how do u find the surface area of a prism...andcan someone name a space figure that does not have any vertice?
374445
sphere has no vertices\n\narea of prism is:\n (area of base)*2 + (perimeter of base)*height.
254189
cube = a 3  \n\nrectangular prism = a b c  \n\nirregular prism = b h  \n\ncylinder = b h = pi r 2 h  \n\npyramid = (1/3) b h  \n\ncone = (1/3) b h = 1/3 pi r 2 h  \n\nsphere = (4/3) pi r 3  \n\nellipsoid = (4/3) pi r1 r2 r3
549797
enclose the ireegular shape with a rectangle and find the empty spaces. then subtract that with the area of the enclosed rectangle to get ur answer.
192399
In the front?  You have to consider the whole cube to find volume.  Volume = base area * height  You could find the surface area of one side, but you'd need to know the side length.
70150
so you have to find the surface area of a cylinder that is 35diameter and 40 ft tall. or \n((3.14)17.5^2)40)
701955
You could use graph paper to get a good approximation of the surface area. Using a large circle would increase the accuracy.\nCount the whole squares, then count the half squares. \n\nAlternatively would be to use graph paper, then precisely cut out the circle, then cut the graph paper. On another sheet of graph paper, try to fill squares with the incomplete pieces from the circle.\n\nModifying Mike's answer, using a sensitive microbalance, find an average mass of one graph square then measure the mass of the cut out circle.
87971
The area of a parallelogram is area = base x height. Now, drawing a diagonal of the parallelogram splits the shape into two congruent triangles. Therefore, each triangle will have an area equal to (1/2)base x height.\n\nThis is true for any triangle; that is, any triangle has a corresponding parallelogram associated with it, which can be cut in half.
827739
Not quite sure what you are asking for....\n\nThe surface area of a right cone (i.e. a cone with its vertex above the center of its base) is:\npi*R*sqrt(R^2+H^2)\n\nwhere R is the radius of the base and H is the height of the cone.
250763
you split it into 2 ,a cone and a cylinder. first u need the slanting height of the cone and you find this by pythagorus theory for triangles ie L2 = h2 + r2, so height=6(found by subtracting the 15 from 21) radius=10 (20/2)therefore L2=6x6+10x10=136 therefore L =sqr root of 136 =11.6 surface area is A=pi*r*L\npi=3.142 so 3.142*10*11.6 = 364.5sq ft now the cylinder the surface area of a cylinder is circumference × h, or 2πr × h\nso 2*3.142*10*15 =942.6 so 942.6+364.5=1307.1 sq ft to find quantity of tins 1307.1/250=5.2 so they need 6 full tins you must check the formulas and calculations as I may have made a mistake ok
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.
572270
triangle area = 1/2(height * width)\n\ncircle area = pi * radius^2
186995
There is no specific word as far as i know, but we used to use the term "Total Surface Area.'\n\nThere are different formulas for calculating the total surface area depending upon the shape of the three-dimensional figure.  If you could elaborate your question, one can give you an exact answer; or is this enough?
679609
Surface Area=sqrt{3}a^2\n    Volume =(1/12)sqrt(2)a^3\nhttp://en.wikipedia.org/wiki/Tetrahedron
857420
Area of what?\n\nRectangle = b×h\nSquare = l^2\nTrapezium = ½(l+b)h\nEllipse = πlb\nCircle = πr2 \nTriangle = ½b×h
825601
its surface area q=q1+q2+q3+q4; area of any triangle =sqrt(p*(p-a)*(p-b)*(p-c)), a, b & c being edges of triangle & p=(a+b+c)/2. Here q1 is area of LSS facet, q2=q3 is area of two LLS facets each, q4 is area of LLL facet.\nIts volume is not as easy: first let’s find the height of facet LSS h1=2*q1/L=srqt(S^2-(L/2)^2), and that of LLL facet h4=2*q4/L=L*sqrt(3)/2, now the area of section Lh1h4=q5=(as above), and the height of this section:\nh5=2*q5/h4=L*q5/q4 that is the height your Tetrahedron.\nV=(1/3)*q4*L*q5/q4=L*q5/3 – oh poor kids!
890633
1. total area of a box of sides a,b,c is 2(ab+bc+ca)\na=4,b=3,c=6 => surface area = 108 sq.ft\n..\n2.Lateral area = pi*radius*slant ht\npi=3.1415,radius=2m(4m/2), slant ht=15m\nans is 188.49sq.metres\n..\n3.   volume of pyramid = (1/3)* height * base area\n= (1/3)* (6) * (6*6)\n= 72 cu.metre\n..\n4. lateral area of cylinder = 2*pi* radius* height\n=2*3.1415*12*60\n=4523.76cu. mm\n..\n5. total surface area = outer surface + inner surface + top surface + bottom surface\n= 2*pi*8*height + 2*pi*5*height + 2 times ( pi*(8-5)^2)\n[the last term is the sum of top and bottom surfaces which are circular rings]\nheight is not given :(
547584
the surface area is (1/2 X perimeter of base X slant height)\nthe total surface area is ((1/2 X perimeter of base X slant height) + (area of base))