# TrigonometryArea of parallelogram

#### HallsofIvy

##### Well-known member
MHB Math Helper
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
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
B= b
angle x = angle alpha in the figure

#### Opalg

##### MHB Oldtimer
Staff member
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
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
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. 