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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,

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

Thus,

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

7 comments

  1. Eric says:

    Pardon my poor English, but I gave this problem some thought. Let’s try to use those area axioms from this section to prove that the area of a right triangle whose smaller sides are the dimensions of a rectangle predicted in axiom 5.

    Let us draw such a rectangle R with vertices A, B, C and D such that CD lies over the X-axis and CD=a, while AC is the vertical side and AC= h. Axiom 5 says that

    a(R)= h.a [1]

    Note that in this rectangle, the right triangles DCB (let’s call set A1) and ABC (set A2), which share the same hypotenuse are congruents, which means that we can use axiom 4, or:

    a(A1) = a(A2) [2]

    Lastly, by axiom 2, we can state that:

    a(R) = a(A1) + a(A2) – a(A1∩A2)

    Since A1 and A2 are disjoint, a(A1∩A2) = 0; using [1] and [2]:

    h.a = 2.a(A1) a(A1) = 1/2.h.a

    Which corroborates with the solution given here, I think.

    • Eric says:

      My mistake: sets A1 and A2 are not disjoint, as like I said they share a hypotenuse, which is a line segment. But, as exercise 1 made clear, the area of segment lines are 0 so, still, a(A1∩A2) = 0.

  2. 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,

      • Anon says:

        Sorry but from where did you deduced the subtraction in your equality? I can see that you used the sides of a rectangle ab to deduce a side c but the subtraction of 1/2ab + sqrt(a² +b²)d

        I simply couldn’t get it.

        In this subtraction, aren’t you assuming that the “remainer” of the area is 1/2 of ab?

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