An infinite number are possible.
3
bobs ur uncle
It is possible to divide a hexagon into 4 or more - up to infinitely many - triangles.
Well, to get that answer, there's about 30 inches in a meter, and about 2.5 cm in an inch. So, to get how many meters 99cm goes into, just multiply 30 by 2.5 and divide 99 by what you get.
A polyhedron with two congruent triangles. There are very many possible configurations.
Infinitely many.
To determine the number of triangles with a perimeter of 15cm, we need to consider the possible side lengths that can form a triangle. The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. With a perimeter of 15cm, the possible side lengths could be (5cm, 5cm, 5cm) for an equilateral triangle, (6cm, 5cm, 4cm) for an isosceles triangle, or (7cm, 5cm, 3cm) for a scalene triangle. Therefore, there are 3 possible triangles that can have a perimeter of 15cm.
3
bobs ur uncle
It is possible to divide a hexagon into 4 or more - up to infinitely many - triangles.
There are 16 possible triangles.
Well, to get that answer, there's about 30 inches in a meter, and about 2.5 cm in an inch. So, to get how many meters 99cm goes into, just multiply 30 by 2.5 and divide 99 by what you get.
A polyhedron with two congruent triangles. There are very many possible configurations.
There is only one equilateral triangle with a perimeter of 60 units. Its side lengths are integers.
33 triangles
There are an infinite number of each. I can't draw them right now, as I am busy.
The least number of obtuse triangles, if all possible triangles are drawn for n points in a plane, is zero. If all the n points lie in a straight line, no triangles are possible and so no obtuse triangles are able to be drawn; thus for any number n, there is a possibility that no obtuse triangles can be drawn, so the least possible number of obtuse triangles drawn is zero.