What is the surface area formula of a pyramid?
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
218464
If by area you mean "Surface area" then\nSA(sphere) = 4(pi)(r)^2\nand\nV(sphere) = 4(pi)(r)^3/3\n\nSince SA = 196 cm^2, then\n\n196cm^n = 4(pi)r^2\nr^2 = 49/pi\nr = 7/sqrt(pi) or approx. 3.949 cm\n\nfor the volume,\nV = 4(pi)(r)^3/3\nV = 4(pi)(7/pi^0.5)^3/3\nV = (4pi/3)(343/pi^1.5)\nV = (1372/(3pi^.5)) or approx. 258.02 cm^3\n\n^_^ Hope this helps ^_^
572751
Lateral Area\nThe lateral area is the surface area of a 3D figure, but excluding the area of any bases. Lateral Area is often abbreviated L.A. Imagine a soup can. Now cut down the side of the can and roll it flat. What was the circumference of the base is now the length of a rectangle. To find the area of this rectangle which is the same as the lateral area, multiply this length by the width, which was the height of the can.\nLateral Area of a prism: perimeter × height \nLateral Area of a cylinder: circumference × height. \nLateral Area of a regular pyramid: ½ perimeter × slant height. \nLateral Area of a right cone: ½ perimeter × slant height. \n\n\nThe lateral area of a regular pyramid or right cone is similar to that of prisms, but since each face is a triangle (or triangle-like), there is a factor of one half. The lateral area is thus half the slant height times the perimeter. The slant height is the distance from the vertex to the edge of the base where it is halfway between the base's vertices. If the pyramid is irregular and certainly if the cone is oblique, the surface area might not be calculatable using elementary techniques (which is a fancy way to say you may need calculus). It depends on if you can obtain the altitude (slant height) of each triangular face.\n\n\nSurface Area\nThe surface area of a figure is the sum of the area of all surfaces of a figure. Surface Area is often abbreviated S.A. The surface area of a prism or cylinder is the lateral area plus the area of each base. Since the bases for a prism or cylinder are congruent, this is often expressed as twice the area of the base.\nSurface Area = Lateral Area + n × Bases\nn = 2 for prisms/cylinders; n = 1 for pyramids/cones; n = 0 for spheres. \n\n\nThe surface area of a pyramid or cone is the lateral area plus the area of the single base.\n\nThe surface area of a sphere is equal to 4r2. Analogous to the unit circle is the unit sphere. Similarly, just as there are 2 radians of angle in one revolution, there are 4 steradians of solid angle in all directions.\n\nExample: Consider a right pyramid A-BCDE with vertex A and square base BCDE of length 20" on each side and a slant height of 26". What are its lateral and surface areas?\n\nAnswer: We don't need the height for this calculation, but we will calculate it anyway to stress the difference between slant height and height. The slant height is the hypotenuse of a right triangle where the height is one leg and 20"/2 = 10" is the other leg. Thus 102 + h2 = 262 or 100 + h2 = 676. Thus h2 = 576 or h = 24". The lateral surfaces are all triangles with a base of 20" and a height (the slant height) of 26". There are four of them. Thus the lateral area is 4×½×20"×26" = 1040 in2. The base is 20" square or 400 in2. Thus the [total] surface area is 1440 in2.\n\n  Understanding surface area may be clearer if you refer back to the net associated with the object. At left is a net for a cube and at right a portion of a net for a sphere. Each of these portions of a sphere is called a gore.\n\n Now is a good time to review something learned in algebra, namely (x + y)2 = x2 + xy + xy + y2 = x2 + 2xy + y2. The diagram at the right should clarify this further, help you remember the FOIL method, as well as give a physical basis for this relationship. (Remember also, the square root of (x2 + y2) does NOT equal x + y.) Consider extending the FOIL method first into trinomials: (a + b + c)(d + e + f) = ad + ae + af + bd + be + bf + cd + ce + cf. The distributive property is another way to consider this situation. Here the box method is useful.
