parallel bars dimensions?
397686
Height 195 cm  Length 350 cm  Distance between the bars 42 cm to 52 cm (adjustable)
783613
Area = length x width
374839
273\nMultiply .42 by 650
1064803
http://www.epanorama.net/circuits/parallel_output.html
405166
Try this:\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut9_bar.htm
720499
7.3m (width) by 2.4m (height)
238652
Assuming you have a rectangular object, then L x H x W will work. For other shapes you need something else. See sources.
218561
A parallel line must have the same slope.\nSo you just need to find slopes of the lines that  you have.\n1) 5y= -4x - 11\ny = -4/5x -11/5\nthe slope m = -4/5\n\n2) 4y= 5x-7\ny= 5/4x -7/4\nthe slope m = 5/4
110772
0.58 (with a horizontal bar right above the 58)
613917
Height times Width times Length
599733
I believe it would be one-half times the length of each diagonal.\n\nIn other words, if the diagonals are 4 & 6 the area would be:\n\nA = 1/2 * 4 * 6 = 12\n\nIf you don't know the diagonals, but you know the length of the sides, you could construct the diagonals and calculate their dimensions using the Pythagorean theorem:  c^2 = a^2 + b^2, and then solve from there.  Keep in mind that you'll have to do some extra geometric analysis to find the length of each leg of the right triangles that are created by the intersection of the diagonals before you can use the theorem.
783072
7ftx 3ft, 6inch. But it depends on the league that the pub plays in some are six by three.
281759
3454.3615403812755165920718499567 centimetres cubed
438309
You are taking away x feet on each side.\nThus, the length is 6-2x and the width is 8-2x and the height is x.\nV = Lwh\nV = (6-2x)(8-2x)x
185652
since your ddoing coordinate geometry you should at least tell what shape you're trying to get.\nI'm assuming it's a parallelogram so u should make it (1,8)
921006
Area = πab where a and b are the semi-major and semi-minor axis length
1009332
Convert everything to inches, then multiply.  LxHxW.  They need to be the same units, though.