61.90(:
166.30
166.30
For an equilateral triangle with side length a, area = (a²√3)/4, which for a= 6cm is 15.6 cm² [rounded to 1 decimal place]
Each exterior angle is 360/13 = 27.7 degrees rounded to 1 decimal place
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)
27.50
166.30
166.30
4.5
Mltiply the 3.3 cm by the number of sides 5,then place your decimal and your awnsers 16.5
The answer depends on whether r is the inradius (apothem) or circumradius. It is not clear from the question which.
34.2497 rounded to 1 decimal place equals ... 34.2! I think ^.^
The correct answer is 0.74
Kind of hard to show in here but... Use five circles to form a regular pentagon, then place the other five circles outside the pentagon - so that each is the point of a triangle (with the other points of the triangle formed by the points of the pentagon.
187.10
187.10
0.9