In a regular triangular pyramid with all eges 6 cm, what is the lateral area and total area.  Full answer not needed.
134116
See previous answer--\n\nGives vertical height as rt27\n\nTherefore 1/2*6*rt27 is area of one side\n\nBase is 6*6 = 36
559013
Let the breadth of each red rectangle be x\n\nSo, area of red portion = 4x + x(3-x)\n\nSo, area of white portion is (4-x)(3-x)\n\nSo, 4x+x(3-x) = (4-x)(3-x)\nor 4x + 3x -3x^2 = x^2 -7x+12\nor 4x^2-14x+12 = 0\nor 2x^2-7x+6 = 0\nor 2x^2-6x-x+6=0\nor(2x-1)(x-3)=0\n\nx=1/2 or 0.5 foot is the reasonable answer.
387532
The area of a parallelogram is equal to the base times the height.\nA = (b)(h)\n\nThe way to get maximum area is a square (which is a parallelogram). In a square, all sides are the same. Since 69/2 + 69/2 = 69...\n\nA = (69/2)(69/2) = 4761/4 = 1190.25 cm²
324783
1.  Area of rt triangle whose leg is 6ft and whose\n    hypotenuse is 10ft?\n\n    Area = 24 sq. ft.  \n\n  Explanation:\n\n    First we need to find the other leg's length...\n    Using Pythagorean Theorem,\n    \n    c**2=a**2+b**2 \n       \n    where c is the hypotenuse, a is one leg, and \n    b is the other leg.\n\n    So, 100 = 36 + b**2;  b**2 = 100-36 = 64; therefore b=8.\n    Now, Area = 1/2(base*height) since the base is one leg\n    (we'll say it is a) and the height is the other leg \n    (therefore we'll name it b) \n    the Area = 1/2(8*6) = 1/2(48)= 24 sq. ft.\n\n2.   The hypotenuse is approx. 4.9 meters.\n    \n  Explanation:\n     \n     Using Pythagorean Theorem: c**2=a**2+b**2 we find that\n     c**2 = (3.5)(3.5) + (3.5)(3.5)\n     c**2 = 12.25 + 12.25\n     c**2 = 24.25\n     c = sq. rt. (24.25)\n     c = 4.9244 meters\n     hypotenuse is approx. 4.9 meters.\n\n3. hypotenuse is approx. 20 inches long.\n\n  Explanation: Area = 1/2(x)**2 \n               where x is length of one leg.\n\n     x**2 = 2*Area;\n     x**2 = 2*98\n     x**2 = 196\n     x = sq. rt. (196)\n     x = 14 inches\n\n     Now, to find length of hypotenuse, we use the\n     Pythagorean Theorem:\n\n     c**2 = a**2 + b**2; since a=b this can be rewritten so\n     c**2 = 2(a)**2\n     c**2 = 2(14)**2 -- Remember 14 is the length of a leg.\n     c**2 = 2(196)\n     c**2 = 392\n     c    = sq. rt. (392)\n     c = 19.7989\n     c is approx. 20 inches long.\n\n\n4.   The researcher will climb approx. 10 meters \n     to reach the crack.\n\n   Explanation:\n      \n      Using the Pythagorean Theorem: c**2 = a**2 + b**2\n      If we set a = 24, then c**2 = (24)**2 + b**2,\n      but we also know that c + b = 36 \n      which means c = 36 - b.\n\n      So to solve for this we need to set\n      (36-b)**2 = (24)**2 + b**2\n      1296-72b + b**2 = (24)**2 + b**2\n      1296 - 72b = (24)**2\n      1296 - 72b = 576\n      -72b = 576 - 1296\n      -72b = -720\n         b = -720/-72\n         b = 10 \n\n      And now we have our 10 meters the researcher \n      must climb to reach the crack in the tree.\n\n\n6.    Approximately a 42 inch stick is the longest stick\n      that will fit in a box that is 24 in X 30 in X 18 in.\n\n  Explanation:\n\n      First use Pythagorean Theorem to find the diagonal\n      inside the rectangle of 24 in X 30 in.\n\n      c**2 = (24)**2 + (30)**2\n      c**2 = 576 + 900\n      c**2 = 1476\n      c = approx. 38.418 in. -- we will call it 38 in.\n\n      Now, use Pythagorean Theorem to find the diagonal\n      inside the "new and invisible" rectangle that is \n      38 in. X 18 in.\n\n      c**2 = (38)**2 + (18)**2\n      c**2 = 1476 + 324\n      c**2 = 1800\n      c = sq. rt. (1800)\n      c is approx. 42.43 in.  \n      \n      We will call "c" 42 inches long.
