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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.
The area of a triangle is its half of the product of the base times the height (length of the altitude). For a triangle with opposite sides ,,, if the three altitudes of the triangle are called ,,, the area is: = = =. Given a fixed base side and a fixed area for a triangle, the locus of apex points is a straight line parallel to the base.
Let ABC be a triangle with side lengths a, b, and c, with a 2 + b 2 = c 2. Construct a second triangle with sides of length a and b containing a right angle. By the Pythagorean theorem, it follows that the hypotenuse of this triangle has length c = √ a 2 + b 2, the same as the hypotenuse of the first triangle.
Set square shaped as 45° - 45° - 90° triangle The side lengths of a 45° - 45° - 90° triangle 45° - 45° - 90° right triangle of hypotenuse length 1.. In plane geometry, dividing a square along its diagonal results in two isosceles right triangles, each with one right angle (90°, π / 2 radians) and two other congruent angles each measuring half of a right angle (45°, or ...
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.
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 (or any one side) approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
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 ...
Any triangle subdivides its bounding box into the triangle itself and additional right triangles, and the areas of both the bounding box and the right triangles are easy to compute. Combining these area computations gives Pick's formula for triangles, and combining triangles gives Pick's formula for arbitrary polygons. [7] [8] [13]