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For example, the perimeter of a rectangle of width 0.001 and length 1000 is slightly above 2000, while the perimeter of a rectangle of width 0.5 and length 2 is 5. Both areas are equal to 1. Proclus (5th century) reported that Greek peasants "fairly" parted fields relying on their perimeters. [ 2 ]
The formula for the perimeter of a rectangle The area of a rectangle is the product of the length and width. If a rectangle has length and width , then: [11] it has area =; it has perimeter = + = (+); each diagonal has length = +; and
Arc length – Distance along a curve; Area#Area formulas – Size of a two-dimensional surface; Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric ...
The apothem a can be used to find the area of any regular n-sided polygon of side length s according to the following formula, which also states that the area is equal to the apothem multiplied by half the perimeter since ns = p.
Shoelace scheme for determining the area of a polygon with point coordinates (,),..., (,). The shoelace formula, also known as Gauss's area formula and the surveyor's formula, [1] is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by their Cartesian coordinates in the plane. [2]
That is, the area of the rectangle is the length multiplied by the width. As a special case, as l = w in the case of a square, the area of a square with side length s is given by the formula: [1] [2] A = s 2 (square). The formula for the area of a rectangle follows directly from the basic properties of area, and is sometimes taken as a ...
The only equable rectangles with integer sides are the 4 × 4 square and the 3 × 6 rectangle. [4] An integer rectangle is a special type of polyomino, and more generally there exist polyominoes with equal area and perimeter for any even integer area greater than or equal to 16. For smaller areas, the perimeter of a polyomino must exceed its area.
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 = that avoids the usual procedure of doubling the area of the triangle and then halving it.