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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 line through the triangle's incenter bisects the perimeter if and only if it also bisects the area. A triangle's semiperimeter equals the perimeter of its medial triangle. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter.
An area cannot be equal to a length except relative to a particular unit of measurement. For example, if shape has an area of 5 square yards and a perimeter of 5 yards, then it has an area of 45 square feet (4.2 m 2) and a perimeter of 15 feet (since 3 feet = 1 yard and hence 9 square feet = 1 square yard).
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]
Triangles have many types based on the length of the sides and the angles. A triangle whose sides are all the same length is an equilateral triangle, [3] a triangle with two sides having the same length is an isosceles triangle, [4] [a] and a triangle with three different-length sides is a scalene triangle. [7]
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]
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 .
A version of the isoperimetric inequality for triangles states that the triangle of greatest area among all those with a given perimeter is equilateral. That is, for perimeter p {\displaystyle p} and area T {\displaystyle T} , the equality holds for the equilateral triangle: [ 8 ] p 2 = 12 3 T . {\displaystyle p^{2}=12{\sqrt {3}}T.}