Search results
Results from the WOW.Com Content Network
Construction of a square with the same area as a given oblong Proof without words of the AM–GM inequality: PR is the diameter of a circle centered on O; its radius AO is the arithmetic mean of a and b. Using the geometric mean theorem, triangle PGR's altitude GQ is the geometric mean. For any ratio a:b, AO ≥ GQ.
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 :
Quarter-circular area [2] ... a = the radius of the base circle h = the height of the semi-ellipsoid from the base cicle's center to the edge Solid ...
Let R be the radius of the arc which forms part of the perimeter of the segment, θ the central angle subtending the arc in radians, c the chord length, s the arc length, h the sagitta of the segment, d the apothem of the segment, and a the area of the segment.
Heron's formula for the area of a triangle is the special case obtained by taking d = 0. The relationship between the general and extended form of Brahmagupta's formula is similar to how the law of cosines extends the Pythagorean theorem .
The formula for the area of a circle (more properly called the area enclosed by a circle or the area of a disk) is based on a similar method. Given a circle of radius r, it is possible to partition the circle into sectors, as shown in the figure to the right. Each sector is approximately triangular in shape, and the sectors can be rearranged to ...
The area of a regular polygon is half its perimeter multiplied by the distance from its center to its sides, and because the sequence tends to a circle, the corresponding formula–that the area is half the circumference times the radius–namely, A = 1 / 2 × 2πr × r, holds for a circle.
The perimeter of a stadium is calculated by the formula = (+) where a is the length of the straight sides and r is the radius of the semicircles. With the same parameters, the area of the stadium is A = π r 2 + 2 r a = r ( π r + 2 a ) {\displaystyle A=\pi r^{2}+2ra=r(\pi r+2a)} .