Explain how to find the surface area of a rectangular prism using a formula. Include an example.?
913838
A rectangular prism is like a box. So it has 6 sides in 3 matching pairs: top and bottom, right and left sides, and front and back. You can find the area of each pair and add them together:\nABxBC+ABxBC+BCxCD+BCxCD+CAxCD+CAxCD = PA (Prism Area)\n\nOr you could get the areas of one of each pair and double it:\n2x(ABxBC+BCxCD+CAxCD)=PA (Prism Area)\n\n;-D AB is the length of the front left side, with A at the bottom left corner, B at the top left corner. BC is the length of the front top with C at the top right corner, and CD is the length of the depth, with D at the back right corner.
374445
sphere has no vertices\n\narea of prism is:\n (area of base)*2 + (perimeter of base)*height.
6731
Area & Perimeter of Parallelogram\n\nA Parallelogram is a four-sided quandrilateral with both pairs of opposite sides parrallel.\nA area of parallelogram is calculated using the following formula:\n\nA=(base)*(height)=bh\n\nThe perimeter of a parallelogram is calculated using the following formula:\n\nP=2a+2b               (3-4)\n\nThe area of a parallelogram is always expressed in square units, and the perimeter of a parallelogram is always expressed in the original units.\n\nExample:\n\nCalculate the area and perimeter of a parallelogram with base (b) = 4', height (h) = 3', a = 5' and b = 4'. Be sure to include units in your answer.\n\nSolution:\nA= bh \nA=(4)(3)\nA=12 square feet \n\nP= 2a+2b\nP=2(5)+2(4)\nP=10+8\nP=18 feet
779149
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.
992822
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.
801034
Here's the answer: (I'm assuming the top of the box is counted in the area)---------   x = side of square base, h = height, S = surface area, P = perimeter length\n\nS = 2x^2 + 4xh\n\nP = 8x + 4h ---->  h = (P-8x)/4  Substituting,\n\nS = 2x^2 + 4*x*(P-8x)/4 = Px - 6x^2\n\nFor your particular case where P = 120 cm,\n\nS = (120x - 6x^2) cm^2
123707
Let:\n\nL = length of the rectangle\nW = width of the rectangle\nP = perimeter of the rectangle\n\nGiven:\n\nW = x\nL = 2x - 4\nP = 2L + 2W from geometry\n\nSolution:\n\nP = 2L + 2W\n64 = 2[2x - 4] + 2x\n\n2[2x - 4] + 2x = 64\n\n[4x - 8] + 2x = 64\n\n4x - 8 + 2x = 64\n\n4x + 2x = 64 + 8\n\n6x = 72\n\nx  =  72/6\n\nx  =  12 cm.\n\nTherefore:\n\nW = width = x = 12 cm. ANS.\n\nL = length = 2x - 4 = 2(12) - 4 = 24 - 4 = 20 cm. ANS.
137325
This is a lot to ask here. Why dont you look it up? \n\nAnd a lot of these are easy to figure out. \n\nSphere surface area is A=4 *pi *R^2 where R is the radius\n\nbecause A={int 0 to 2 pi} {int 0 t0 pi} R^2 sin(a)dadb\n\n\nThe cube surface area is A=6*S^2, where S is the length of an edge because there are 6 sides, each of area A^2.\n\nThe cylinder surface area A=2*pi*(R^2)+2*pi*R*H, where R is radius and H is height, because it has two circular end caps of area pi*(R^2) and a side of dimensions 2*pi*R by H, when it is unrolled. \n\nCone surface areas can be found the same way. Just unroll the side, and you get a triangle with height sqrt(H^2+R^2) and base 2*pi*R. The area of a triangle is .5*base*height. Then add in the area of the round bottom. \n\nAnd so on. I am sure you can do this yourself.
