Geometry Homework HELP Please!?
953424
Yes. Right triangles have them all the time. it's called the orthocenter i think....it's been a while since Geometry.
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)
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
243852
Hope this helps\n\n\nhttp://argyll.epsb.ca/jreed/math9/strand3/triangle_congruent.htm\n\nhttp://mathforum.org/geopow/fullsolutions/102596.solutions.html or www.mathforum.org
897859
I find this an interesting question, and I suspect there are several possible methods for solving it.  However, I recommend we deal with it straightforwardly by solving for, say, the base length, height ... any 'variables' found within these polygons' area formulas.\n\nThe understanding here is that both polygons are regular polygons (the triangle is therefore equilateral, the hexagon has 6 equal-length sides, etc.).\n\nThe most creative solutions (and least mathematical) would probably involve cutting up one of the shapes and "re-gluing" it to form the other, thereby gaining some hint at the relationship between their perimeters.\n\nI normally prefer straight math, and would recall:\n\n A(triangle) = (1/2) base * height\n A(hexagon) = ([3sqrt(3)]/2) (side)^2\n\nBut let's "logic" our way through this one first.  Think of the hexagon as being composed of 6 equilateral triangles, each containing 1/6 of the overall area.  Because the original hexagon had an area in 1:1 ratio with the given triangle, we now can state:\n\n Area (SmallTriangle) = (1/6) (BigTriangle)\n\nRecalling a particular theorem concerning similar polygons ("The ratio of their areas will be the square of the ratio of their sides"), we can take this further:\n\n Ratio of BigTri:SmallTri Areas = 6:1\n\nSo,\n\n Ratio of BigTri:SmallTri Sides = sqrt(6):1\n\nPicture now how the small triangles would be "glued together" to form their parent hexagon.  Only the "outside side" would contribute to the hexagon's perimeter ... SIX such sides would be needed.  In short:\n\n Ratio of hexagon perimeter:BigTri perimeter = 6sqrt(6):1\n\nHope this (is correct and) helps!
997132
Draw a picture of the two congruent sections, which are connected at B. The exact angle between them doesn't matter so long as it's not greater than 90 degrees. You will see that if AD=CD, then angles A and C are equal. So what happens as angle A gets larger and angle C gets smaller? You should be able to see the answer for yourself.
908071
Coordinate Geometry\nThe Distance Between two Points\nThe length of the line joining the points (x1, y1) and (x2, y2) is:\n\nsqure root of (x1-x2)^2+(y1-y2)^2\n\nExample\nFind the distance between the points (5, 3) and (1, 4).\n(So in this case, x2 = 1, x1 = 5, y2 = 4 and y1 = 3).\n\nDistance = Ö[(1 - 5)2 + (4 - 3)2]\n= Ö[16 + 1] = Ö17\n\nThe Midpoint of a Line Joining Two Points\nThe midpoint of the line joining the points (x1, y1) and (x2, y2) is:\n\n[½(x1 + x2), ½(y1 + y2)] \n \nExample\nFind the coordinates of the midpoint of the line joining (1, 2) and (3, 1).\n\nMidpoint = [½(3 + 1), ½(2 + 1)] = (2, 1.5)\n\n \nThe Gradient of a Line Joining Two Points\nThe gradient of the line joining the points (x1, y1) and (x2, y2) is:\ny2 - y1\nx2 - x1\n\nExample\nFind the gradient of the line joining the points (5, 3) and (1, 4).\nGradient = 4 - 3 =  1 =  -0.25\n               1 - 5    -4\n\n \nParallel and Perpendicular Lines\nIf two lines are parallel, then they have the same gradient.\n\nIf two lines are perpendicular, then the product of the gradients of the two lines is -1.\n\nExample\na) y = 2x + 1\nb) y = -½ x + 2\nc) ½y = x - 3\n\nThe gradients of the lines are 2, -½ and 2 respectively. Therefore (a) and (b) and perpendicular, (b) and (c) are perpendicular and (a) and (c) are parallel.\n\nThe Equation of a Line Using One Point and the Gradient\nThe equation of a line which has gradient m and which passes through the point (x1, y1) is:\n\ny - y1 = m(x - x1) \nExample\nFind the equation of the line with gradient 2 passing through (1, 4).\n\ny - 4 = 2(x - 1)\ny - 4 = 2x - 2\ny = 2x + 2\n\nSince m = y2 - y1/\n              x2 - x1\n\nThe equation of a line passing through (x1, y1) and (x2, y2) can be written as:\ny - y1 / y2 - y1=\nx - x1 /   x2 - x1\n\n\n--------------------------------------------------------------------------------
821450
go to a math website off yahoo. it should help.
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
87753
Just add up the lengths of the three sides. Works for any triangle - wether it's equilateral, isosceles or obtuse.
458187
find the area of the base and multiply it by the height\n\n(1/2)*b*h1*h2  for a triangular prism h1 is height of the triangle, h2 is the height of the prism\n\npi*(r^2)*h for cylinder, h is height, r is radius of the base
401455
Sorry.  This is going to be a little rough, typing the answer.  Here it goes:  \n(x-2)2(squared)+ (y-4)2 (squared)= 25\n\nYou find the midpoint between the two points for the center.  You find the distance between the two points for the diameter and divide that in half for the radius which equals 5.  The center is (2,4).  You square the radius and that is what comes after the equals sign.\n\nThe guy who entered before me is wrong.  I promise.
788352
1) Rearrange so\ny = -(5/3)x + (2/3)\nThis means the perpendicular to this line (which the point must lie on) is of the form\ny = (3/5)x + n \nlet (a,b) = the point we require, so\nb = -(5/3)a + (2/3)   equation 1\nand -5 = -4(3/5) + n so n = -(13/5)   equation 2\nand b = a(3/5) + n  equation 3\nsubstitution of equations 1 and 2 into 3 give:\n-(5/3)a + (2/3) = a(3/5) - 13/5\nso a = 49/34\nand consequently b = -59/34\nso the point is (49/34, -59/34)\n\n2) let h = height of box and l = width of box = length of box (square base)\nso volume, v = hl^2\nand material area, a =l^2 + 4lh = 1300 cm^2\nso h = (1300-l^2)/(4l)\nso now v = (1300l - l^3)/4\nfor max volume \ndv/dl = 0 \nso dv/dl = (1300 - 3l^2)/4\n(1300 - 3l^2)/4 = 0 when l = sqrt(1300/3)\nwe take only the positive solution as the negative is meaningless in this problem.\nto check this is indeed a maximum value (as oppose to a minimum) we derive d^v/dl^2:\nd^v/dl^2 = -3l/2 so for l = sqrt(1300/3) this is always negative confirming this is a maxima.\nso now, v  = (1300l - l^3)/4\nsubstitute l = 20.817 cm\ngives volume, v = 4510.28 cm^3
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.
192062
Maybe this link will help:\nhttp://www.mathinvestigations.com/
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.
659736
Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!