Evaluate tan15 ° / (tan215 ° - 1) To process

Evaluate tan15 ° / (tan215 ° - 1) To process

tan15º=tan(60º-45º)=(tan60º-tan45º)/(1+tan60ºtan45º)=(√3-1)/(√3+1)=(√3-1)^2/2=2-√3∴tan15°/（tan215°-1）=(2-√3)/[tan(180º+15º)-1]=(2-√3)/[tan15...

Known: circle inscribed square area is 32, find the same circle inscribed hexagon area As shown in the figure

The area of the square is 32
The side length of the square is 4 times the root 2
The diagonal is 8, that is, the diameter of the circle is 8
The diagonal of the inscribed regular hexagon is 8
The inscribed regular hexagon is divided into 6 triangles with 3 diagonals
The side length of each triangle is half of the diagonal length = 4, the triangle area is 4 times the root sign 3, and the hexagon area is 24 times the root sign 3

Area of regular hexagon If the side length is 0.5, how to calculate the area? Side length x side length x 6 if we use two isosceles trapezoid to calculate the area, how to calculate the height of isosceles trapezoid? What about the area of N deformation?

It can be divided into six small regular triangles
Area formula of regular triangle
S = root 3 * a 2 / 4
(a is the side length of an equilateral triangle)
Therefore, s total = (root 3 * 0.5) / 4 * 6
=(3 pieces 3) / 8
It is very difficult to find the area of n-polygon directly. We can use the method of partition and use trigonometric function to calculate the height

In this paper, the circumferences of circumscribed regular hexagons, circles and inscribed regular hexagons are proved,

Let the radius of the circle be r. then the length of the inscribed hexagon of the circle is R. the length of the six deformed sides of the circle is 2 / 3 times of the root sign 3R
The circumference length is 2 pails R
The inner circumference is 6R
The circumscribed perimeter is 4 times root 3R
So the ratio is: Pie: 3:2 times root 3
therefore

What is the ratio of the side length of an inscribed regular triangle to an inscribed regular hexagon of the same circle?

Let the radius of a circle be r. for an inscribed regular hexagon of a circle: to connect the center of a circle with each vertex of a regular hexagon, it can be seen that a regular hexagon is divided into six exactly the same positive △, so the side length of a regular hexagon a = R. for an inscribed regular triangle of a circle, the radius of the circle is

The ratio of the area of an inscribed regular triangle of the same circle to an inscribed regular hexagon is

Let the radius of the circle be r, for the inscribed regular n-polygon, and if the center of the circle is connected to each vertex, then it is divided into n isosceles triangles (the waist is r, the vertex angle is 360 / N), and the area Sn = n * 1 / 2 * r * R * sin (360 / N), then the area of the regular triangle is 3 * 1 / 2 * r * sin120 = 3 √ 3 / 4 * R ^ 2, and the area of orthogonal polygon is 6 * 1 / 2 * r * r * sin60 = 3 √ 3 / 2 * R ^ 2

The ratio of the side length of a circle circumscribed regular hexagon to a circle inscribed regular hexagon is________

The ratio of the side length of a circle circumscribed regular hexagon and a circle inscribed regular hexagon is (root 6:3)
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Given that the side length of inscribed regular hexagon of circle O is 3cm, try to find the area of inscribed regular triangle square of circle o

It can be seen from the figure that the radius of the circle is
The area of the triangle is
The square area is

A is a point on ⊙ o, the inscribed square ABCD and the inscribed regular hexagon of ⊙ O. in the figure above, if the point E is on the arc ad, De is one of the inscribed regular dodecagons of ⊙ o A is a point on ⊙ o, the inscribed square ABCD and the inscribed regular hexagon of ⊙ o

Let the center of the circle be o
Connect Ao, EO, do
∵ AE is the side of a regular hexagon
∴∠AOE=60°
∵ ad is a square side
∴∠AOD=90°
∴∠EOD=30°
360÷3=12
⊙ De is the side of ⊙ o inscribed with a regular dodecagon

It is known that the square ABCD is the inscribed square of circle O. its side length is 2. Find the radius and the distance between the edge centers RT

If you don't have a picture, you can listen to it and prove: do on perpendicular to BC, perpendicular foot is n, and extend n to circle O to point m, do OE perpendicular to CD, perpendicular foot to e, connect OC. Because quadrilateral ABCD is square, so quadrilateral once is square, so OC is square, once diagonal is square, and because angle DCB = 90 ° so angle OCM = 45 °