A circle is the cross-sectional shape of a cylinder, which is spun out by a rotating bed. Positive 6×2? Prism is the cross-sectional shape of a prism, which is cut by cutting edges (n is a natural number). Add 6×2 to the infinity of n? Edge and circle are only close, approximate or equivalent, but never equal.
Because the cylinder is a cylinder, the prism is a prism, and the infinite side of the prism is still a prism. Therefore, people summed up in practice that "tangent circle does not rotate". Therefore, workers are not allowed to wash the axle by chamfering when processing the axle. How can you say that a regular hexagon becomes a circle on an infinite side?
In fact, the so-called pi is originally positive 6×2? What is the circumference of a polygon plus 6×2? The ratio of diagonals of polygons passing through the center point should be called plus 6×2? Edge rate. Therefore, whether a regular hexagon is a tangent circle or an inscribed circle, π deduced on the infinite side has nothing to do with the length and area of the circle.
The reason is: 2πR equals inscribed circle plus 6×2? The perimeter of a polygon must be less than the perimeter of a circle; πR？ Is equal to the tangent of a circle 6×2? The area of a polygon must be greater than the area of a circle. There is π. Deviate from πR if you want to satisfy 2πR? ; What if π satisfies πR? , it will deviate from the contradiction of 2π r, if π r? As a circular area, "gains and losses" are inevitable.
When the circle is divided into several infinitesimal sectors, because the area of infinitesimal sectors is greater than zero, the length of the rectangle is πR and the width is limited to r, and each sector is not allowed to exceed the width r of the rectangle when it is put into the rectangle. So we can only use these sectors to form a "sawtooth shape" with teeth up and down. Only when the tooth peaks on the "saw shape" are connected in a straight line to form a rectangle with parallel opposite sides, is the area of this rectangle πR? The area of.
"Saw" is different from rectangle. The upper and lower lengths of the "saw shape" are respectively wave curves formed by connecting arcs and polylines that are not on two straight lines (the endpoints at both ends of the radius are side by side). The length πR of the upper and lower sides of a rectangle refers to two parallel straight lines. Because the meaning of curve is different from that of straight line, "saw shape" does not have the meaning of rectangle. For this reason, the circular area is only a "zigzag" area, not a rectangular area. Conversely: only this "saw-shaped" area can be simplified into a circular area by equal area. Because πR? It is a rectangular area, and the equal area of a circle is a "saw" area. ) When the saw shape overlaps the length and width of the rectangle, it will display: πR? The reason why it is larger than the circular area s is because the area of the "vacancy angle" between the arc outside each sector and the length of the rectangle in the "saw shape" does not belong to the circular area and is defined by πR? It's all in the circular area. Using the value of π: the sector is infinitesimal, and the "vacancy angle" is also infinitesimal, but the number of copies increases correspondingly, and the total "vacancy angle" area does not decrease, but the "crescent" area between the arc and the chord on each sector decreases, which has nothing to do with dividing the infinitesimal sector into "vacancy angle" areas. Furthermore, the area of each infinitesimal "vacancy angle" is always greater than the limit of the area (zero area). Therefore, the "vacancy angle" greater than zero area will never disappear, and it will bring a permanent increase to the circular area.
That is to say: for the rectangular area πR, only the circular area S plus the area of all "vacancy angles" is enough? .
When overlapping rectangular areas and zigzag areas are reduced together, sectors and sectors are spliced into circular areas; Every sector has a "vacancy angle". Is it really the tangent of the circle 6×2? Edge area.
Because "any tangent is 6×2? The area of that edge is lar than the area of its inscribed circle ". So πR? Greater than the circular area s
For this reason, the circular area s is equal to πR? Subtract all "empty corner" areas.
But what about πR? In the early stage, there is still the problem that the area of circle is smaller than that of circle S. The reason why the area of circle S is smaller than that of circle S is that π is infinite, and 2πR is inscribed with positive 6×2? The perimeter of a polygon is "arbitrarily positive 6×2? The circumference of a polygon is less than the circumference of its circumscribed circle, πR must be less than half the circumference of a circle, which will lead to the loss of a sector. The more bits occupied by π, the less sector loss; The less bits occupied by π, the more sectors are lost. When π takes one or two digits, πR? Less than the circular area s. This indicates that the lost sector area is larger than all redundant "vacancy angle" areas. The loss of sector area can be recovered by the infinite value of π. It is natural to find the missing area on the circle. But the more you look for πR? The larger the circular area s is. When π takes more than three digits, the area of the circle will increase due to the redundant "vacancy angle", and the missing sectors will be completely recovered. πR？ It began to get bigger and bigger, larger than the circular area s, so πR? For a circular area: "You gain and you lose". Lost sectors that should not be lost; I got the "vacant corner" I didn't deserve. What is πR? & gts .
For this reason, the circular area s is equal to πR? Subtract all "vacant corner" areas and add all missing sectors.
Add 6×2 for inscribed circle? The area of the polygon πr? Generally speaking, because the infinity of chord center distance R is always smaller than the radius R, R is an unknown number in practical operation. So πr? Do not have the known conditions for calculation.
Because πR? So the circle is the circumscribed circle of 6x2? The area of the side must be greater than the area of the circle. According to the axiom of area "softening" equal area deformation, it is found that if the circular area is 7a? So its circumscribed circle area is 9a? Therefore, the formula of "circular area is equal to 7 times of 65438 +0 square and 3/3 of its diameter" is derived: s=7(d/3)? .