3
bobs ur uncle
It is possible to divide a hexagon into 4 or more - up to infinitely many - triangles.
A polyhedron with two congruent triangles. There are very many possible configurations.
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.
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.
Infinitely many.
3
bobs ur uncle
There are 16 possible triangles.
It is possible to divide a hexagon into 4 or more - up to infinitely many - triangles.
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.
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.
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.