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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 triangle with sides a, b, and c. In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of the three side lengths , , . Letting be the semiperimeter of the triangle, = (+ +), the area is [1]
This formula generalizes Heron's formula for the area of a triangle. A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as d approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
The area formula for a triangle can be proven by cutting two copies of the triangle into pieces and rearranging them into a rectangle. In the Euclidean plane, area is defined by comparison with a square of side length , which has area 1. There are several ways to calculate the area of an arbitrary triangle.
A version of the isoperimetric inequality for triangles states that the triangle of greatest area among all those with a given perimeter is equilateral. [35] The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a given circle is equilateral. [36]
The area a of the circular segment is equal to the area of the circular sector minus the area of the triangular portion (using the double angle formula to get an equation in terms of ): a = R 2 2 ( θ − sin θ ) {\displaystyle a={\tfrac {R^{2}}{2}}\left(\theta -\sin \theta \right)}
In particular, to find the quadrilateral, or the triangle, or another particular figure, with the largest area amongst those with the same shape having a given perimeter. The solution to the quadrilateral isoperimetric problem is the square , and the solution to the triangle problem is the equilateral triangle .
Another proof that uses triangles considers the area enclosed by a circle to be made up of an infinite number of triangles (i.e. the triangles each have an angle of d𝜃 at the centre of the circle), each with an area of 1 / 2 · r 2 · d𝜃 (derived from the expression for the area of a triangle: 1 / 2 · a · b · sin𝜃 ...