254331
The surface area:\nS = 0.5*(B1+B2)*Hs  +  A1 + A2\nWhere B1 & B2 are the perimeters of the 2 bases\n      Hs = Slant height\n      A1 & A2 are the areas of the 2 bases\n\nThe volume:\nV = (1/3)*h*(A1+A2+ sqrt(A1*A2))\nWhere h = the height of the truncated tetrahedron
558462
Assuming you mean a regular hexagon, then with the 30-60-90 relationship, you should get that half the length of a side is 2.\n\nSo, the area of that yellow triangle in the link below is\n.5 * 2 * 2 * sqrt(3) = 2 * sqrt(3)\n\nThere are 12 triangles, so the area is 24 * sqrt (3).
176967
Well... a slightly advanced method is that you should fist realize that a triangle drawn through these three points will have an area = 0\n\nformula for area of a triangle is:\n        \n| a1 b1 1 |\n| a2 b2 1 |  x 1/2 = A\n| a3 b3 1 |\n\nWhere (a1,b1) ; (a2,b2) and (a3,b3) are the 3 vertices of the triangle.\n{Sorry --- no better way to represent a matrix here...}\n\nor; A = 1/2 [ a1(b2-b3) - b1(a2-a3) + (a2.b3 - a3.b2) ]\n\nHere; A = 0 ; a1=-5; b1=1 : a2=1; b2=p and a3=4; b3=-2\n\ntherefore; 0 = 1/2 [ -5(p+2) - (1-4) + (-2 - 4p)]\nor;        0 = [-5p -10 + 3 -2 -4p]\nor;        0 = -9p -9\nor;       9p = -9\n\n=> p = -1\n\nAnother method would be to find the line through two given points and pass the line through the third point... but that is slightly lengthy.... \n\nIf you have any doubt; feel free to ask me.
395425
The triangle formed by the three centres has area\nsqrt(3)/4  (Standard result for equilateral triangle).\n\nEach of the little bits between that triangle and the arcs that surround it is\n(area of sector with angle pi/3) - area of triangle\n= pi/6 -sqrt(3)/4, and so the total area of that inside bit (a sort of curved triangle formed by the three centres and the arcs joining them) is\n3*(pi/6 -sqrt(3)/4)  + sqrt(3)/4\n=pi/2 - sqrt(3)/2   [I'm sure there must be a slick trick to avoid all this!]\n\nNow look at the region common to any two of the circles -- I think it's called a "lune".  If you join the points of intersection of the circles, so as to halve the lune, each half has area\n=(area of sector with angle 2*pi/3) - (1/2)*1*sin(2*pi/3)\n= 2*pi/3 - sqrt*(3)/4\n\nDouble this to get the area of the "lune":\n4*pi/3 - sqrt(3)/2\n\nNotice that if we add the areas of two circles, we've included the area of a "lune" twice, so to get the total area of two circles we'd have to subtract the area of a "lune".  So:\n\nNow add the areas of the three circles, subtract the areas of three "lunes" because each of them is included twice.  However the bit which is common to the three lunes has then been subtracted three times, so we should add twice its area.  i.e.\n\n3 circles (3*pi) - 3 lunes (3*(4*pi/3 - sqrt(3)/2) + 2*middle bit  (2*(pi/2) - sqrt(3)/2)\n\nYou can finish the calculation, but I bet I've made a mistake somewhere.  Haven't time to check.  Sorry.
