how would u figure out a quadrilateral's other side if only 2 of the 4 sides are given?
341464
A quadrilateral is a four-sided polygon with four angles. There are many kinds of quadrilaterals. The five most common types are the parallelogram, the rectangle, the square, the trapezoid, and the rhombus. \n The sum of the angles of a quadrilateral is 360 degrees. \n Angle CB  would equal 2 because it is congruent to angle AD.
306739
4 times the length of one side
139968
false. a quadrilateral is a four sided polygon. so a square and rectangle are also quadrilaterals. a trapezoid is a four sided polygon with 2 opposite sides that are parallel to each other.
785535
Multiply both sides by a. Then divide both sides by 9. You should get a=5.
303035
If the triangle is a right triangle, you can find the other side by Py's Th and then by the formula,(A=1/2bh).\nIf it is not, you should know the angle they make each other or an adjacent angle. Only if, you can find the area by sine laws or cosine laws.
337862
go do your homework by yerself next time.\n\nBut becos im feeling nice today...\n\na triangle = 1 triangle , duh\na quadilateral can be divided into two triangles with vertices (points) all at the points of the quadilateral\n5 sided --> 3 triangles\n6 sided --> 4 triangles\n....\n12 sided --> 10 triangles\n1 triangle -->180 degrees\n10 triangles --> 1800 degrees
763972
Let there be a right-angled triangle.\nnow, \nwe know that, \n(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 2(4n^3 + 2n+ 2n^2)\n=>(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 8n^3 + 4n+ 4n^2\n=>(2n^2+2n+1)^2=(4n^4 + 4n^2 + 8n^3) + (1 + 4n+ 4n^2)\n=>(2n^2+2n+1)^2=(2n^2 + 2n)^2 + (2n+1)^2\nThis gives solutions\n(3,4,5); (5,12,13); (7,24,25); (9,40,41)........................\nthe above relation is true for all n belonging to natural numbers\nthis way you can find solution for any pair of sides the 3rd side been given.\nput the given side equal to one in equation and find positive n by solving the quadratic and find all the sides.\nEg:If hyputenuse is given as 5\nSolve 2n^2 + 2n + 1= 5\n=>n=1 therefore other sides are 3 and 4\n\nChoose this as best, if correct.
23529
that is one of the answers, if the length of one side is 5. Use the Phytagorous theorem.
725566
The other way is to balance out the equation so only x is left on one side. So here we go.\nFirst, add 6 to both sides\n3x=6\nThen, divide by three on both sides and get\nx=2\nGood Luck
671808
4 sides and corners in a square
237985
Cut a 2x2 piece off. Cut the remaining 2x3 piece in half lengthwise.  Put the two resulting 1x3 pieces one atop the other and cut the stack diagonally.  You now have 4 triangles with sides = 1, 3 and sqrt(10).  Arrange them with their right angles touching the corners of the 2x2 square piece, 3-meter sides parallel to and touching the sides of the square piece, and all "pointing" in the same direction, clockwise or counterclockwise.  The result is a square that is sqrt(10) on a side.  Area = 2x5 = sqrt(10)^2 = 10 square meters.  Three cuts were used to make five pieces.\nTo solve this I started with the requirement of the large square being sqrt(10) on a side.  Sqrt(anything) suggests the hypotenuse of a right triangle.  A little mental fiddling produced the sides of 1 and 3 meters.  Luckily a 2x2 piece remained.
825411
first side=x\nsecond side=2*x\nthird side=2*x+3\n1. + 2. + 3. => x+2x+2x+3=33 => x=6 \n\nfirst side: 6cm\nsecond side: 12cm\nthird side: 15cm\ntotal: 33cm
375748
All squares are rectangles,\nall parallelograms are quadrilaterls\nand all squares are parallelograms
1166171
The easy way to remember this is to memorize the \n\n3 4 5 \n\nright triangle.  You will notice that any two multiples of the above will give you the other leg as a multiple.  The theorem is\n\nA^2 + B^2 = C^2,  so  3^2 + 4^2 = 5^2,  hence  9 + 16 = 25\n\nnow you have\n\n2*3 2*4 thus your other leg is 2*5\n\n6^2 + 8^2 = 10^2,  which is 36 + 64 = 100\n\nas for the angles, use the law of sines\n\nsin(90)/5 = sin(a)/3 = sin(b)/4\n\nso a = asin(sin(90)*3/5) = ~37 deg\n\nhence angles of 90 53 37
803501
1.  add 1 toboth sides, then multiply by 3\n\n(x+4)^2=18  now take the square root of both sides\n\nx+4= +/- 3sqrt2\nx=-4 +/- 3sqrt 2\n\nyou also could use the quadratic formula\n\n\n2.  multiply num and denom by 3 - 2 sqrt3\n\n12-8sqrt3 +2sqrt3 -2sqrt6  all over -3\n\n\n3.  mult num and denom by 2 - sqrt 12 and get \n\n12-6sqrt12 all over -8\n\n12-12sqrt2  all over -8\n\n3-3sqrt2 all over -2
610972
considering both y values are -2 for A and C....\n\nwe know that a straight line exists between these two points, which means that it HAS to be 1 of the 4 sides.\n\ngiven that, the coordinates or K and L HAVE to be (3, infinity (EXCEPT -2)) and (-4, infinity (EXCEPT -2))
848127
When you added x to the right side, you should have gotten:\n\n4 = x^2\n\nThen take the square root of both sides to get\n\nx = 2, -2