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initial surface area=6a^2\nnew surface area=2[2a^2+4(a^2/2)]\n=2*4a^2=8a^2\nit will be 4/3 times the original surface area
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Surface Area=sqrt{3}a^2\n    Volume =(1/12)sqrt(2)a^3\nhttp://en.wikipedia.org/wiki/Tetrahedron
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Since each has eight edges it is a square pyramid\nAnd also the Lemght if the edge is given as 8 unit\nNow the altitude or axis of pyramid is sqrt(8^2+128)\nV(Volume)=1/3 * a^2 * height\nVolume of two pyramidas will be twice this that is 2V\nNow since we know the volume and base edge(a=8)\nwWe can find the axis by volume of pyr=1/3 * a^2 * axis\nconsidering a right triangle with axis and half of base making the right angle and slant height as hypotenuse you can find slant height using H^2=B^2+ALT^2\n\n\nWhile doing the problem dont calculate the values You can just fid an approximate value ad cancell terms thus making ita lot simple
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T= L + 2B\n\nT is total surface area\nL is Perimeter x Height\nB is base area
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There is a delightful article by the late Dr. John Kenny on that, back a few years ago, looking at the formulae for surface area and volumes of spheres; I'll see if I can find it for you--it's in my files SOMEWHERE--and give you the reference, and the formulae he generated. It turns out that they're not precisely the same as what you've got--but not terribly far from it.\n\nAs I said, I'll see if I can find it, and I'll get you the reference and the formulae.\n\nAbout 9pm CST...\n\nOK, here is the formulae, done the best I can on this thing...\n\n\nFor a sphere of dimension 2*k, radius R\n\nVolume: (pi^k)*(R^(2*k))/k!\n\nFor a sphere of dimension 2*k + 1 and radius R\n\nVolume:\n(2^(2*k+1))*k!*(pi^k)* (R^(2*k+1))/(2*k+1)!\n\n\nTo get the surface areas for each of these, all you do is differentiate with respect to R...\n\nAs I said, it was a little more complex than you thought--the surface areas come out a little bit differently for odd and even numbered dimensions.  I do not recall why, frankly.
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A cone and pyramid have one thing in common, the base area is proportional\nto the square of the height.  If you keep the angles and overall shape the\nsame and just increase the size (like a slight zoom of a camera lens), the\nbase area is some constant times the square of the height:  A=cH^2.  The "c"\nis a constant that will eventually divide out.  Also, the volume is\nproportional to the cube of the height:  V=kH^3.  We want to find out how k\nmust relate to c.\n\nConsider two cones with identical shapes: one has height H, one has height\nH+h.  "h" is extremely small.  If H were 10 meters, h would be a millimeter.\nThe base area and volume for the first are A1=cH^2 and V1=kH^3.  The other's\narea and volume are A2=c(H+h)^2 and V2=k(H+h)^3.\n\nIf we subtract the larger smaller volume from the larger volume, we get the\nvery thin sheet at the bottom.  If h is small enough, the thin sheet will\nhave area A1=cH^2 and height h.  The thin sheet will have a volume (cH^2)h.\nIf you don't say that h is very small, then the sheet will have a more\ncomplex volume formula.  We now have V2-V1=(cH^2)h.  Let's do the algebra:\nk(H+h)^3- kH^3=(cH^2)h\nkH^3+(3kH^2)h+(3kH)h^2+kh^3  -  kh^3  = (cH^2)h\n(3kH^2)h + (3kH)h^2 + kh^3 =(cH^2)h\n\nNow, the limits:\n\n"h" is tiny, so the second and third terms are extremely small compared to\nthe first.  They are like adding a penny to a million dollars.  They really\ndon't count enough to keep.  Limit theory lets us drop them under the\ncircumstances.  This yields:  (3kH^2)c=(cH^2)h.  Divide out common terms and\nget:  3k=c, or k=(1/3)c.  Put this in the original formula for volume:\nV=kH^3=(1/3)cH^3=(1/3)(cH^2)H=(1/3)AH.  Volume equals (1/3)Base area times
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The triangle area formula is\n\nA = 1/2 b h\n\na = 1/2(5)(2)\n\nA = 5/2(2)\n\nA = 10/2\n\nA = 5\n\nThe area of the triangle is 5 feet\n\nThe answer is " a "\n\n- - - - - - - - - - - -s-
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This is the formula for the volume of a triangular prism.  Volume = Area of base x height.  ab/2 is the area of the triangular base, and multiplying by h (height) gives you the volume.
