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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.
The process of drawing the altitude from a vertex to the foot is known as dropping the altitude at that vertex. It is a special case of orthogonal projection. Altitudes can be used in the computation of the area of a triangle: one-half of the product of an altitude's length and its base's length (symbol b) equals the triangle's area: A = h b /2 ...
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 ...
In this example, the triangle's side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. As long as they obey the strict triangle inequality, they define a triangle in the Euclidean plane whose area is a positive real number.
Apothem of a hexagon Graphs of side, s; apothem, a; and area, A of regular polygons of n sides and circumradius 1, with the base, b of a rectangle with the same area. The green line shows the case n = 6. The apothem (sometimes abbreviated as apo [1]) of a regular polygon is a line segment from the center to the midpoint of one of its sides.
Its surface area is four times the area of an equilateral triangle: = =. [7] Its volume can be ascertained similarly as the other pyramids, one-third of the base times height. Because the base is an equilateral, it is: [ 7 ] V = 1 3 ⋅ ( 3 4 a 2 ) ⋅ 6 3 a = a 3 6 2 ≈ 0.118 a 3 . {\displaystyle V={\frac {1}{3}}\cdot \left({\frac {\sqrt {3 ...
The difference in area for an equilateral triangle is small, just over 1%, [2] but as Howard Eves pointed out, for an isosceles triangle with a very sharp apex, the optimal circles (stacked one atop each other above the base of the triangle) have nearly twice the area of the Malfatti circles. [3] In fact, the Malfatti circles are never optimal.