Put one angle of each triangle at the center of the hexagon.
Yes, draw the lines from the obtuse angles to the center of the line opposite from it.
27 triangles.27 triangles.27 triangles.27 triangles.
No, there are two circles (incircle, circumcircle) associated with triangles and in general the locations of their centres are different.
Triangles without right angles are:- Scalene triangles Obtuse triangles Isosceles triangles Equilateral triangles
You can see triangles in stairs, stras, bridges, and Disney Epcot Center
Arrange the toothpicks to form a square with a diagonal crossing through the center. Then, create triangles by connecting each corner of the square to the center of the diagonal. Each of the four resulting triangles will be congruent to each other.
20 isosceles triangles with each base being a side of the 20-gon, and the opposite vertices at the center of the polygon.
Pick a point around the center of the polygon. Join each vertex to this point. You will now have 10 triangles. The total angle of these 10 triangles is 10x180 degrees. Now you need to subtract the angles of the ten triangles at the center (which is 360 degrees). Therefore total internal angles is (10x180)-360 = 1800-360 which is 1440 degrees.
No.
If you draw a single square, and then put one X through it, you have 4 triangles. Put another X through it and you have 8. Now, if you want to make four triangles, simply remove the two diagonal lines in the center, and it will create four triangles. Now, since they're separated by the other triangles, it would mean removing four lines, but that is still your answer.
Put one angle of each triangle at the center of the hexagon.
True
359 if the widest part if the triangle is allowed to be 1 degree and the apexes start in the center of the square.
The base length of both triangles. The base using the shadow method would be the shadow. The base using the mirror method would be from the object to the center of the mirror.
All isosceles triangles are not equilateral triangles
How the center of irregular shapes I am unaware of, but for triangles, where any one side is flat on altitude 0, the altitude of the center is 2 x Area divided by Perimeter That formula is used for getting the in-radius of a circle. The center of the circle of best fit is the center of the triangle.