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The volume of the pyramid is 1/3 the volume of the prism.\n\nTo prove this, look at a horizontal cross section of pyramid. Let A = the area of the base.\n\n\nIf the the height of the pyramid is h, look at a cross section of the pyramid at a height t above the base. Any one dimentional measurement x in the base, will have size x(h-t)/n in the cross section. Any two dimensional measurement, (area) will be multiplied by ((h-t)/h)^2, i.e. A((h-t)/h)^2.\n\nNote\nNow integrate for t = 0 to h, A((h-t)/h)^2 dt\n\nYou get -A/3(h - t )^3/h^2.\nEvaluating at t = 0 and t = h you get 1/3Ah.\n\nThe volume of the prism is Ah.
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Imagine you unraveled the cylinder, the middle would turn into a rectangle, right? so to find the surface area, take the height of the cylinder and multiply it by the circumference of the circles at the bases, then add in the area of the circles. Circumference is 2(pi)r and area is (pi)r^2.
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If you are looking for the surface area or volume of a pyramid, check this out:
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to find the surface area of a pyramid take 1/2 of the perimeter of the base times the slant height and add on the area of the base.  First lets find the area of the base with 3 sides of 17.32 each. You need more information for this, unless you have been taught some simple trig functions. To get the area of the base we need to know the height of this base(if it were standing up).If I use the tangent function I find that the height of the triangle is approximately 15 inches. So the area of the base is 1/2 times 17.32 times 15. I did the tangent of 60 degrees = the height of the triangle divided by 1/2 the side of the triangle.  I hope this helps solve the rest of the problem.
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The surface area of any prism equals the sum of the areas of its faces, which include the floor, roof and walls. Because the floor and the roof of a prism have the same shape, the surface area can always be found as follows:\ncall the 'floor' the 'base'. The height of the prism = 'H'.\nThe surface area of a prism = 2 × area of base  +  perimeter of base × H\n\nexamples:\n\nRectangular based prism:\nBase shape: Rectangle, length 'L' and width 'W'\nArea of base: L  × W\nPerimeter of base: 2(L+W)\nSurface area = 2LW + 2(L+W)H\n \nCircular based prism\nBase shape: Circle, radius 'R'\nArea of base: pR²\nPerimeter of base: 2pR\nSurface area = 2pR² + 2pRH\n\nTriangular based prism\nBase shape: Triangle: base 'b', height 'h', and sides S1, S2 and S3 \nArea of base: ½b×h\nPerimeter of base: S1+ S2 + S3\nSurface area = bh + (S1+ S2 + S3)H\n\nAs for lateral area of a cone , If r is the radius of the base, h is the altitude, and l is the length between the vertex and a point on the base circle \nlateral area pi rl=pi r x sq.root of r to the power 2 + sq.root of h to the power of 2\n or try this site- http://www.mathepower.com\n\nThe lateral area of a prism is the sum of the areas of the lateral faces (all the ... How many lateral faces does this prism have?\nWhat is the area of each \n\nelse there are many sites to giv better details matching ur specific figure ...try google for search..all luck..
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Surface Area = 4*PI*r^2\n\nr = SQRT(SA/4/PI)