solving triangles using ratios, and a given side?
339327
Corresponding sides of similar triangles are proportionate.
823408
Your ratio is 1:8.15          You fid this out by starting at 13:106 and work yoiur way down to 1. \n\nyou divide both sides by 13 to get your answer.
905611
If u r saying that u have three sides of a triangle as a,b,c\nAnd u want to determine angle b/w them, u can use the formula\ncosA=(b^2+c^2-a^2)/2bc where A is the angle formed by the sides b and c.similarly if B is the angle formed by a and c, it can be calculated as\ncosB=(a^2+c^2-b^2)/2ac\nonce u have calculated A and B, subtract their sum from 180, u will get third angle.\nRemember u will have to use inverse table for cos to find A and B
103557
8/5 the initial iteration (triangle)
331507
sorry hunny you need at least 2 side lengths in order for us to help you figure it out... plus once you have that the formula for figuring it out is using the pathagorean theorem! a2+b2=c2 (s = squared) say for instance... the 2 sides of your triangle are 7 and 3inches. 3>2+7>2=c>2\nwhich would turn into 9+49=c>2~~~~~~~ then 58=c>2 then to get c by itself(which is needed) you would take the square root of 58! your hypotenuse would equal 7.6 or rounded, 8!\nnow if you use my steps you should be able to do your problem! I'm a 10th grader in geometry and it's the most lovely subject!
748308
Need to say what you are doing with the problem: solving for x, solving for y, other?\ny=(3/2)x+(7/4)
268070
Check out this site. It gives you a few examples. It was helpful for my cousin when she was leaning how to do them.\nhttp://www.purplemath.com/modules/solvelin.htm
972787
Heron's formula can only be used when the sides are given. The formula is:\nArea = square root of[(s)(s - a)(s - b)(s - c)]\n\ns = (a + b + c)/2\nwhere a,b and c are the sides of the triangle.\n\nSorry, but there really isn't enough information to work with.
811153
Hi, \n\nThe answer lies in the pythagorean theorem. \n\nFor any right (one of the angles is 90 degrees) triangle with sides a,b and c where c is the side opposite the 90 degree angle:\n\na^2 + b^2 = c^2\n\nso for your first excercise:\n\na = 4\nb = 4 Sqrt(3)\nc = ?\n\na^2 + b^2 = c^2\n4^2 + {4 Sqrt(3)}^2 = c^2\n16 + 4^2 * 3 = c^2\n16 + 16 * 3 = c^2\n64 = c^2 \nc = +- 8 (since we're dealing with a triangle, -8 doesn't make sense, so just ignore it.)\n\nc = 8. \n\nFollow that same pattern, and you should be all set. \n\nHope that helps, \nMatt
792000
I suppose the triangle is right-angled.\nLet the unknown side be 'x'\nIn that case,\n5^2 + x^2 = [9*(sqrt3)]^2\n25 + x^2 = 243\nx^2 = 243 - 25\nx^2 = 218\nx = sqrt(218)\n   = 14.7648230602334(approx)
190089
use heron's formula: \nArea=SQRT(s(s-a)(s-b)(s-c)), \nwhere s=(a+b+c)/2 or perimeter/2. \n\nyou only need to know the sides of the triangles. \nfrom (0,9) to (4,6), the distnace is a=5\nfrom (0,9) to (-1,12) the distance is b=sqrt(10)~3.16\nfrom (4,6) to (-1,12) the distance is c=sqrt(61)~7.81\n\nso, s=(5+3.16+7.81)/2=7.985\n\nplugging these 4 numbers into the formula gives \nSQRT(7.985(7.985-5) (7.985-3.16) (7.985-7.81))=4.486\n\nso the area is approximately 4.486
730817
Define the lengths of the smallest triangle:\nAP = ½ AC\nAQ = ½ AB\n\nSquare the equations:\nAP^2 = ¼ AC^2\nAQ^2 = ¼ AB^2\n\nPythagorean theorem for smallest triangle:\nAP^2 + AQ^2 = PQ^2\n\nSubstitute:\n ¼ AC^2 + ¼ AB^2= PQ^2\n\nMultiply by 4:\nAB^2 + AC^2 = 4PQ^2\n\nPythagorean theorem for triangles ABP and ACQ:\nAP^2 + AB^2 = BP^2\nAQ^2 + AC^2 = CQ^2\n\nAdd the two equations:\nBP^2 + CQ^2 = AP^2 + AQ^2 + AC^2 + AB^2 \n\nSubstitute:\nBP^2 + CQ^2 = PQ^2 + AC^2 + AB^2\n\nSubstitute:\nBP^2 + CQ^2 = PQ^2 + 4PQ^2\n\nTherefore:\nBP^2 + CQ^2 = 5PQ^2\n\n\nIf you get stumped with this type problem, sometimes it helps to start at the end and work backwards...
905903
There are three medians in a triangle, not just one.\n\na) is your answer.
265237
3s = -6\ndivide each side by 3\ns = -2
551883
The answer is C\nwell, one side's length of a triangle should be shorter than the other 2's plus (2 sides plus should be longer than one side)\nthen: 3x+4+8x+2>6x-1\n      8x+2+6x-1>3x-4\n      6x-1+3x+4>8x+2\n--> x>-7/5\n    x>3/11\n    x>-1\nThus, the answer should be x>3/11\n:)
925629
polyeig(-5,-2,4,2) \n\nwill solve the equasion.  To learn about interpelations type:\n\nhelp interp1
721808
If you are trying to make a triangle out of three equal lengths, all you do is devide 180 by 3.....and taddaaa you get your angles.\n\nIf does not matter how long each side is.