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Prove that every triangular region is measurable

Given that every right triangular region is measurable since it can be obtained as the intersection of two rectangles, prove that every triangular region (not necessarily right triangular) is measurable and has area \frac{1}{2} bh, where b is the base and h is the altitude (or height) of the triangle.

Proof. From geometry, we know that every triangular region is the union of two-nonintersecting right triangles with one leg equal to the height of the triangular region, and the sum of the other leg from each right triangle equal to the base of the triangular region.

Since every right triangle is measurable (given), their union is measurable (Axiom 2 of area). Further, denoting the two right triangles A and B, and the triangular region T, we have

    \[ a(T) = a(A) + a(B). \]

(Since A and B are disjoint, a(A \cap B) = 0.)
Letting the altitude of the triangular region be denoted by h, and its base by b, we have,

    \[ a(A) = \frac{1}{2} (hb_1), \qquad a(B) = \frac{1}{2} hb_2, \qquad \text{with } b_1 + b_2 = b. \]


    \[ a(T) = \frac{1}{2} hb_1 + \frac{1}{2} hb_2 = \frac{1}{2} h(b_1 + b_2) = \frac{1}{2} hb. \qquad \blacksquare \]


  1. Jacare says:

    Hello. Since the area of a right angled triangle is not given, should we not find that too ( using axiom 4 and axiom 2 ), before we find the area of a general triangle?

    Thanks again for all these solutions. They are very helpful to me.

    • luis says:

      Look this is from the complement of the book. Exercise 2. Let A, B be rectangles. By Axiom 5, A, B are measurable. By Axiom 2, A ∩ B measurable.
      a(A ∩ B) = square root(a^2 + b^2)d + ab − (1/2ab +square root(a^2 + b^2)d = 1/2ab,

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