Four, in order to form a tetrahedron; this is alos the simplest possible 3-dimensional object constructed of regular polygons.
It looks like four equilateral triangles that have been put together so that one vertex of each meet at a point.
Each angle in an equilateral triangle is 60 degrees. In order to create a regular tessellation of an area, we need for the angles of the polygons we are putting near each other to sum to 360 degrees. If you place six equilateral triangles so that all of them share a vertex, and each triangle is adjacent to two others, you get 60*6 = 360 degrees in that vertex. Please see related link for a demo of a triangular tessellation.
They're similar triangles.
18 triangles
It can form 7 triangles
10
It looks like four equilateral triangles that have been put together so that one vertex of each meet at a point.
Equilateral triangles have 3 equal sides, but isosceles only have two equal sides. Also, equilateral triangles have three 60° angles. Isosceles triangles have two congruent base angles and a vertex angle.
Equilateral triangles can tile a plane, but regular heptagons cannot; nor can they tile the plan together. Where vertices meet (at a point on the plane) there is a complete turn of 360°. Each vertex of an equilateral triangle is 60°; 360° ÷ 60° = 6, a whole number of times, so a whole number of equilateral triangles can meet at a vertex of the tiling. Each vertex of a regular heptagon is 128 4/7°; 360° ÷ 128 4/7° = 2 4/5 which is not a whole number, so a whole number of regular heptagons cannot meet at a vertex of the tiling, so there will be gaps. With one regular heptagon there are 360° - 128 4/7° = 232 3/7°, but this cannot be divided by 60° a whole number of times, so one regular heptagon and some equilateral triangles cannot meet at a vertex of the tiling without gaps. With two regular heptagons there are 360° - 2 x 128 4/7° = 102 6/7°, but this cannot be divided by 60° a whole number of times, so two regular heptagons and some equilateral triangles cannot meet at a vertex of the tiling without gaps. With three or more regular heptagons, they will overlap when trying to place them on a plane around a point - leaving no space for any equilateral triangles.
There are two ways to think of this question, if the triangles don't have to intersect then the answer is zero. If the triangles have to intersect, then the minimum number of points is one, if the triangles meat at vertex to edge or vertex to vertex.
equilateral triangles
That shape is a unit equilateral triangular dipyramid, and it has 6 faces.
Well a Regular Octahendron is simply a platonic solid composed of 8 equilateral triangles, four of which meet at each vertex.
It depends on the type of triangle: -- scalene triangles have three sides of different length, and no lines of symmetry -- isoceles triangles have one line of symmetry that includes the apex -- equilateral triangles have three lines of symmetry, all bisectors through a vertex
Put four equilateral triangles so that each one of them has a vertex at a single point and the triangles abut one another. The shape will be 4/6 (= 2/3) of a regular hexagon.
Since every triangle can be split into two triangles simply by drawing a line from any vertex to the side opposite that vertex, the answer to the question is that there can be no rule to find a number that is infinite.
In two dimensions 6 triangles meet at a vertex. In 3-dimensions any number of triangles (greater than 2) can meet at a vertext - a pyramid with the base in the shape of an n-gon will have n triangles meeting at its apex.