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- Thread starter leprofece
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- #1

- Jan 29, 2012

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1) there is no "B" in your picture

2) there are "a" and "b" but they are lengths, not triangles.

You

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B = b type error sorry friends

1) there is no "B" in your picture

2) there are "a" and "b" but they are lengths, not triangles.

Youdohave 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.

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- #5

- Feb 7, 2012

- 2,785

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.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

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- #7

- Feb 7, 2012

- 2,785

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.Sorry In the book is : (a) (sinalpha)/(b)

Do you see now??

This is Nominator:a*sin (alpha)

and denominato;r b