Find out base & perpendicular?
918825
Yes. When 'c' is given, find the value of c^2.  Then, find out the possible numbers, whose sum = that number. Find out the root of all these numbers, till you arrive a perfect square root( if possible ).\n\nEg: when given, c = 10, find c^2. Here, it is 100. \n   therefore, a few possibilities are : 50 + 50 ( none are perfect squares)\n                                                         25 + 75 ( only 1 of them is a perfect square)\n                                                         36 + 64 ( both are perfect squares)\n\n Therefore, the answers being looked for are 6 and 8.
851772
1/2 bh or bh over2.\nType triangle is inconsequential.\nb=base  h=height\nMeasure both, multiply, divide in half.
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
847030
Create a triangle whose sides are x,y and z and whose angles opposite those sides are X, Y and Z respectively.  Draw a perpendicular line from one angle to the opposite side.  This segment is the height(h) of the of the triangle and the opposite side is the base.  If you picked /_ Z to draw the height then side z is your base and the area of triangle XYZ is 1/2 the base times the height or 1/2(z)(h).  There are 2 right triangles formed in triangle XYZ by the height segment to side z.  The point on side z I labeled W.  Triangle XZW and YZW have shared side ZW(h).  W divides side z into two segments XW and WY.  I let XW = d and WY = (z-d).  Using the Pythagorean Theorem we get two equations:\nh^2 + d^2 = y^2  and  h^2 + (z-d)^2 = x^2  I expaneded the second equation to h^2 + z^2 -2dz + d^2 = x^2.  Next I solved the equation for h^2 + d^2 = x^2 + 2dz - z^2 .  Using the two equations with \nh^2 + d^2 you can set  y^2 = x^2 + 2dz - z^2.  You can now substitute x = (a^2 + c^2)^.5, y=(a^2+b^2)^.5 and z=(b^2+c^2)^.5 which were given as the sides fo the original triangle into the last equation solved for d as (y2+z^2-x^2)/2z.  This will give you a value for d in terms of a,b an c.  I got d = b^2/(b^2 + c^2)^.5 \n\nAfter getting this value you can substitute for d and y in the equation h^2+d^2=y^2 to solve for h.  \n\nUsing the value of h and z both in terms of a,b and c you can calculate the area of triangle XYZ in terms of a, b and c using \n\narea of a triangle = 1/2(height)(base) \n\nwhere the height = h and the base = z.  \n\nWhen I solved for h, I got \nh = [(a^2b^2 + a^2c^2 + b^2c^2)/(b^2 + c^2)]^.5\nz = (b^2 + c^2)^.5 was given\n\nArea Triangle XYZ = (1/2)(h)(z) = (1/2) times \n[(a^2b^2 + a^2c^2 + b^2c^2)/(b^2 + c^2)]^.5 times\n(b^2 + c^2)^.5)\n\nThe numerator and denominator have a common factor\n(b^2 + c^2)}^.5.  when the ratio is reduced you get\n\nArea = [(a^2b^2 + a^2c^2 + b^2c^2)^.5]/2
310302
using @ for angle theta:\n\nx = r cos @, y = r sin @, use cos^2 @ + sin^2 @ = 1\n\nr^2 = 1/4*(1-5/4*sin^@)
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
91628
Also, when working out the surface area of the triangle sides, ensure that you are NOT usng the perpendicular height to calculate this. In some instances, you may need to use pythagoras' theorem to get the height measurement of the triangle sections
924939
yes take the slope of both lines  \nthen if same slope, they are parallel\nif negative reciprical, they are perpendicular,\nif neither, they are neither....\n\nbtw.... m = change in y divided by change in x.....\n    or (y2-y1)/(x2-x1)
541010
1.  Find the line that goes through d that is perpendicular to line l.  The slope of the perpendicular line is -1/m, where m is the slope from line l.  Since you have the slope for that line, then just plug in d to get the intercept, which I will call c.\n\n2.  Find the point (x,y) where the two lines meet by solving the system of equations.\n\ny = mx + b\ny = (-1/m)x + c\n\nI will call this point e.\n\n3.  Find the distance between d and e.  If d = (x1,y1) and e = (x2,y2) then the distance between d and e is\n\nsqrt((x1-x2)^2 + (y1-y2)^2)
68795
this looks fairly accurate to me\n\n x = b*h/(b + h)\n\nx=side of the square\nwhere b is the base of the triangle\nh is the height of the triangle
1064341
Remeber that perpendicular lines have negative reciprocal slopes, so the slope of the line you are trying to find is 1/3 (that is the negative reciprocal of -3).  You know that the y intercept is 5, remember that slope intercept form is y = mx + b where m is slope and b is y intercept.  You know both of these things now, the m = 1/3 and the b = 5.  Your equation is y = 1/3x + 5.
605775
Are you talkin about a trapezium or a trapezoid.\nif its a trapezium,here's how : \nlet the height be "h".\ndrop two perpendiculars equal to h on the base.the lengths b&#92;w the perps.is 6 units.therefore the rest of the length is 10.each individual length to the left of the perp is 5 units.\nin the triangle with the diagonal,the side adjacent to the angle with the base and diagonal is 6+5=11\ntherefore the angle is cos^-1(11/12)=a\ntherefore tan a=h/11\ntherefore h=11tan a\nanswer is height,h=4.795831523
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
980603
I think the first is wrong because the Y point is not ZERO in the firsst equation and thats the point you need to find the perpendicular not (12.5, 0) because that point is not part of the first one...\n\nFirst, we get "m" of the first equation, which we can write this way\ny= -4/5x - 2\nand m= -4/5\nso, the m we need for the new equation would be 5/4\nFinding the intercept of the second equation\n2x=25\nx=12.5\nAnd finding y in the first one, y = -12\n\nNow we find the new equiation with m= 5/4 in (12.5, -12)\ny+12 = 5/4 (x-12.5)\n4y + 48 = 5x - 62.5\n4y - 5x + 110.5 = 0  (this is the equation you are looking for)
385116
2x + y = 7\ny = -2x + 7\n\nThe slope of this is -2, so the slope of the perpendicular is 1/2.  Using the point-slope formula, you can find the equation of the other line:\n\ny - y1 = m(x - x1)\ny - 4 = (1/2)(x - 4)\ny - 4 = x/2 - 2\ny = x/2 + 2
554282
A = base * height\n\nbase = 8\n\nslant height = 6\nto get the actuall height, we use trig\n\nthink of 6 as the hypotenuse\n\nsin 30 = height / slant height\n\nor height = sin30 * 6\n\nheight = 3\n\ntherefore area = 8 * 3 = 24
174825
(-1,4), perpendicular to 2x+3y=8\n\n2x+3y=8\n3y = -2x+8\ny = (-2/3)x + (8/3)\nThen slope is: (-2/3)\nA rect with slope m perpendicular to the first:\nm(-2/3) = -1\nthen m = 3/2\nand the equation is \n(y-4) = (3/2)(x+1)\n2y-8 = 3x+3\n2y + 3x = 11