Basic Level

Areas

The application allows arriving to the formula of the area of ​​a triangle using the construction of a parallelogram.

The triangle [A’B’C ‘] is a copy of the triangle [ABC] obtained through a translation of the triangle [ABC] along the line containing the vertices B and C, so that the vertex C’ coincides with the vertex B.

Through a 180º rotation in the direct direction, the vertex B ‘coincides with the vertex C and the two triangles are juxtaposed, forming a parallelogram with twice the area of ​​the triangle [ABC].

So, which formula can we obtain for the calculation of the area of ​​the triangle [ABC]?

The area of ​​the triangle [ABC] is given by $$b \times h\over 2$$, where $$b$$ is the length of the segment [AB] and $$h$$ is, simultaneously, the height of the parallelogram [ABA’C] resulting from the juxtaposition of the two triangles and the height of the triangle [ABC] relative to the [AB] side.