981115
that way you can remember the formula with generic terms like L and W then later insert the corresponding numbers.\n\nlength times width equals area\nL * W = A\nl2" * 12" = 144 square inches
282503
Let's break the area into many vertical rectangular strips between x=1 and x=2, each strip so narrow that we can assume that the top of the strip is horizontal. Thus each strip will be a very narrow and tall rectangle.\n\nLet's say the number of vertical strips we use is n. The first strip will have its left-hand side at x = 1, the next will have its left-hand side at x = 1 + 1/n, the third will have its left-hand side at x = 1 + 2/n, and in general the kth strip will have its left-hand side at x = 1 + (k-1)/n.\n\nNow, what is the y value at each of these x points? Using f(x)=x² we get, for the kth point,\n\ny = (1 + (k-1)/n)² = 1 + 2(k-1)/n + ((k-1)/n)² \n\nNow the area of each strip is equal to its base times its height. The base of each strip is 1/n. So the area of the kth strip is\n\nA = 1/n * (1 + 2(k-1)/n + ((k-1)/n)²)\n\nNow we have to add the areas of all these strips from k = 1 to k = n and rearrange the terms.\n\n.....k=n\nA = Σ (1/n + 2(k-1)/n² + (k-1)²/n^3)\n.....k=1   \n\n.....k=n\nA = Σ (1/n + (2k - 2)/n² + (k² - 2k + 1)/n^3)\n.....k=1\n\nGiving all fractions the common denominator of n^3,\n\n.....k=n\nA = Σ ((n² + 2kn - 2n + k² - 2k + 1)/n^3)\n.....k=1\n\n.....k=n\nA = Σ ((n² - 2n + 1)/n^3  + (2kn + k² - 2k)/n^3)\n.....k=1\n\nThe first term in the summation sign appears n times in the sum, so we can take it out by multiplying it by n. \n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ (2kn + k² - 2k)/n^3\n\nNow let's clear the parentheses in the summation:\n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ 2k/n² + Σ k²/n^3 - Σ 2k/n^3)\n\nThe first summation contains k=1, 2, 3, ... n, so we use the formula for the sum of the arithmetic series:\n\nS = n(n+1)/2\n\nso the first summation becomes\n\nn(n+1)/n² = (n² + n)/n² = 1 + 1/n\n\nThe last summation also contains k=1, 2, 3, ... n, so we use the same formula for the sum of a series, and we get\n\nn(n+1)/n^3 = (n² + n)/n^3 = 1/n + 1/n²\n\nFor the middle summation, we have to compute the sum of k² for k=1, 2, 3, ... n, or that is the sum of 1, 4, 9, 16, ... n².\n\nThe site\n\nhttp://mathworld.wolfram.com/PowerSum.html\n\ngives the formula for the sum of the squares of a series as\n\nS = (2n^3 + 3n² + n)/6\n\nso the middle summation becomes\n\n1/3 + 1/2n + 1/6n²\n\nNow, we are almost done.\n\nLet's let the number of strips get very large, go to infinity as a limit. Let's see what each term becomes:\n\n(n² - 2n + 1)/n² becomes 1 as n gets very large,\n1 + 1/n becomes 1 as n gets very large,\n1/n + 1/n² becomes 0 as n gets very large, and\n1/3 + 1/2n + 1/6n² becomes 1/3 as n gets very large.\n\nThe area under the curve is 1 + 1 + 1/3, or 7/3.\n\nYou can see why integral calculus was invented. Using integral calculus, this problem can be solved in two steps. The integral of f(x)=x² is x^3/3. Inserting the limits of x=1 and x=2 gives\n\nA = 2^3/3 - 1^3/3 = 7/3
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)
1127924
Area of rectangle = length x width \nArea of a circle = pi x R x R = pi R^2 where pi~3.14 and R - radius\nArea of an ellipse = pi x a x b see http://www.tpub.com/content/engineering/14069/css/14069_33.htm \n\nVolume = area of the base times hight\nVolume of a rectangular prism= area of rectangle x height = length x width x height\nVolume of a sphere = (4/3)pi x R x R x R = (4/3)pi x R^3\n \nCheck math tables or the web for equation of other geometric figures \n\nPercentage = 100 x (partial amount) / (total amount)\n\nHave fun
140062
L x W = Area\n\nCheater
985015
I would give an example if I could understand the question.  Perhaps you meant, "Why are measurement formulas expressed with variables?"\n\nIt is so that you can plug in the values that you observe and calculate the formula.  \nExample:  Area of rectangle = B * H\nB = base, H = height\nSo you can use this formula whenever you know the value of these two variables, for ANY rectangle.\n\nI hope this has answered your question; again, it wasn't clear.
