# Thread: Parallellogram - Euclidean Geometry

1. ## Parallellogram - Euclidean Geometry

Okay I'll just ask the question.

If I know that the sides of a parallellogram is 1 (l.u = length unit ) and 2 l.u and that one angle is 60degrees, can I using plain Euklides Geometry calculate the diagonals exactly?

No need for you guys to solve it completly but a tip would be nice!

I can get all angles easy. This would be a breeze if it was a rectangle, but cuz it's leaning the diagonal must become somewhat longer. I tried to get it by creating it into a 90degree traingle so I could use phytagoras, but for that I'd have to make a million small triangles and I hardly belive it's the easiest way.

2. Hello, Hanga!

If I know that the sides of a parallellogram is 1 and 2, and that one angle is 60°,
can I, using plain Euclidean Geometry, calculate the diagonals exactly? . . . . Yes!

We have parallelogram ABCD with: .$\displaystyle AB = CD = 2,\;AD = BC = 1,\;\angle B = \angle D = 60^o$

This parallelogram can be divided into four equilateral triangles!

Code:
          A * - - - - - * - - - - - * B
/ \         / \         /:
/   \       /   \       / :   _
1 /     \     /     \     /  : ½√3
/       \   /       \   /   :
/ 60°     \ /         \ /    :
* - - - - - * - - - - - * - - +
D     1           1     C  ½  E

Draw diagonal $\displaystyle AC.$

It can be seen that $\displaystyle \Delta DAC$ is a right triangle
. . with: $\displaystyle \angle DAC = 90^o,\:AD = 1,\: hyp = 2$
Hence: .$\displaystyle AC \,=\,\sqrt{3}$

Draw diagonal $\displaystyle BD.$

In right triangle $\displaystyle BED\!:$

. . $\displaystyle BD^2 \:=\:DE^2 + BE^2 \:=\:\left(\tfrac{5}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2 \:=\:\tfrac{25}{4} + \tfrac{3}{4} \:=\:\tfrac{28}{4} \:=\:7$

Therefore: .$\displaystyle BD \:=\:\sqrt{7}$

3. Originally Posted by Hanga
Okay I'll just ask the question.

If I know that the sides of a parallellogram is 1 (l.u = length unit ) and 2 l.u and that one angle is 60degrees, can I using plain Euklides Geometry calculate the diagonals exactly?

No need for you guys to solve it completly but a tip would be nice!

I can get all angles easy. This would be a breeze if it was a rectangle, but cuz it's leaning the diagonal must become somewhat longer. I tried to get it by creating it into a 90degree traingle so I could use phytagoras, but for that I'd have to make a million small triangles and I hardly belive it's the easiest way.

Here comes a different attempt:

Use the Cosine rule in the triangle CDA to calculate the diagonal AC.

Use the Cosine rule in the triangle DCB to calculate the diagonal DB.

Since $\displaystyle \cos(60^\circ) = 0.5$ and $\displaystyle \cos(120^\circ) = -0.5$ the calculations are really simple.

4. Thanks alot, both of you!

The latter explenation I could've done, problem was that I had to proove it using euclides geometry only!

I understand this now, thx alot