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I'm searching different ways to prove

"If $\triangle ABC$ and $\triangle PQR$ congruent triangles then areas of $\triangle ABC$ and $\triangle PQR$ are equal."

This is the "best" way to me.

Let $a$, $b$, $c$ be sides of the $\triangle ABC$.

then there exists sides $a^\prime$, $b^\prime$, $c^\prime$ in $\triangle PQR$ such that $a=a^\prime$, $b=b^\prime$, $c=c^\prime$.

Hence, with $s=\frac12(a+b+c)=\frac12(a^\prime+b^\prime+c^\prime)$, by Heron's Formula, $$\text{area of $\triangle ABC$} = \sqrt{s(s-a)(s-b)(s-c)} =\sqrt{s(s-a^\prime)(s-b^\prime)(s-c^\prime)}= \text{area of $\triangle PQR$}$$

Thus, the area of $\triangle PQR$ equals the area of $\triangle ABC$. $\square$


I can think of another one or two ways (that may or may not be mathematical proof). but I like to see your opinion, as a motivation to math.

What are the ways you can think-of? Comment or answer below. Thanks.

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  • $\begingroup$ Isn't that an axiom and implicit in the definition of congruent? After all, how do you know congruent squares have the same area? $\endgroup$ – fleablood Sep 3 '18 at 5:10
  • $\begingroup$ @fleablood Two triangles are congruent if their corresponding sides are equal in length, in which case their corresponding angles are equal in measure-from wikipedia link- en.wikipedia.org/wiki/Congruence_(geometry) $\endgroup$ – Bad English Sep 3 '18 at 5:24
  • $\begingroup$ The congruence of two triangles means they are of the same shape and of the same size. If that is the case, their areas must be the same. $\endgroup$ – Mick Sep 3 '18 at 14:37
  • $\begingroup$ My point is doesn't being congruent imply that area is equal? If not, how is the concept of area defined? If you have two squares and they both have same length of sides (and same angles of course) how do you know the two squares have the same area? And what does area mean? You can't prove two congruent triangle have the same area because that is tacitly assumed. Your proof is circular because your formula for the area of a triangle assumes congruent shapes have the same area. $\endgroup$ – fleablood Sep 3 '18 at 15:44
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    $\begingroup$ It's axiomatic. The axioms of area declare that congruent regions (in particular, triangles) have equal area. See, for instance, the Wikipedia entry for "Area". $\endgroup$ – Blue Sep 3 '18 at 17:22
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You can't prove it. In defining what is meant by have a consistent and measurable concept of "area" it is assumed that congruent shapes (having equal measurements and "able to fit ontop of each other") will have the same area. That is what "area" means.

So your proof is circular. Hurons formula (or the standard Area = half base times height) are all derived with the assumption that congruent shapes have the same area.

Consider the basic concept of Area = $\frac 12$ base $\times$ height. Why? Well, because if we have to congruent triangles we can form a parallelogram out of the them. The parallelogram has area base $\times$ height (why?). And so each triangle has half that area (Wait! Why do we assume the to congruent triangles have the same area?)

And you prove that a parallelogram has the area base times height because if you cut of a right triangle from one side to and move it to the other you have a rectangle that has sides that are length base and length height. So the area of the rectangle is base $\times$ height. (But 1; why does cutting the parallelogram and rearranging the pieces keep the area the same?; 2; why is the area of a rectangle base $\times $ height.)

Well, cutting a parallelogram and moving the pieces should keep the areas the same. (Why? Why shouldn't a triangle have one area when it is in one position and a different one in another?) And rectangle can be cut into a grid of squares. A square of $1 \times 1$ must have an area of $1$. (Why? Why should two congruent squares have the same area? And why does a $1 \times 1$ square have an area of $1$ unit?)

ALL of this is based on a single concept: That the quality that we call "area" is an aspect of dimensional lengths and angles. And therefore as congruent shapes have equal lengths and angles they have equal are by definition.

That's it. There is NOTHING to prove.

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  • $\begingroup$ oh, yes my definition for congruent is wrong. thanks $\endgroup$ – Bad English Sep 3 '18 at 16:46

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