# Parallellogram - Euclidean Geometry

• Feb 4th 2009, 01:22 AM
Hanga
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.

• Feb 4th 2009, 10:07 AM
Soroban
Hello, Hanga!

Quote:

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: . $AB = CD = 2,\;AD = BC = 1,\;\angle B = \angle D = 60^o$

This parallelogram can be divided into four equilateral triangles!

Code:

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

Draw diagonal $AC.$

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

Draw diagonal $BD.$

In right triangle $BED\!:$

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

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

• Feb 4th 2009, 11:56 AM
earboth
Quote:

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 $\cos(60^\circ) = 0.5$ and $\cos(120^\circ) = -0.5$ the calculations are really simple.
• Feb 4th 2009, 02:08 PM
Hanga
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 :)