how can i find the orthocenter of a triangle?
955686
The orthocentre of a triangle is the point where all the altitudes meet\n\nSo if you have ΔABC one altitude is the line through A perpendicular to BC, another is the line through B perpendicular to AC and the last is the line through C perpendicular to AB\n\nUse these facts with the coordinates of A, B, and C and the slopes of AB, AC and BC (Find where two altitudes meet and if you are correct that point will also lie on the 3rd altitude)\n\neg Find the orthocentre of the ΔABC where A ≡(1, -3), B ≡(-1, -5), C ≡(3, 1)\n\nSlope AB = Δy/Δx\n= 2/(2) \n= 1\n\nSlope of perpendicular = -1 (-1 * 1 = -1)\nAltitude to AB goes through C\nSo equation of altitude of form y = -x + b\n(3, 1) lies on altitude\nTherefore 1 = - 3 + b\nb = 4\nThus altitude through C is y = -x + 4\nie x + y = 4 ............... (1)\n\nSlope AC = Δy/Δx\n= 4/2\n= 2\n\nSlope of perpendicular = -1/2 (2 * -1/2 = -1)\nAltitude to AC goes through B\nSo equation of altitude of form y = -1/2 x + b\n(-1, -5) lies on altitude\nTherefore -5 = -1/2 * -1 + b\nb = -11/2\nThus altitude through C is y = -1/2 x - 11/2\nie x + 2y = -11 ............... (2)\n\nSlope BC = Δy/Δx\n= 6/(4) \n= 3/2\n\nSlope of perpendicular = -2/3 (3/2 * -2/3 = -1)\nAltitude to BC goes through A\nSo equation of altitude of form y = -2/3 x + b\n(1, -3) lies on altitude\nTherefore -3 = -2/3 * 1 + b\nb = -7/3\nThus altitude through C is y = -2/3 x - 7/3\nie 2x + 3y = -7 ............... (3)\n\nie we have:\n x + y = 4 ................. (1)\n x + 2y = -11 .................. (2)\n2x + 3y = -7 ............... (3)\n\nEquation (2) - equation (1)\ny = -15\nFrom equation (1) x = 19\n\nCheck in equation (2) \nx + 2y = 19 + -30 = -11 YESSSSSSSSS!!!!!!!!!!!!!\n\nCheck in equation (3) to ensure that this point lies on this altiude also\n2x + 3y = 38 + -45 = -7 YESSSSSSSSS!!!!!!!!!!!!!\n\nSo the orthocentre of ΔABC is (19, -15)
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
640114
Ellipse first:\nThe center is at the origin, because you have x^2 and y^2. If there is a translation, you would see something like (x+5)^2, that would be a translation of 5 units LEFT. The "a" value is always the bigger value. If it is under the x, then the ellipse is long the horizontal way, if it is under the y, then long the vertical way. From the center, count "a" units the long way, and "b" units the short way to find the shape of the ellipse. There is a formula for finding the focus points also. It is c^2=a^2-b^2, solve for c, and go that distance from the center along the long axis (or the major axis) of the ellipse.\n\nHyperbola:\nNote that the same translation holds for ellipse, but for the formula you put down, the center is at the origin. If the x^2 term comes first, the hyperbola opens left/right. If the y^2 term comes first, the hyperbola opens up/down. For a hyperbola, the "a" term is ALWAYS first, and "b" is next. Count "a" units left and right (if x^2 term is first) to find the vertices. Also, it is helpful to count "b" units up and down, and label those. Then, I have my students draw a rectangle between the a and the b markings, these are helper lines, and the diagonals of the rectangle are the asymptotes. Use basic slope-intercept to find the equations of the asymptotes. The focus has a formula, it is c^2=a^2+b^2, you can find "c" and label it in the bowl of the hyperbola.\n\nHope all that helped!
749836
The 14 foot line is called a secant.  Bisect it, and draw a radius through the point of bisection.  You now have two identical right triangles, with hypotenuses of 12 feet, and known sides of 7 feet.  Use trigonometry to calculate the angle at the center of the circle of either triangle, and then a simple proportion (involving the circle's total circumference) will give you the arc length you want.
740750
easy \n\nU make a triangle between the centers of the 3 circls \nmeaning it would have three lines = 2*R = 2m\nso the hight of the triangle would be root of ( R*R - 0.5R * 0.5R)\nso it would be the root of (3) which is( 1.73)\nthe hight would be 1 + 1.73+ 1 = 3.73 m
174683
The simple way to find the area of any shape is to break it up in to trangles and squares.  \n\n\nOr use integration... but i do not think that is what you are looking for
960360
To do this accurately, you need to find the midpoint of each side of the triangle. Connecting each vertex to the midpoint of the opposite side will give you the Centroid(the place where all the medians meet).  I graphed it out and it appears to be (-7,-3).