188358
if u kno the base and height.. then area= (base x height)/2\n\nif the triangle is scalar,\nthen :\n\nfirst find s, using hero's formula\n\ns= (a+b+c)/2   .... where a,b,c are the sides of triangle.\n\nthen , area = squareroot(s x (s-a) x (s-b) x (s-c)
186357
From your question, we know the 3 points of the triangle are:\n(0,0), (0,y) and (x, y).\n\nThis points us to the fact that the right angle is at point (0,y).\n\nHence, area A of triangle is given by the product of base and height divided by 2. Therefore, we have:\n\n Area A of triangle = 1/2 [base.height]\n                    = 1/2 y.x\n\nWe are told y = x2, therefore:\n Area A of triangle = 1/2 y.x\n                    = 1/2 x3 (read as half x-cube)
29943
The answer is 13.5 in.\n\nFirst we have that the total area is 178 in^2, and we have the lenght and width, so we just need to write the formula for the area of each side and sum them. We would have an unknown variable: height.\n\n178 = (2*4) + (2*4) + (4*X) + (4*X) + (2*X) + (2*X)\n\n178 = 16 + 8X + 4X\n\n(178-16)/12 = X\n\nX = 13.5
592325
The Lourve pyramid is made of 4 triangles and a square base.\nSo the surface area of this pyramid\n= 4*(b*h)/2 + l^2\n= 2lh + l^2\nwhere the base of the triangle, b = length of a side of the square and h is the perpendicular height of the triangle.\n\nVolume = (1/3)(base area)(height)
225741
the formula for the area of any regular shape is \n(pa)/2, so there is no multipling the base by 2 involved.\np=perimeter\na=apothem
576605
Ok, you've got a good one here.\nFor a right triangle, height and base are the short sides; and area of the triange is (bh)/2. Since we know the area, and the length of the sides, we can plug it in to an equation, and find x. So...\nA = (bh)/2\n60 = ((5x)(3x-1))/2\n120 = (5x)(3x-1)\n120 = 15x^2 - 5x\n15x^2 - 5x -120 = 0\nDivide both sides by 5 to get\n3x^2 - x -24 = 0\nOk, there's a number of ways to do this, but the simplest way is to list the combinations of integers that will multiply to get -24 (the constant in the above equation). That gives us the combinations of (1, -24; 2, -12; 3, -8; 4, -6; 6, -4; 8, -3; 12, -2; and 24, -1). Now, we need to figure out the one combination that, when the second number is multiplied by three (the coefficient of x^2), you can add it to the first number to become (-1). It happens to be (8, -3) {-3*3=-9, -9+8 = -1}. There; you've factored the equation, leaving you with...\n(3x+8)(x-3)=0\nSince multiplication rules say that when ab=0, either a=0 or b=0, we can test by setting both numbers equal to zero; giving us...\n3x+8 = 0 ; x-3 = 0\nx = -8/3; x = 3\nSince x is an absolute length, and lengths can't be negative, we can disregard (-8/3) for now. That gives us x=3, and putting that back into the original lengths gives us sides of 15 and 8. To find the hypotenuse, we use Pythagorean Theorum: a^2 + b^2 = c^2, substituting in the known lengths gives us:\n15^2 + 8^2 = c^2\n225 + 64 = 289  = c^2\nc = √289 = 17\nSo, the hypotenuse of your triangle is 17.
490663
16.72 inches squared.\n\n\nDivide an equilateral triangle into two equal right triangles (meaning, draw the height).  The right triangles will have angles of 30, 60, 90, right (an equilateral triangle has 3 60 degree angles)?\n\nThe ratio of the sides in such a right triangle (the smallest leg being a) is a:a(sq(3)):2a, where 2a is the hypotenuse and a(sq(3)) is the other leg.\n\nThe "other leg" in this case is the height.  So a, half of the length of the side of the whole equilateral triangle, is 5.382/sq(3).\n\nMultiply this by 2 to get the whole length of the side.  We have our base and height.\n\n\nThe area of an equilateral triangle is 1/2bh.  b=2(5.382)/sq(3)\nh=5.382\n\nPlug those in and you get 16.72.
199535
The area of a triangle is\n\n1/2 * Base * Height\n\nSo it the max base is 2 ft, and the max area is 20 ft^2, then the max height would have to be 20ft
400981
you can figure out the hypotenuse using the pythagorean theorem (a^2 + b^2 = c^2), so the hypotenuse is 10.\n\nAfter that, you just add up the areas of all the surfaces - For the triangles, its 6x8 divided by 2 = 24\n\nFor the rectangular faces, you do the following\n4x8 = 32\n4x6 = 24\n10x4 = 40\n\nAdd all these up and you get as surface area of 120.