Geometry help please....?
826923
The easiest way to do this would be to put the equations into slope-intercept form and compare the slopes.  If the slopes are the same, the lines are parallel.  If the slopes are the negative reciprocals of each other (like 2/3 and -3/2), they are perpendicular.  If neither, then neither.\n\n3x + 6y = 30\nx + 2y = 10\n2y = -x + 10\ny = (-1/2)x + 5\n\nSlope:  -1/2\n\n4y + 2x = 9\n4y = -2x + 9\ny = (-1/2)x + 9/4\n\nSlope:  -1/2\n\nYour call here.
563121
a right angled isosceles triangle\n\nplease give me 10 points now
849885
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
967287
The triangle will have 2 sides of length x, and the remaining side will be length P-2x making the parameter x + x + P - 2x = P.\nBegin by drawing a line from the vertex where the two sides of length x meet, to the center of the side which is length P-2x. The line should be perpendicular to the side P-2x thereby splitting the triangle in half, into 2 right triangles. One of the right triangles has hypotenuse = x and one leg = (1/2)(p-2x). So the remaining leg has length S.R. (x^2 - (p/2 -x)^2). We calculate the area of this right triangle, and double it since there are two triangles:\nArea = (2)(1/2){(1/2)(p-2x)S.R.(x^2 - (p/2-x)^2)\n         = (1/2)(p-2x)S.R.{px - (p^2)/4}\n         = (1/4)(p-2x)S.R.(4px-p^2)\nNote that your given answer is wrong - it has an x^2 where it should have a p^2....
47718
So you've got an eight-sided figure and each side is 72-feet.  If you draw horizontal and vertical lines from each vertex (the angles where sides meet) to the other side of the octagon, you'll see that the octagon can be split up into one large square (in the center), four rectangles (up, down, left and right), and four triangles (in the corners).  You following me?  Get out a pencil and get to it!  Okay.  So what are lengths of your square's sides?  You'll note that the square has 72-foot sides, just like the original octagon.  Since you know how to find the area of the square, you're on your way.\n\nNext.  you have four rectangles, each with a length of 72-feet, but the short end - the width, how do we figure out the width of the rectangle?  Okay.  Look at your triangles.  Each triangle has one side of 72 feet - the side which is the outside of the octagon - and then two equal sides - it's an isosceles triangle with a 90-degree angle.\n\nOkay, time to call on pythagoras.  Pythagorean theorem states that a^2 + b^2 = c^2.  What that means is this: If you take the length of sides a and b (the ones we don't know) and square them, and add them together, the answer will be 72-squared (side C).  Too bad I'm not there with you to make this part easier to understand.  We know that a and b are equal in this case because our triangle is isosceles.  Therefore, 2x^2 is equal to 72^2.  This gives us a messy equation wherein our answer is, x is ROUGHLY equal to 51 feet.\n\nSo.  You have a square with sides 72 feet long.  You have 4 rectangles each with length 72 and width 51.  You have 4 triangles each with base 51 feet and height 72 feet.  You know that squares and rectangles have an area equal to length times width, and triangles have an area equal to one-half base times height.\n\nCalculate the area for your 9 figures and add them together.  THAT will be the square footage of your octagon.\n\nWhat... did you think I was going to do it ALL for you?
400701
what you want to do is draw an isosoles triangle with the 2 congruent angles the size you want. \n\nStart with a base with a certain length (use the compas to measure, say 4 inches).  Next determine the midpoint by marking arcs nboth above and below your line from each endpoint using the same compass diameter, then drawing a line through the two intersections.  You now have a right abngle at the midpoint.  You can now determine how tall your triangle must be using the sin table and use your compass to mark the point on the vertical line to complete the triangle with the angle you want.
890883
1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.\nFind the sides of the triangle. (Pythagorean Theorem)\n\nWhat we can take from this problem is that one LEG is 12 cm long, while the other LEG is unknown (we will call it X) and the HYPOTENUSE -- the longest side of a right triangle -- is also unknown (we will designate it X+2, since it differs 2 cm from the unknown leg).\n\nSetting up Pythagorean Theorem as always:\n\n "Leg squared" + "Leg squared" = "Hypotenuse squared"\n\n  12^2 + x^2 = (x+2)^2\n\nExpanding:\n\n  144 + x^2 = x^2 + 4x + 4\n\nCollecting like terms:\n\n  4x = 140\n\nDivide by 4:\n\n   x = 35\n\nThe unknown leg is 35 cm, while the hypotenuse is 37.  Thus, we can also call the triplet (12,35,37) a "Pythagorean Triple" (of which there are infinitely many!).\n\n\n2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides\nis 2 cm. Find the sides of the triangle. (Pythagorean Theorem\n\nIn this case, we know the hypotenuse, but do not know the lengths of the legs.  For convenience, we will again use the expressions X and (X+2) for each side's length.\n\nBy P.T.:\n\n x^2 + (x+2)^2 = 58^2\n\nExpanding:\n\n x^2 + x^2 + 4x + 4 = 3364\n\nCollect like terms:\n\n 2 x^2 + 4x - 3360 = 0\n\nDivide by the GCF (2):\n\n  x^2 + 2x - 1680 = 0\n\nCan we factor this trinomial easily?  We could, if we knew integer factors of -1680 (one positive, one negative) that combine to make 2.\n\nLuckily, -1680 can be factored into -2*2*2*2*3*5*7, and these can be grouped into two sets:\n      2*3*7 = 42\n -2*2*2*5 = -40\n\nFactor:\n\n (x+42) (x-40) = 0\n\nAlthough two solutions come from these factors (-42 and 40), we may only consider the positive answer, since lengths of triangle's legs are always positive!\n\n Therefore, x=40 (Leg 1), and Leg 2 is 42 cm.