978664
Orthocentre = point where altitudes intersect (An altitude is the line from the vertex to the opposite side (extended if necessary))\n\nIncentre = centre of the incircle = point where perpendicular bisectors of each of the sides intersect. \nRadius of incircle = √[(s-a)(s-b)(s-c)/s] where s = semiperimeter\n\nCentroid = point where medians meet (Median is the line joining a vertex to the midpoint of the opposite sides) The centroid is the centre of gravity of the triangle and the centroid divides each median in the ratio 2:1 On a numberplane the centroid of a triangle is the point which is at the average of the sum of the three x- and y-coordinates of its vertices)\n\nOne other is:\nCircumcentre = centre of the circumcircle = point where angular bisectors meet.\nDiameter of circumcircle = a/sinA = b /sinB = c/sinC
935236
The answer depends on what point you use as the center.\nanyway,L=r(cross)p where r is the radius vector for the path, and p is the linear momentum. \n\nBut, to use this formula, we need to find the linear momentum of the bob. Since we know its mass, we need to find its velocity. There are   three forces of the string. tension, gravity, and centripetial force. since the object is not accelerating, the sum of these is 0. So, \nFt+Fc+Fg=0\n\nOr, Ft=-Fc-Fg\n\nIf y is the vertical axis and x is the horisontal axis, then the y compontent of the force of tension is\n\nFt*cos(angle)=-mg\n\nand the x component is\n\nFt*sin(angle)=-mv^2/r\n\ndividing the x component by the y component gives\n\ntan(angle)=v^2/rg\nso, v=sqrt(r*g*tan(angle))\nso, the linear momentum is p=mv=m*sqrt(r*g*tan(angle))\nwhere m is the mass of the bob, r is the length of the string, g is the acceleration from gravity, and angle is the angle of the string to the central axis. \n\nNow, you know the linear momentum, and you can find out the radius vector, so the angular momentum is easy enough to figure out.
653474
Pick an angale, not the 90 degree angle and call it Theta.  The short side that forms this angle is called the x side of the triangle and the long side that forms this angle is the hypotenuse or 'r'.  The third side is called the opposite side or 'y'.\n\n\n.... /|\n..../.|\n.../..|\n../...|\n./....|\n/___|  Bottom is x, slanting side is r, and right side is \n          the y side.  The angle at the bottom left is Theta.\n\n\nSin(Theta) will = y/r\nCos(Theta) will = x/r\nand\nTan(Theta) will = y/x side.\n\nThis all presumes you don't already know 2 of the sides.  If you know 2 sides already, you use the pythagorean thm.\n\nExampe: Theta = 30 degrees, r = 10 feet, what are the others?\n\nSin(Theta) = y/r  and Sin(30 degrees) happens to be 1/2.\n\nso, 1/2 = y/10.  Solving for y you get y=5.
1106432
Diameter == 10.\nRadius == 5.\nCenter == (1,7)\n\nMethod: use Pythagorean Theorem.
520768
First you need to find the slope of the line. Use the info from the slope (rise over run) to make a right triangle with hypotenuse is the line. Use inverse tangent on info from slope to find angle.
506648
use a protractor - thats' the semi circle type thing...\n\nmeasure the angle on a straight line which is 180 degrees in total, the obtuse (larger) angle and the acute (smaller) angle will equal 180 degress so calc the obtuse one and subtract this by 180 leaving the acute angle.
841334
Easiest way would be to drop a bisector from two of the angles.  The intersection will be the centroid.  This is assuming a regular pentagon.\n\nIf it is an irregular pentagon, you could tie a weight to a string.  Hold the pentagon by a vertex while also holding the string.  This line passes through the center of gravity.  Do the same for another vertex.  The lines will intersect at the centroid (aka center of gravity).  This method would work for a regular pentagon also since the lines would be the angle bisectors.
952581
http://planetmath.org/?op=getobj&from=objects&id=3808
552905
The AREA of an isosceles triangle is found by taking 1/2 the sum of the long and short parallel sides, times the height. One way to compare this with the parallelogram is to take two copies, turn one 180 degrees and glue them along the equal (nonparallel) edges. \nYou now have a larger parallelogram - with one side composed of the sum of the original short and long parallel edges. Use the usual formula for the area of this parallelogram, then divide by 2 (since it is composed of two equal copies).
27656
I used the acronym sohcahtoa to remeber that sin=opp/hyp cos=adjacent/hyp and tan=opp/adj, so you have a angle you want to measure the degree of that angle, using the lengths of the sides of the triangle opposite, adjacent and hypotenuse and then using your calculater you divide the lengths and hit the button sin-1, cos-1 or tan-1 key, also cot=1/tan, hope that helps