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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 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.
Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, h, is less than the circle radius. Also, let each side of the polygon have length s; then the sum of the sides is ns, which is less than the circle circumference. The polygon area consists of n equal triangles with height h and base s, thus equals nhs/2.
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]
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.
Some troops leave the battlefield injured. Others return from war with mental wounds. Yet many of the 2 million Iraq and Afghanistan veterans suffer from a condition the Defense Department refuses to acknowledge: Moral injury.
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