175713
Your question is too complicated to appreciate the math in it, which is supposed to be the real beauty in the first place. Anyway, I have tried figuring it out and come with equations. I do not even want to try and begin substitution and end up with more complications. I just thought once you saw the relation, you would throw those 10 points my way.\n\nAssuming a to be sides of the squares of the prism and b to be the length of the rectangles, we know that the sum of all dimensions is 14a+5b. Now I do not even want to get started why you brought it the story of the equilateral triangle. You can just say the sum  of the dimensions is 61. Anyway, since any side of the trianle, as it is equiangular hence equilateral, is twice 14a+5b and 122, the sum of the dimensions will be 61. So your first equation is 14a+5b=61\n\nYou know the volume of the prism is the length X width X heiht = a x a x b = (a^2)b This is equal to the surface area. The surface area is the sum of all the areas on the prism and that is a^2+4ab. (How about you calculate how that came up?)\n\nSo there is your second equation, (a^2)b = a^2+4ab So then from 14a+5b=61, express a in terms of b or the other way around and plug it into the second equation and experience brutal complications and figure out what the variables are. I think you know it is too easy to calculate each dimension and area after that.
335338
geometry too? try your damndest, dont EVER give up, kiss ass, lol
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
598737
I wish I remembered terms for this stuff.\nFind the length of the base, divide that by two.\nUse this number to find the length of a line drawn from the point of the triangle to the base. Congrats, that number is the height.\nTriangle ABC.\nLength from point A to C is 3.\nFrom A to B is 4.\nSo B to C is 5.\nTake the base, we'll say line AC, and divide that by 2, so you get 1.5.\nUse that number, now, as the length from A to D.\nSo, 1.5 squared plus 4 squared is 18.25. Square root of 18.25 is 4.27, so your height is 4.27.\nPythagorean Theorem.\nA^2 + B^2 = C^2
328129
just draw a triangle has right angle which dimensions 1,sqrt(3),and hypothenus 2. if you want it, i can send to you
100993
how do you din the height of a trapezoid if one of the non parallel sides is straight, and the other is slanted. Which makes the two parallel sides different lengths? Also, how would you find the width of a trapezoid in this situation?\n\nIf the length of the shorter parallel side is A and the length of the longer parallel side is B and the length of the straight side is C and the length of the slanted side is D then:\n\nthe height is:\nh = C (the length of the straight side.\n\nThe width depends on where you measure the width.\n\nI am imagining that your trapezoid looks like this:\n.__\n/__|
441947
pentagon DUH! lol sry (Hexagon is SIX!)im in geometry rite now so yea
901368
You need to take triangles with leg length 2-sqrt(2) from each corner. That is about 0.586".\n\nThe length of each side of the octagon will be 2 * sqrt(2) - 2. This is about 0.828"\n\nThis was calculated by setting the length of each leg of each triangle to x.\n\nhypotenuse = remaining side of square\n\nx * sqrt(2) = 2 - 2x. And then solve for x.
621583
What type of triangle is it?\nIf it is an equilateral triangle, all sides will be the same length, so p=(1.25 X 3)m\n\nHope this helps
689719
The mode of presentation of data in the question is not sufficient to draw a single quadrilateral, there are two such quadrilaterals and hence two different answers.\n\nFIRST ANSWER: \nEqual sides i.e. sides of length 5*sqrt(2) are adjacent.\nLet the quadrilateral be ABCD. Then AB = BC = 5*sqrt(2), CD = 6, <DAB and <ABC are right angles. Drop a perpendicular to side AD from C intersecting AD at P.\n\nTotal area = area of rectangle ABCP + area of triangle CPD\n= [5*sqrt(2) x 5*sqrt(2)] + [0.5*{5*sqrt(2)-6}*5*sqrt(2)]\n= 50 + 3.7868 sq units\n= 53.7868 sq units\n\nSECOND ANSWER: \nEqual sides i.e. sides of length 5*sqrt(2) are opposite to each other.\nLet the quadrilateral be ABCD. Then AB = CD = 5*sqrt(2), BC = 6, <BCD and <ABC are right angles. \nClearly, quadrilateral ABCD is a rectangle. \nSo area = product of lengths of adjacent sides = AB x BC = 5*sqrt(2) x 6\n= 42.4264 sq units