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Is there a formula for calculating area of a parallelogram using only the length of diagonals? if so what is it?

Alex Wertheim
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user77273
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  • No You need the angle. Each of the diagonals bisect each other. So no matter what the angle between them, you always get a parallelogram (a different one). So, given the diagonals, it is this angle that fixes the parallelogram. Hence you need this angle. –  Jan 16 '15 at 03:44

3 Answers3

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No. Imagine two fixed diagonals that are attached and can rotate in the center. As they pivot, different parallelograms are induced; if they are almost parallel the area will be close to zero.

vadim123
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Imagine the two diagonals are sticks of wood, with a nail sort of holding them together around the middle, except that you can rotate one of the diagonals around the nail. By making the angle between the diagonals small, you can make the area as small as you wish. By "opening up" the angle so that it is $90^\circ$, you can maximize the area.

Remark: The area of the parallelogram is $\frac{1}{2}pq\sin\theta$, where $p$ and $q$ are the lengths of the diagonals, and $\theta$ is the angle between the diagonals. But we do not have to know that to see that the area is not determined by the lengths.

André Nicolas
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  • Well, beat me to this... Another way of saying this is that if the two non-parallel sides are expressed by the vectors $\overrightarrow{a} \ \text{and} \ \overrightarrow{b}$, then the area is found from $| \overrightarrow{a} \cdot \overrightarrow{b} |$ . But the diagonals are $ \overrightarrow{a} + \overrightarrow{b} \ \text{and} \ \overrightarrow{a} - \overrightarrow{b} $ . There would be no way to express these diagonals (or their lengths) in terms of the area without including a factor related to $\cos \theta$ . – colormegone May 11 '13 at 02:07
  • @RecklessReckoner, minor edit. Cross product, not dot. – Ted Shifrin May 11 '13 at 04:15
  • D'oh : not so minor! (And so, $\sin \theta$ , not cosine.) Thanks for spotting that -- it's been a long math week for me... – colormegone May 11 '13 at 05:27
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If the diagonals are given in vector form, then as @RecklessReckoner said, one diagonal is vector sum $a+b$ and the other is vector sum $a-b$. So you have two equations and two variables, you can find out vectors $a$ and $b$, then simply find the magnitude of their cross product.

If the diagonals are vectors $p$ and $q$, the area can be represented as the magnitude of the cross product: $\frac 14 (p+q) \times (p-q)$.

Win Vineeth
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Catchyviddy
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