If Two Triangles Have The Same Area, What Else Is Needed For Congruence?

skipjack

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Dec 2006
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If a triangle has area of 6 and sides of 3 and 4, what length other than 5 could the other side have?
For that example, sin(θ) = 1 where θ is the angle between the given sides, so θ = 90°, which implies the third side has length 5, but there are typically two possible values for θ.

ASS doesn't prove congruence
It doesn't in some cases. It does if the given angle isn't acute.
 
Oct 2013
717
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New York, USA
$\triangle DEF$ has sides $8,12,18$ and $\triangle PQR$ has sides $12,18,27$. We have $12=12$, $18=18$ and since the triangles are similar all their angles are the same.
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The congruent parts are not corresponding. You taught that AAS and ASA are an oversimplification because the congruent parts need to be corresponding. I don't know if there's any way to add "C" to the three letters to indicate that the parts must be corresponding. I wasn't taught examples of how triangles can have AAS in common that are not corresponding and not be congruent. I wasn't taught how to attempt to prove triangles congruent when they may or may not be congruent and use the attempt to prove if they're congruent. I previously said (not here) that the way I was taught proofs was flawed, and this shows my point. The proofs say not to use the appearance of the triangles, and sometimes say they are not drawn to scale, but the proofs do not state what the corresponding parts are, so students are implicitly identifying corresponding parts from the appearance. They teach proofs where it's impossible to know that congruent parts are corresponding without using appearance. Obviously they are not going to ask to prove triangles are congruent when they're not, but assuming they're congruent because that's what you're told to prove is circular logic.

It shows that math can have details that seem irrelevant and can be irrelevant for the level of math they're being taught, but the details are necessary to avoid false conclusions in higher math.

For that example, sin(θ) = 1 where θ is the angle between the given sides, so θ = 90°, which implies the third side has length 5, but there are typically two possible values for θ.
So are there two possible values in all acute triangles, all acute or obtuse triangles, or something else? I'm not saying you're wrong, but I'm still interested in numbers for two congruent sides and equal areas with the third sides not congruent.

Saying ASS proves congruence if and only if the triangles are not acute makes sense because the visual example of ASS being congruent for triangles that are not congruent shows two acute triangles. If the triangles are right, then they would be congruent hy-leg.
 
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skipjack

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I'm still interested in numbers for two congruent sides and equal areas with the third sides not congruent.
If the area is 210 and two sides are 17 and 28, the third side can be 25 or 39.

Saying ASS proves congruence if and only if the triangles are not acute makes sense . . .
No it doesn't, because ASS never tells you that the triangle is acute. If the given angle isn't acute, the triangle can't be acute, and congruence follows (if a triangle is possible). If the given angle is acute and a triangle is possible, there are two possible triangles, one acute and one obtuse.
 
Oct 2013
717
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New York, USA
If the area is 210 and two sides are 17 and 28, the third side can be 25 or 39.


No it doesn't, because ASS never tells you that the triangle is acute. If the given angle isn't acute, the triangle can't be acute, and congruence follows (if a triangle is possible). If the given angle is acute and a triangle is possible, there are two possible triangles, one acute and one obtuse.
If you take decrease one of the base or height and leave the other alone, the area decreases. If you decrease one of the sides and leave two alone, the area can increase. I didn't know that. How many shapes is that possible for? Can you decrease at least one side of a quadrilateral, leave the rest alone, and get a larger area? I'm confident it's not possible for parallelograms.

For a 17-25-28 triangle and a 17-28-39 triangle, I tried using the law of cosines to solve for the angles. Then I tried using the law of cosines for one angle and using the law of sines for the other two. I got the answers both ways, and they're both wrong because they add up to under 180 degrees.

I wasn't saying that ASS determines if the triangle is acute. I was saying that if you know without ASS that a triangle is acute, you can use ASS to prove congruence.
 
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skipjack

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I'm confident it's not possible for parallelograms.
It is possible for parallelograms.

