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Trigonometry Area of parallelogram

leprofece

Member
Jan 23, 2014
241
Scan2.jpg

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

triangle a = triangle B maybe this is the main premise
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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

Member
Jan 23, 2014
241
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

Member
Jan 23, 2014
241
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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
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

Member
Jan 23, 2014
241
Sorry In the book is : (a) (sinalpha)/(b)
Do you see now??
This is Nominator:a*sin (alpha)
and denominato;r b
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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)