can anyone help me with how to find the outer edges of a parallelogram when you know the lengths of the lines inside that intersect and one angle in the center where the lines intersect?
igloo myrtilles fourmis
Probably a .gif or .jpg would load better, John. They tend to be a lot smaller and faster and more likely to be compatible with people's browsers.
I do have an experimental image gallery, but I disabled it after the hacking attack for fear that was how they got in (example: they may have found a way to uploaded a program that they could then run). When the image gallery is stable and secure, then I will try to get it integrated with the forum so people can upload images here. That'd be nice!
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
In a parallelogram,
opposite sides are parallel, and equal; opposite angles are equal, and the diagonals (lines inside that intersect) bisect each other.
When you know the length of the diagonals, half of them would be the sides of a traingle they form with one of the sides of the parallelogram.
Use the theorems that (i) when two lines intersect each other, the vertically opposite angles are equal and (ii) sum of the total angles is equal to 360 degrees. This way, all the angles can be known.
Now, use the Cosine Theorem
a² = b² + c² - 2bcCosA
(where A, B, C are three angles of a traingle and a,b,c are the three sides opposite to angles A,B,C respectively)
for knowing the third side of the triangle, which forms a side of the parallelogram. Following this method, the adjacent side too can be found! Since opposite sides of a parallelogram are equal, we know all the four sides!
I know this reply is long, and may not be of much help; that's because some Mathematical problems are difficult to explain without a diagram!
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
And Image gallery would be awsome mathisfun!!!!!
A place were we could store images we post here!
You could create a small quota for each user, lets say, 15 or 20 mb, wich is more than enought for gif/jp*g images.