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

Oct 2013
713
91
New York, USA
By the semiperimeter formula, given the area and two sides, the third side can be calculated, so area and two sides is sufficient to prove triangles congruent. If two pairs of angles are congruent, the third pair must be congruent. With all the angles known, the law of sines can give the ratio of the sides and be used to express all sides in terms of one of them. Then that and the area can be used in the semiperimeter formula to solve for one side, which can solve for the other two. Are area, one side, and one angle sufficient to prove congruence? Phrased another way, can you make two triangles with an area of x, one pair of congruent sides of length y, and one pair of congruent angles with measure z that are not congruent? If it is possible, can you give an example?
 
Feb 2010
737
162
"so area and two sides is sufficient to prove triangles congruent"

No way. Let $A=$ area. $A = \dfrac{1}{2}ab \sin \theta$ where $\theta$ is the included angle between sides $a$ and $b$. Since $\sin 30^{\circ} = \dfrac{1}{2}= \sin 150^{\circ}$, then a triangle with sides $a,b$ and included angle $30^{\circ}$ has the same area as a triangle with sides $a,b$ and included angle $150^{\circ}$. These two triangles are clearly not congruent.
 
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skipjack

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Dec 2006
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Area, one side, and one angle suffice to prove congruence.
 
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Oct 2018
9
1
Arizona
"so area and two sides is sufficient to prove triangles congruent"

No way. Let $A=$ area. $A = \dfrac{1}{2}ab \sin \theta$ where $\theta$ is the included angle between sides $a$ and $b$. Since $\sin 30^{\circ} = \dfrac{1}{2}= \sin 150^{\circ}$, then a triangle with sides $a,b$ and included angle $30^{\circ}$ has the same area as a triangle with sides $a,b$ and included angle $150^{\circ}$. These two triangles are clearly not congruent.
But your two triangles don't have a common angle.
 
Feb 2010
737
162
"so area and two sides is sufficient to prove triangles congruent"

Nowhere do you say "common angle".
 
Feb 2010
737
162
Two sides and three angles don't even guarantee congruent triangles.
 
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skipjack

Forum Staff
Dec 2006
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By the semiperimeter formula, given the area and two sides, the third side can be calculated
There would typically be two distinct possible values for the third side.
 
Oct 2013
713
91
New York, USA
"so area and two sides is sufficient to prove triangles congruent"

No way. Let $A=$ area. $A = \dfrac{1}{2}ab \sin \theta$ where $\theta$ is the included angle between sides $a$ and $b$. Since $\sin 30^{\circ} = \dfrac{1}{2}= \sin 150^{\circ}$, then a triangle with sides $a,b$ and included angle $30^{\circ}$ has the same area as a triangle with sides $a,b$ and included angle $150^{\circ}$. These two triangles are clearly not congruent.
I haven't seen the formula of area using the included angle, but it looks like area = (base*height)/2 for an included angle of 90 degrees is a specific case of that just like the Pythagorean Theorem is a special case of the law of cosines. I took the square of the area formula, changed theta to c, and expressed (sin c)^2 as 1-(cos c)^2 to use the same terms as the law of cosines. Then can the law of cosines and area formula with two sides and the included angle be used to derive the semiperimeter formula?

Take a 3-4-5 right triangle with area of 6. Let's say you have the area and two sides and try to solve for the third side. First, let's say you know the sides of 3 and 4 but don't know they're perpendicular to be the base and height. The square of the area is 36. Let the semiperimeter be p, so p = (a+b+c)/2, a is 3, and b is 4. p = (7+c)/2. p-a = (7+c)/2 - 3. p-b = (7+c)/2 - 4. p-c = (7+c)/2 - c. You know the product of those four terms. The only way I can think of for the triangles to not be congruent is for it to produce an equation of a higher degree that makes multiple solutions. Is that what happens? If a triangle has area of 6 and sides of 3 and 4, what length other than 5 could the other side have?

Two sides and three angles don't even guarantee congruent triangles.
I don't understand. One side and three angles proves congruence ASA or AAS.

There would typically be two distinct possible values for the third side.
That made me think of the drawing that shows two possible triangles if you know ASS, which doesn't prove congruence. However, my example is different because I'm talking about equal areas, and the example showing that ASS doesn't prove congruence uses triangles of unequal areas.
 
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Feb 2010
737
162
I haven't seen the formula of area using the included angle,
It is a standard high school trig area formula. Google it.

In the attached picture, $\triangle ABC$ and $\triangle ABD$ both have $AB=AB$ and $BC=BD$ and both have area $\dfrac{1}{2}(8)(3)=12$.
pic.png
I don't understand. One side and three angles proves congruence ASA or AAS.
$\triangle DEF$ 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.
pic2.png
 
Oct 2018
9
1
Arizona
It is difficult to see who is agreeing with whom in this thread. So I will throw in my additional 2 cents worth. Consider the following picture:
triangles.jpg
Let's say the original triangle is ABC, and the base AB is equal to the base of the second triangle. Place the vertex C above the base, which sets the height h. Place the second triangle with its equal base on AB and its third vertex above. In order to have the same area its vertex must be on the dotted horizontal line to have the same height. I have indicated the given angle at A in a green arc. The OP's question said nothing about the angles being corresponding angles. It's pretty clear if the second triangle's angle A is the same, the triangles are the same. Consider the case when the second triangle's equal angle is at B, so it is ABC' (the red triangle). Then ABC is congruent to BAC' by symmetry. But what if the equal angle in the second triangle is at the vertex on the dotted line? Look at triangle ABC'' in blue. Clearly the angle at C'' is less than the angle at A. Now look at triangle ABC''' in violet. In the figure the angle at C''' is slightly larger than a right angle and clearly larger than the angle at A. At C'' the angle is less, at C''' the angle is more, so there must be a point X between them (the brown triangle) where the angle at X equals the angle at A. So triangle ABX has an equal angle, an equal side, and the same area. From the sketch they clearly aren't congruent because ABX has an obtuse angle at B and ABC has no obtuse angles.
 
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