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The area A of any triangle is the product of its inradius (the radius of its inscribed circle) and its semiperimeter: A = r s . {\displaystyle A=rs.} The area of a triangle can also be calculated from its semiperimeter and side lengths a, b, c using Heron's formula :
As proved by Archimedes, in his Measurement of a Circle, the area enclosed by a circle is equal to that of a triangle whose base has the length of the circle's circumference and whose height equals the circle's radius, [11] which comes to π multiplied by the radius squared: =.
is the radius, C = 2 π r {\displaystyle C=2\pi r} is the circumference (the length of any one of its great circles ), S {\displaystyle S} is the surface area ,
Following Archimedes' argument in The Measurement of a Circle (c. 260 BCE), compare the area enclosed by a circle to a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius. If the area of the circle is not equal to that of the triangle, then it must be either greater or less.
Proposition one states: The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference of the circle. Any circle with a circumference c and a radius r is equal in area with a right triangle with the two legs being c and r.
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.
Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle. [15] The incircle radius is no greater than one-ninth the sum of the altitudes. [16]: 289
If is the radius of the incircle of the triangle, then the triangle can be broken into three triangles of equal altitude and bases , , and . Their combined area is A = 1 2 a r + 1 2 b r + 1 2 c r = r s , {\displaystyle A={\tfrac {1}{2}}ar+{\tfrac {1}{2}}br+{\tfrac {1}{2}}cr=rs,} where s = 1 2 ( a + b + c ...