Theorem. The area of a parallelogram is equal to the product of its base by its altitude.
The best way to introduce this theorem is to cut a parallelogram from paper, and then, with the class, separate it into two parts by a cut perpendicular to the base. The two parts may then be fitted together to make a rectangle. In particular, if we cut off a triangle from one end and fit it on the other, we have the basis for the proof of the textbooks. The use of squared paper for such a proposition is not wise, since it makes the measurement appear to be merely an approximation. The cutting of the paper is in every way more satisfactory.
Theorem. The area of a triangle is equal to half the product of its base by its altitude.
Of course, the Greeks would never have used the wording of either of these two propositions. Euclid, for example, gives this one as follows: If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle. As to the parallelogram, he simply says it is equal to a parallelogram of equal base and "in the same parallels," which makes it equal to a rectangle of the same base and the same altitude.
The number of applications of these two theorems is so great that the teacher will not be at a loss to find genuine ones that appeal to the class. Teachers may now introduce pyramids, requiring the areas of the triangular faces to be found.
The Ahmes papyrus (ca. 1700 B.C.) gives the area of an isosceles triangle as ½ bs, where s is one of the equal sides, thus taking s for the altitude. This shows the primitive state of geometry at that time.
Theorem. The area of a trapezoid is equal to half the sum of its bases multiplied by the altitude.
An interesting variation of the ordinary proof is made by placing a trapezoid T', congruent to T, in the position here shown. The parallelogram formed equals a(b + b'), and therefore
T = a · (b + b')/2.