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The chord is longer than a side of the inscribed triangle if the chosen point falls within a concentric circle of radius 1 / 2 the radius of the larger circle. The area of the smaller circle is one fourth the area of the larger circle, therefore the probability a random chord is longer than a side of the inscribed triangle is 1 / 4 .
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 small object nearby may subtend the same solid angle as a larger object farther away. For example, although the Moon is much smaller than the Sun, it is also much closer to Earth. Indeed, as viewed from any point on Earth, both objects have approximately the same solid angle (and therefore apparent size). This is evident during a solar eclipse.
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 ] The ratio of the area of the incircle to the area of an equilateral triangle, π 3 3 {\displaystyle {\frac {\pi }{3{\sqrt {3}}}}} , is larger than that of ...
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
Given a circle, let u n be the perimeter of an inscribed regular n-gon, and let U n be the perimeter of a circumscribed regular n-gon. Then u n and U n are lower and upper bounds for the circumference of the circle that become sharper and sharper as n increases, and their average ( u n + U n )/2 is an especially good approximation to the ...
Thales's theorem can also be used to find the centre of a circle using an object with a right angle, such as a set square or rectangular sheet of paper larger than the circle. [7] The angle is placed anywhere on its circumference (figure 1). The intersections of the two sides with the circumference define a diameter (figure 2).
Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle. [15] The incircle radius is no greater than one-ninth the sum of the altitudes. [16]: 289