It is a regular polygon as for example an equilateral triangle
First make sure you understand that concyclic simply means the points lie on a common circle. We are not told it is a regular pentagon but we will assume it is. We could create pentagons that are not even convex and would not be concyclic.Let's start with a regular pentagon. You can split it up into 5 congruent triangles with the points meeting at the middle. Any side of one of these triangle is connected from each of the vertices of the pentagon to the center of the pentagon. Since all 5 triangles are congruent, this distance must be the same for each of the vertices. So, we see that each of the vertices is equidistant from a given point. Now if we drew a circle centered at that point with a radius equal to the distance between the point and any vertex, that circle would pass through all 5 vertices. Therefore any four ( really all 5), vertices of a regular pentagon are concyclic.A nice proof would use Ptolemy's theorem. I will place a line to an answers.com page that helps with that.Another solution:The pentagon has to be regular. Otherwise, the question is impossible to prove.Consider a regular polygon ABCDE.Take triangles BCD & ECDBC=ED (sides of a regular polygon are equal)CD=CD (common side)
A convex polygon.
Four, as in a tetrahedron. Any three points lie in a plane, so if you only had three vertices they couldn't make a three-dimensional object.
A non-degenerative triangle has 3 vertices which does not lie on a stright line. i.e the are not collinear
triangle
It is a regular polygon as for example an equilateral triangle
For any number greater than or equal to 3, a suitable polygon can be found.
No.
false
chord
Assuming all the vertices of the segmentation lie on the circle, then you can choose any three of them as the corners of a triangle circumscribed by the circle. The perpendicular bisectors of the sides of that triangle intersect at the center of the circle.
A chord is a line segment whose endpoints lie on a circle. A secant is a line (or line segment) that intersects a circle in two places, endpoints NOT on the circle.
First make sure you understand that concyclic simply means the points lie on a common circle. We are not told it is a regular pentagon but we will assume it is. We could create pentagons that are not even convex and would not be concyclic.Let's start with a regular pentagon. You can split it up into 5 congruent triangles with the points meeting at the middle. Any side of one of these triangle is connected from each of the vertices of the pentagon to the center of the pentagon. Since all 5 triangles are congruent, this distance must be the same for each of the vertices. So, we see that each of the vertices is equidistant from a given point. Now if we drew a circle centered at that point with a radius equal to the distance between the point and any vertex, that circle would pass through all 5 vertices. Therefore any four ( really all 5), vertices of a regular pentagon are concyclic.A nice proof would use Ptolemy's theorem. I will place a line to an answers.com page that helps with that.Another solution:The pentagon has to be regular. Otherwise, the question is impossible to prove.Consider a regular polygon ABCDE.Take triangles BCD & ECDBC=ED (sides of a regular polygon are equal)CD=CD (common side)
A convex polygon.
A chord of a circle is a line segment whose two endpoints lie on the circle. The diameter, passing through the circle's centre, is the largest chord in a circle. So the answer is 6 m
A chord of a circle is a geometric line segment whose endpoints both lie on the circumference of the circle. A secant or a secant line is the line extension of a chord.