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

[leprofece]

View attachment 2089

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

triangle a = triangle B maybe this is the main premise

 

[HallsofIvy]

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.

 

[leprofece]

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

 

[leprofece]

Please help me it is a demostration you have the diagrama oo figure book here
B= b
angle x = angle alpha in the figure

 

[Opalg]

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.

 

[leprofece]

Sorry In the book is : (a) (sinalpha)/(b)
Do you see now??
This is Nominator:a*sin (alpha)
and denominato;r b

 

[Opalg]

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)