By cutting off two of its corners which will leave you with 2 triangles and 1 pentagon.
1 triangle is formed.
Consider the pentagon ABCDE. By drawing diagonals from B, we get: 1. Triangle ABE 2. Triangle BDE 3. Triangle BCD -Ashwin Hendre
A regular pentagon has five (5) equilateral triangles within it. Find the area of each triangle (1/2bh where b is the base of the triangle or the length of a side of the pentagon, and h is the height of the triangle or the apothem of the pentagon) and multiply the area of the triangle times five (5).
You didn't say it was a regular pentagon. For an arbitrary pentagon, you would calculate its area as you would for any polygon: divide it up into triangles, and add up the areas of the triangle. The area of a triangle is 1/2 times the base times the height, the height being the length of the perpendicular dropped to the base from the opposite vertex.
With a good ol' fashion pen. Impossible
You'd have to have fractional parts of triangles. Each group would have 1 and 1/3 triangle.
there are 27 triangles in a triangle
One possible way to divide a pentagon into five parts is (assuming this is a convex pentagon) to start by placing a dot directly in the center. Then, draw a 5 lines from that center dot connecting to the 5 points around the edge of the pentagon. You should now have 5 triangles instead of 1 pentagon.
a hexagon * * * * * Depending on the shape and sizes of the triangles and where they are attached, you can also get a kite, a rectangle, a parallelogram, a decagon, octagon, pentagon.