822332
Since you have to divide the area into five rectangles, you will need rectangles from:\n\n0 to 1, 1 to 2, 2 to 3, 3 to 4, 4 to 5\n\nSince you are using the right endpoints, you need to be using the second value of each division; that is, in the "x to y" examples, you'll be using the y endpoints 1,2,3,4, and 5.\n\nThe area of each rectangle will then be 1*f(x), since the width of each rectangle is 1.  Therefore, the total sum formula should be as follows (with * representing multiplication).\n\n1 * f(1) + 1*f(2) + 1*f(3) + 1*f(4) + 1*f(5)\n\nOr, quite simply, f(1) + f(2) + f(3) + f(4) + f(5) to solve using the right endpoints.  Calculating this, we get\n\n(25 - 1^2) + (25 - 2^2) + (25 - 3^2) + (25 - 4^2) + (25 - 5^2)\n(24) + (21) + (16) + (9) + 0 = 70\n\nTo calculate using the left endpoints, recall that our intervals for the rectangles were:  0 to 1, 1 to 2, 2 to 3, 3 to 4, and 4 to 5.  In this case, we use the LEFT endpoints and take the FIRST value of each interval; that is, 0,1,2,3, and 4.  The width of the rectangle doesn't change at all (the width is 1), so our left endpoint summation formula because\n\n1 * f(0) + 1 * f(1) + 1 * f(2) + 1 * f(3) + 1 * f(4)\n\nor, quite simply\n\nf(0) + f(1) + f(2) + f(3) + f(4).  We calculated f(1) through f(4) already above, so we'll copy those values.  But f(0) = (25 - 0^2) = 25, so our total now becomes:\n\n25 + 24 + 21 + 16 + 9 = 95.\n\nArea using right endpoints = 70\nArea using left endpoints = 95
87402
a: width\nb: length\nperimeter = 2 x (a+b)= 420 ---> a+b = 210\nwe need to find max (a x b)\n\na= 210-b ---> area = a x b = b x (210-b) = 210 - b2 = - b2 + 210b = - b2 + 2 x 105 x b - 105squared + 105squared\n= - (b2 - 2 x 105 x b + 105squared) + 105 squared\n= - (b - 105)squared + 11025\n\nSince (b - 105) squared is always >= 0,\n - (b - 105)squared is always <= 0\n  - (b - 105)squared + 11025 is always  <= 11025\n\nthe largest area will be 11025 squared feet. This happens when - (b - 105)squared = 0, which means b = 105.\n\na = 210 - b = 105\n\n\nthe shortest way is to say that a rectangle has the largest area when it's a square, so each side would be 420/4 = 105.
38235
---------Y cm----------\n==========\n    1 cm\n_______________________\n          "           "   '\n    1.5cm "    B      "   '\n         _"__x cm_____"   ' \n  A       "           "   '3.5 cm\n      2cm "           "   ' \n__________"___________"   '\n\n\n7/2= 3.5\n\n3.5 cm- 2 cm= 1.5 cm\n\nY cm- X cm = 1 cm (from statement "The length in the hole is 1 centimeter less than the length of metal sheet. Length of metal sheet is Y cm, therefore, Y cm- 1 cm= X cm or Y cm - X cm= 1 cm).\n\nIf you compute the area of Rectangle A, that would be..\n\nlength x width = 3.5 cm x 1 cm = 3.5 cm sq.\n\n"After the whole is cut, are of remaining metal is 11 cm squared"... therefore\n\nArea of Rectangle A + Area of Rectangle B = 11 cmsq.\n\n11 cm sq. - 3.5 cm sq= 7.5 cm sq.\n\n\nGiven: Area of Rectangle B is 7.5 cm sq.\n       Length x width  (rect. B) = 7.5 cm sq.\n\nWe know width is 1.5 cm (from 3.5 cm- 2 cm).\n\nSolve: Length x width = Area\n       Length x 1.5 = 7.5\n       Length = 7.5 / 1.5\n       Length = 5\n\nTherefore:\n\nLength of x (hole) = 5 cm.\nLength of Y (metal sheet) = 5 cm + 1 cm. = 6 cm.\n\nI hope this helps. Let me know if you have other questions.