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Prove that every right triangular region is measurable and its area is 1/2bh 
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#1
Jul711, 01:01 PM

P: 320

1. The problem statement, all variables and given/known data
Prove that every right triangular region is measurable because it can be obtained as the intersection of two rectangles. Prove that every triangular region is measurable and its area is one half the product of its base and altitude. (Apostol's Calculus Vol1. 1.7 Exercises) 2. Relevant equations We assume that there exists a class M of measurable sets in the plane and a set function a whose domain is M with the following properties: 1) A(S) >= 0 for each set S in M. 2) if S and T are two sets in M their intersection and union is also in M and we have: A(S U T) = A(S) + A(T)  A(S ∩ T) 3)If S and T are in M with S ⊆ T then T − S is in M and a(T − S) = a(T) − a(S). 4) If a set S is in M and S is congruent to T then T is also in M and a(S) = a(T). 5) Every rectangle R is in M. If the rectangle has length h and breadth k then a(R) = hk. 6) Let Q be a set enclosed between two step regions S and T. A step region is formed from a finite union of adjacent rectangles resting on a common base, i.e. S⊆ Q ⊆ T. If there is a unique number c such that a(S) <= c <= a(T) for all such step regions S and T, then a(Q) = c. 3. The attempt at a solution Well, I first tried to define a right triangle as the intersection of two rectangles as Apostol suggests but I failed to do that. I think It's impossible to define a right triangle as the intersection of two rectangles unless we rotate one of them which makes it hard to describe the triangle by an algebraic equation as simple as the one we have for rectangles. I then tried to define a right triangle by writing down three equations and then intersect them. I wrote down 0 <= x <= k, 0 <= y <= h and y<= h/k x + h where k and h are two positive numbers. the intersection of those three equations gives us the set that consists of all pairs like (x,y) s.t 0 <= x <= k & 0 <= y <= h/k x + h. so let's take T to be the set defined in the following way: T = {(x,y) (0 <= x <= k) & (0 <= ky <= hx + kh)}. This is supposed to be a triangle that looks like _\. Now assume S to be the intersection of the three equations: 0<= x <= k & 0<= y < h & y >= h/k x + h. so we define the set S to be: S = {(x,y) (0 <= x <= k) & (hx + kh <= ky <= kh)}. This one is supposed to be a triangle that looks like \. It it easy to show that the union of these two algebraic equations is the set S U T: S U T = {(x,y) 0<= x <= k & 0 <= y =< h} which is indeed a rectangle. Using the 2nd Axiom we have: A(S U T) = A(S) + A(T)  A(S ∩ T) It's easy to show that the intersection of S & T is the line hx + ky = hk. I have previously proved that the area of a line is 0. so if I show that S and T can be congruent by defining a onetoone correspondence between them that keeps the distance between two arbitrary points the same then using the 4th axiom I can conclude that A(S) = A(T) and I'm done. Well, to prove that the two sets T and S are congruent I define a reflection on the points in T across the point (k/2,h/2) that lies on the center of the line hx + ky = hk. if I define r: T > S by r(t) = 2p  t then r is a reflection of t across the point p. any reflection in the Euclidean plane is isometric, so it definitely keeps the distance between any two points unchanged. as I mentioned earlier, we let p be (k/2,h/2) and let t=(x,y) be any point in T. then r(x,y) = 2(k/2,h/2)  (x,y) = (k,h)  (x,y) = (kx,hy). The function r is bijective because for any t in T: ror(t) = t. that means r has an inverse therefore it is bijective. to show that T and S are congruent we show that r which is a bijective isometric function puts the points in T into onetoone correspondence with the points in S. for any (x,y)∈T : 0<= x <= k & 0 <= ky <= hx + kh if we apply r on (x,y) we'll have: 0<= kx <= k > k <= x <= 0 > 0 <= x =< k. 0<= k(hy) <= h(kx) + kh > 0 <= kh  ky <= hk +hx + kh > 0 <= kh  ky <= +hx > 0 <= ky =< +hx  kh > hx + kh <= ky =< kh which proves that r(x,y)∈S which proves that S and T are congruent. therefore A(S) = A(T) using the 4th axiom. Now, as stated earlier: A(S U T) = A(S) + A(T)  A(S ∩ T). A(S U T) = A(S) + A(T)  A(S ∩ T) => A(S U T) = A(S) + A(T)  0 => A(S U T) = 2A(S) => hk = 2 A(S) => A(S) = hk/2. This proves that the area of a right triangle is one half the product of its base and its height. but to prove that a right triangle is a measurable set seems to be impossible or very hard by my approach so I think Apostol's suggestion works well to prove that a right triangle is a measurable set ( the 2nd axiom proves that a right triangle is a measurable set because It's the intersection of two rectangles and the intersection of two measurable sets is a measurable set). Now to generalize this result to any arbitrary triangle one just need to split the triangle into two disjoint right triangles and then use the 2nd property. Well, It seems that the problem has been solved but the way that I've proved it is a hard way of proving this because although It's rigorous but It's very lengthy. I'm here to ask if my proof is right and if yes, does anyone here know a simpler way of proving this using the six axioms given above? 


#2
Jul811, 08:55 AM

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P: 18,084

Hi AdrianZ!
Here are some comments on your proof: 


#3
Jul811, 11:10 AM

P: 320




#4
Jul811, 11:21 AM

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P: 18,084

Prove that every right triangular region is measurable and its area is 1/2bh
Here's my method: For a triangle ABC with rectangular sides AB and AC, you can define a rectangle X with sides AB and AC. And you can define a rectangle Y with side BC. The intersection of these two rectangles X and Y will be the triangle. This is what Apostol wants you to do. Your method isn't wrong! But I guess it can be shortened. If my method above is not clear, then I'll try to clear it up a bit... 


#5
Jul811, 12:08 PM

P: 320




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