# Find area of parallelogram

I need to find area of parallelogram if diagonals of parallelogram are vectors $2a-b$ and $4a-5b$ and $|a|=|b|=2$ and the angle between $a$ and $b$ is $30^\circ$.

I know it must be really easy, but I struggle a lot...

• By the way, sorry for not using mathematical symbols since I don't have a PC now and have to write this in phone. When I get PC, I'll fix it. – Karagum May 18 '17 at 14:12

EDIT: I mistook the vectors $2a-b$ and $4a-5b$ for the edges of the parallelogram. The method still applies but we must solve the edges. The diagonals are sum and difference of the edge vectors (the order we choose to denote these doesn't affect the area). We have (denoting the unknown edges $u$ and $v$):

$$u+v = 2a-b$$ $$u-v = 4a-5b$$

We get $u =3a-3b$ and $v = -a+2b$. Hence the determinant in the last part is $3\times2-3\times1 = 3$ and the answer $2\times 3 = 6$.

(EDIT ends here)

Think of it as two linear mappings: first mapping the standard coordinate vectors to $a$ and $b$ and then mapping with $L(x_1, x_2) = (2x_1, -x_2, 4x_1 -5x_2)$. The determinant of that total mapping gives the area of the mapped parallelogram. We know the area of the "middle parallellogram", that is the one with sides $a$ and $b$. It is $2\times 2\times \sin(30^\circ) = 2$. Now this only needs to be multiplied by the absolute value of the determinant $\begin{vmatrix} 2 & 4 \\ -1 & -5 \\ \end{vmatrix} = |-6| = 6$

So the answer is $6\times 2 = 12$.

• I upvoted this for its demonstration of how to use the determinant of the transformation. But note that the vectors $2a-b$ and $4a-5b$ were diagonals of the desired parallelogram, not edges. – David K May 18 '17 at 14:40
• Oops, should read the questions more carefully. Thanks for notifying. My edited answer came out half the 12, so should be correct according to the theory already given by other answers :). – ploosu2 May 18 '17 at 15:31

You can find the vectors at the edges of the parallelogram by expressing the diagonals in terms of the edges, giving you equations you can solve. This reduces the problem to finding the area of a parallelogram given its edge vectors.

Alternatively, show that the area of a parallelogram with diagonal vectors $u$ and $v$ is half the area of a parallelogram with edge vectors $u$ and $v.$ This means you can take the two given vectors as edges of a parallelogram, find the area of that parallelogram, and divide by two.

Other answers give multiple ways to get the area of a parallelogram given its edge vectors. Here's a third way (just for completeness of answers; personally, I think the determinant is neater):

To find the area, you can use the fact that the area of a parallelogram with edge vectors $u$ and $v$ is the magnitude of the cross product, $\lVert u \times v \rVert = \lVert u \rVert\lVert v \rVert \sin\phi,$ where $\phi$ is the angle between $u$ and $v.$

Compute the cross product. Since the terms of the cross product will be expressed as linear combinations of $a$ and $b,$ you will probably want to simplify the cross product using such rules as $(u+v)\times w = u\times w + v\times w,$ $v\times v = 0,$ and $u\times v = -v\times u.$

You should be able to simplify the area down to just some multiple of the cross product $a\times b.$ You can compute the magnitude of that cross product from the information given about $a$ and $b.$

• +1 for the hint on how to compute the area directly from the diagonals. – amd May 18 '17 at 20:27

Let sides of the parallelogram be the vectors $x$ and $y$. Then the diagonals are vectors $x+y$ and $x-y$. Rewrite these diagonals in terms of $a$ and $b$ and you will obtain $x$ and $y$ as function ob $a$ and $b$.

The area of the parallelogram writes $S = |x|^2|y|^2 - |x\cdot y|^2$ where $x\cdot y$ is a scalar product. The conditions on $a$ and $b$ say that, $$a\cdot a = 4\\b\cdot b = 4\\a\cdot b = 2\sqrt 3.$$ With this in mind and expressions for $x$ and $y$ in terms of $a$ and $b$, you are now able to find $S$.

The area is: diagonal by diagonal by the sine of the angle all divided by 2.

Sorry no mathjax, using phone as well.

$d_1 d_2 sin(x)/{2}$