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Prove that trapezoids and parallelograms are measurable

by RoRi

Prove that trapezoids and parallelograms are measurable, and that the formulas for their areas are given by

    \[ \frac{1}{2} a (b_1+b_2), \qquad \text{and} \qquad ab, \]

respectively.

Proof. Every trapezoid is measurable since, from geometry, we know it is the union of a rectangle and two right triangles (which are pairwise disjoint, and each of which is measurable by the axioms and by the previous exercise).
Further, its area is the sum of the areas of the right triangles and the rectangle (since they are pairwise disjoint, their intersection has zero area). To calculate this area we specify the lengths of the two unequal sides of the trapezoid to be b_1 and b_2. The height is denoted by a. Then, the area of the rectangle is ab_1. The area of the triangles is \frac{1}{2} a \cdot b_3 and \frac{1}{2} a \cdot b_4 where b_1 + b_3 + b_4 = b_2. So, denoting the trapezoid by T, we have

    \[ a(T) = ab_1 + \frac{1}{2} ab_3 + \frac{1}{2} ab_4 = \frac{1}{2}ab_1 + \frac{1}{2}a (b_1 + b_3 + b_4) = \frac{1}{2}a (b_1+b_2). \]

Next, a parallelogram is just a special case of a trapezoid, in which b_1 = b_2; hence, by the above formula, and denoting the parallelogram by P,

    \[ a(P) = \frac{1}{2}a(2b) = ab. \qquad \blacksquare \]

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