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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
Alternatively, the closely related mean diameter (), which is twice the mean radius, is also used. For a non-spherical object, the mean radius (denoted R {\displaystyle R} or r {\displaystyle r} ) is defined as the radius of the sphere that would enclose the same volume as the object. [ 1 ]
Since the diameter is twice the radius, the "missing" part of the diameter is (2r − x) in length. Using the fact that one part of one chord times the other part is equal to the same product taken along a chord intersecting the first chord, we find that (2r − x)x = (y / 2) 2. Solving for r, we find the required result.
The hydraulic diameter is the equivalent circular configuration with the same circumference as the wetted perimeter. The area of a circle of radius R is . Given the area of a non-circular object A, one can calculate its area-equivalent radius by setting = or, alternatively:
Consider a circle in with center at the origin and radius . Gauss's circle problem asks how many points there are inside this circle of the form ( m , n ) {\displaystyle (m,n)} where m {\displaystyle m} and n {\displaystyle n} are both integers.
The Schwarzschild radius was named after the German astronomer Karl Schwarzschild, who calculated this exact solution for the theory of general relativity in 1916. The Schwarzschild radius is given as =, where G is the gravitational constant, M is the object mass, and c is the speed of light.
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 ...
Let A′ be the point opposite A on the circle, so that A′A is a diameter, and A′AB is an inscribed triangle on a diameter. By Thales' theorem , this is a right triangle with right angle at B. Let the length of A′B be c n , which we call the complement of s n ; thus c n 2 + s n 2 = (2 r ) 2 .