MHB Is the area of a parallelogram equal to a/b times sine alpha?

AI Thread Summary
The discussion centers on the formula for the area of a parallelogram, with participants debating whether it can be expressed as a/b times sine alpha. It is clarified that the correct formula for the area is actually ab*sin(alpha), where a and b represent the lengths of the sides of the parallelogram. Participants point out that the initial assertion misrepresents the relationship between the sides and the angles involved. There is confusion regarding the terms "triangle a" and "triangle B," which do not accurately reflect the components of the diagram. Ultimately, the consensus is that the proposed formula a/b*sin(alpha) cannot represent an area, as it does not conform to the necessary dimensions for area calculation.
leprofece
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demonstrate or show that the figure area is = a/b sen alpha

triangle a = triangle B maybe this is the main premise
 

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You talk about "triangle a" and "triangle B" but
1) there is no "B" in your picture
2) there are "a" and "b" but they are lengths, not triangles.

You do have two triangles with altitudes of length a and b. The areas of those triangles add to half the area of the parallelogram but you still need to know the lengths of the sides to find the areas of those two triangles.
 
HallsofIvy said:
You talk about "triangle a" and "triangle B" but
1) there is no "B" in your picture
2) there are "a" and "b" but they are lengths, not triangles.

You do have two triangles with altitudes of length a and b. The areas of those triangles add to half the area of the parallelogram but you still need to know the lengths of the sides to find the areas of those two triangles.

B = b type error sorry friends
 
Please help me it is a demostration you have the diagrama oo figure book here
B= b
angle x = angle alpha in the figure
 
leprofece said:
https://www.physicsforums.com/attachments/2089

demonstrate or show that the figure area is = a/b sen alpha

triangle a = triangle B maybe this is the main premise
There must be something wrong here. The area of the parallelogram is $ab\sin\alpha$, where $\alpha$ is as in the diagram, but $a$ and $b$ have to be the sides of the parallelogram, not the perpendicular lengths shown in the diagram. In any case, the answer surely has to be $ab\sin\alpha$, not $a/b\sin\alpha$, which could never represent an area.
 
Sorry In the book is : (a) (sinalpha)/(b)
Do you see now??
This is Nominator:a*sin (alpha)
and denominato;r b
 
Last edited:
leprofece said:
Sorry In the book is : (a) (sinalpha)/(b)
Do you see now??
This is Nominator:a*sin (alpha)
and denominato;r b
No, that cannot be right! An area is always a length times a length. A length divided by a length could never be an area. (Shake)
 

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