and a 2 = h 2 + (c − d) 2 according to the figure at the right. Subtracting these yields a 2 − b 2 = c 2 ...