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Yes, any polygon can be composed using only triangles.
How can a you make a hexagon using a trapozid and 2 triangles
Oh, dude, a hexagon has six sides, right? So, you'd need to draw three triangles to make a hexagon. Each triangle would cover two sides of the hexagon, like a sneaky little puzzle piece fitting into place. So, like, three triangles would do the trick.
Because all other polygons are made from triangles as for example an 8 sided octagon has 6 triangles within it or a 6 sided hexagon has 4 triangles within it.
If all the vertices of the regular hexagon are joined to the centre of the hexagon, 6 equilateral triangles are created: the area of the hexagon is 6 times the area of one of these triangles. If the length of the side of the hexagon is m, then the length of each of the sides of these triangles is also m. Using Pythagoras the height of these triangles can be found to be m x sqrt(3)/2. Thus the area of the hexagon = 6 x area triangle = 6 x (m x m x sqrt(3)/2) / 2 = (3/2) sqrt(3) m2 ~= 2.6 x square of length_of_side
Yes, any polygon can be composed using only triangles.
How can a you make a hexagon using a trapozid and 2 triangles
You can do that by simply proving that the hexagon is a regular hexagon. You could do this by dividing the hexagon into 6 equilateral triangles of the same size successfully that tesselate to form a hexagon, thus proving that all sides are equal.
use 4 triangles, make 2 trapezoids then put them into a hexagon
Oh, dude, a hexagon has six sides, right? So, you'd need to draw three triangles to make a hexagon. Each triangle would cover two sides of the hexagon, like a sneaky little puzzle piece fitting into place. So, like, three triangles would do the trick.
Because all other polygons are made from triangles as for example an 8 sided octagon has 6 triangles within it or a 6 sided hexagon has 4 triangles within it.
24
If all the vertices of the regular hexagon are joined to the centre of the hexagon, 6 equilateral triangles are created: the area of the hexagon is 6 times the area of one of these triangles. If the length of the side of the hexagon is m, then the length of each of the sides of these triangles is also m. Using Pythagoras the height of these triangles can be found to be m x sqrt(3)/2. Thus the area of the hexagon = 6 x area triangle = 6 x (m x m x sqrt(3)/2) / 2 = (3/2) sqrt(3) m2 ~= 2.6 x square of length_of_side
30, its a combination. C(6,3) because there are six vertices of a hexagon and three vertices of a triangle
In a six-sided polygon (hexagon), there are a total of four triangles that can be formed. Each triangle is formed by connecting three non-adjacent vertices of the hexagon. This can be calculated using the formula n(n-3)/2, where n represents the number of sides of the polygon. So, for a hexagon, the calculation would be 6(6-3)/2 = 4 triangles.
The question does not state that it is a regular hexagon and so you may not assume that it is. Therefore, there is no simple formula because a hexagon can have very many shapes. One method would be to pick a point in the plane of the hexagon and join it to all the vertices. This divides up the hexagon into triangles. Their areas can be calculated using base and height, or three sides, or two sides and included angle - whichever you like. Finally the areas of the triangles must be combined to get the area of the hexagon.
The hexagon will consist of 6 equilateral triangles of 3 equal sides of 10 cm and the apothem will divide the triangle into 2 right angle triangles with a base of 5 and an hypotenuse of 10 and so by using Pythagoras' theorem the height of the triangle which is the apothem works out as 5 times square root of 3 or about 8.66 cm rounded to 2 decimal places.