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Measurement of tree circumference, the tape calibrated to show diameter, at breast height. The tape assumes a circular shape. The perimeter of a circle of radius R is .Given the perimeter of a non-circular object P, one can calculate its perimeter-equivalent radius by setting
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 value of the two products in the chord theorem depends only on the distance of the intersection point S from the circle's center and is called the absolute value of the power of S; more precisely, it can be stated that: | | | | = | | | | = where r is the radius of the circle, and d is the distance between the center of the circle and the ...
Given a chord of length y and with sagitta of length x, since the sagitta intersects the midpoint of the chord, we know that it is a part of a diameter of the circle. Since the diameter is twice the radius, the "missing" part of the diameter is (2r − x) in length.
The area of an annulus is the difference in the areas of the larger circle of radius R and the smaller one of radius r: = = = (+) (). As a corollary of the chord formula, the area bounded by the circumcircle and incircle of every unit convex regular polygon is π /4
A is the cross-sectional area of the flow, P is the wetted perimeter of the cross-section. More intuitively, the hydraulic diameter can be understood as a function of the hydraulic radius R H, which is defined as the cross-sectional area of the channel divided by the wetted perimeter. Here, the wetted perimeter includes all surfaces acted upon ...
where A is the area of a squircle with minor radius r, is the gamma function. A = ( k + 1 ) ( k + 2 ) π r 2 {\displaystyle A=(k+1)(k+2)\pi r^{2}} where A is the area of an epicycloid with the smaller circle of radius r and the larger circle of radius kr ( k ∈ N {\displaystyle k\in \mathbb {N} } ), assuming the initial point lies on the ...
an object of diameter 1 AU (149 597 871 km) at a distance of 1 parsec (pc) Thus, the angular diameter of Earth's orbit around the Sun as viewed from a distance of 1 pc is 2″, as 1 AU is the mean radius of Earth's orbit. The angular diameter of the Sun, from a distance of one light-year, is 0.03″, and that of Earth 0.0003″. The angular ...
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