Derivation of the Area Formula for Triangles Using Axioms

[vrble]

From Apostol's Calculus Volume I, "Area as a Set Function"

1. Homework Statement :
Right triangular regions are measurable because they are constructed from the intersection of two rectangles. Prove that all triangular regions are measurable and have an area of the product of one-half, their base length, and their height.

Homework Equations

(The class of all measurable sets is referred to as M and the following axioms are relevant to this problem.)
a(x) = Area of x
1. If S and T are both sets in M, then S ∪ T is in M and a(S ∪ T) = a(S) + a(T) - a(S ∩ T).
2. All rectangles are members of M, and their area can be calculated as the product of their base length and their height.

The Attempt at a Solution

Take an arbitrary triangle and take the longest side as the base and extend a parallel line from it's height to it's base, thus forming two right triangles. This can be done to any triangle, thus by axiom 1 and the previously known information that all right triangles are measurable we can conclude that all triangular regions are also measurable. As for the area of this arbitrary triangular region, we inscribe it within a rectangle with base and height equal to that of the triangle. This forms two additional triangles that are congruent to their respective counterparts in the inscribed triangle and the sum of these parts equals the area of the entire rectangle. We find that this inscribed triangle is precisely one half the area of the rectangle. Thus the area of the triangle is equal to one-half times its base times its height.

 

[Mark44]

From Apostol's Calculus Volume I, "Area as a Set Function"

1. Homework Statement :
Right triangular regions are measurable because they are constructed from the intersection of two rectangles. Prove that all triangular regions are measurable and have an area of the product of one-half, their base length, and their height.

Homework Equations

(The class of all measurable sets is referred to as M and the following axioms are relevant to this problem.)
a(x) = Area of x
1. If S and T are both sets in M, then S ∪ T is in M and a(S ∪ T) = a(S) + a(T) - a(S ∩ T).
2. All rectangles are members of M, and their area can be calculated as the product of their base length and their height.

The Attempt at a Solution

Take an arbitrary triangle and take the longest side as the base and extend a parallel line from it's height to it's base, thus forming two right triangles. This can be done to any triangle, thus by axiom 1 and the previously known information that all right triangles are measurable we can conclude that all triangular regions are also measurable. As for the area of this arbitrary triangular region, we inscribe it within a rectangle with base and height equal to that of the triangle. This forms two additional triangles that are congruent to their respective counterparts in the inscribed triangle and the sum of these parts equals the area of the entire rectangle. We find that this inscribed triangle is precisely one half the area of the rectangle. Thus the area of the triangle is equal to one-half times its base times its height.

Do you have a question?

If you're asking whether your proof is reasonable, it looks OK to me except for one part.

extend a parallel line from it's height to it's base

I think I understand what you're trying to say, but what you actually say is pretty confused.
"extend a parallel line" -- I think you mean "extend a perpendicular from the top vertex of the triangle to the base..."

When you say "extend a parallel line" the obvious question is "parallel to what?"

 

[vrble]

Yes, I meant "Take the longest side of the triangle as its base and extend a perpendicular from the vertex opposite the base." Sorry, I must have been thinking about something else while writing up the proof. Is there any improvements that could be made?