Sections of nuts and pencils, bee hundreds and snowflakes have the hexagonal - "hexagonal" - form, for example. The correct geometrical figures of such form have the certain feature distinguishing them from other flat polygons. It is that the radius of the circle described about hexarutting is equal to length of its party - in many cases it considerably simplifies calculation of parameters of a polygon.
Instruction
1. If in statements of the problem the radius (R) of the circle described about the correct hexagon is given, it is necessary to calculate nothing - this size is identical to length of the party (t) of hexarutting: t = R. At the known dimeter (D) just divide it in half: t = D/2.
2. The perimeter (P) of the correct hexagon allows to calculate length of the party (t) simple operation of division. As a divider use number of the parties, i.e. the six: t = P/6.
3. Radius (r) of the circle entered in such polygon is connected with length of its party (t) in a little more difficult coefficient - double radius, and divide the received result into a square root from the three: t = 2*r/√ 3. The same formula with use of diameter (d) of an inscribed circle on one mathematical operation will become shorter: t = d/√ 3. For example, at radius in 50 cm length of the party of a hexagon has to be approximately equal 2*50 / √ 3 ≈ 57.735 cm.
4. Izvestnaya Square (S) of a polygon with six tops allows to calculate length of its party (t) too, but the numerical coefficient connecting them precisely is expressed through fraction from three natural numbers. You divide two thirds of the square into a square root from the three, and take a square root from the received value: t = √ (2*S / (3 * √ 3)). For example, if the area of a figure is 400 cm², length of its party has to be approximately √ (2*400 / (3 * √ 3)) ≈ √ (800/5.196) ≈ √153.965 ≈ 12.408 cm.
5. Length of the circle (L) described about the correct hexagon is connected with radius, so and with a length of the party (t) through Pi's number. If it it is given in statements of the problem, divide its size into two numbers of Pi: t = L/(2*π). Let's tell if this size is equal to 400 cm, length of the party has to be about 400 / (2*3.142) = 400/6.284 ≈ 63.654 cm.
6. The same parameter (l) for an inscribed circle allows to calculate length of the party of a hexagon (t) calculation of a ratio between it and the work of number of Pi on a square root from the three: t = l/(π * √ 3). For example, if length of an inscribed circle is 300 cm, the party of a hexagon has to have the size approximately equal 300 / (3.142 * √ 3) ≈ 300 / (3.142*1.732) ≈ 300/5.442 ≈ 55.127 cm.