I got the answers both ways, and they're both wrong because they add up to under 180 degrees.
The angles for the 17-25-28 triangle are approximately 81.2°, 61.93°, and 36.87°, whose sum is 180°.

I was saying that if you know without ASS that a triangle is acute, you can use ASS to prove congruence.
That would have been true, but it isn't what you posted. You stated "Saying ASS proves congruence if and only if the triangles are not acute makes sense".
 
Oct 2013
717
91
New York, USA
It is possible for parallelograms.


The angles for the 17-25-28 triangle are approximately 81.2°, 61.93°, and 36.87°, whose sum is 180°.


That would have been true, but it isn't what you posted. You stated "Saying ASS proves congruence if and only if the triangles are not acute makes sense".
How is it possible for parallelograms? You can take a triangle off one side and move it to the other side to make a rectangle with same base and height. You can do the opposite to make a rectangle into a parallelogram that isn't a rectangle. Why is it possible to do for a parallelogram that is not a rectangle but not a rectangle?

If two triangles have two pairs of congruent sides and one pair of sides that are not congruent, such as 17-38-25 and 17-38-39 where I wrote the side that varies last rather than in increasing order the first time, how many pairs of angles can be congruent?

Even if I did not phrase it properly, I never believed that ASS proved whether the triangle was acute.

Edit: I thought about if the diagonals and base are sufficient to solve for the height, and if the diagonals and height are sufficient to solve for the base. If the answer is no, that lets one pair of sides increase, the other pair stay the same, and the area decrease.
 
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Feb 2010
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I gave you examples and pictures in post #2 and post #9 showing two triangles with the same area and two sides congruent but the triangles themselves are not congruent. Here is a picture of two parallelograms that have two sides congruent and the same area but they are different.

It has finally dawned on me. The concept you seem to be missing is that for area you need height and if two lines are parallel then the perpendicular height between them is always the same. Think about it.
 

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skipjack

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Dec 2006
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Why is it possible to do for a parallelogram that is not a rectangle but not a rectangle?
By changing just its angles, the area of any parallelogram that isn't a rectangle can be increased or decreased. For a parallelogram that happens to be a rectangle initially, changing just its angles can decrease its area, but not increase it.
 
Oct 2013
717
91
New York, USA
It has finally dawned on me. The concept you seem to be missing is that for area you need height and if two lines are parallel then the perpendicular height between them is always the same. Think about it.
I know that area = base*height and that the distance between parallel lines is constant, but I guess I wasn't posting like that. It looks like given a parallelogram that is not a rectangle, if the perimeter remains the same and the angles get more equal (closer to a rectangle), the area increases, and if the perimeter remains the same and the angles get less equal, the area decreases. Is that true? Given a fixed perimeter, the greatest four-sided area is a square, which is a special case of a rectangle. Therefore leaving all the sides and therefore the perimeter alone and changing the angles so it is no longer a square will decrease the area.

A parallelogram can be divided into two congruent triangles. By reflecting a triangle over one of its sides, you can make any triangle be half of a parallelogram. In that case, any way of making the perimeter and area of a triangle move in opposite directions also works for a parallelogram that is not a rectangle. For a rectangle, the congruent triangles would be right triangles. The sides of the rectangle would be the legs of a right triangle, which leaves only one possibility for the hypotenuse. If you know two sides of a triangle, that it is not a right triangle, and nothing else about the angles, the third side can be any length that satisfies the inequality that the sum of the smallest two sides is greater than the larger side.

Given two sides of a triangle, you can calculate the range of possibilities for the third side. Can you calculate a range of areas? For a right triangle, the area is half the product of the two smallest sides, and for any other triangle, the area is less than half the product of the two smallest sides. Let's say two sides are 2 and 3. The third side ranges from 1 to 5 excluding 1 and 5. The product of the two smallest sides is less than or equal to 6, so the area must be less than or equal to 3.
 
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