Find the area of an isosoceles trapezoid with legs 5 and bases 9 and 17?
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I'm thinking of a trapezoid with a base of 9 on top and 17 on the bottom, with the other two sides being 5.  If I were to drop a verticle line from one corner where the 9 base meets one of the legs to the other base, then I have a right triangle.  The hypotenuse would be 5 and the base would be (17-9)/2 = 4.  So the height would be sqrt(5^2-4^2) = 3.  So you have a trapezoid that has bases 9 and 17 and height 3, so the area is\n\nA = (17+9)3/2 = 39 (Area of trapezoid is (b1+b2)h/2, where b1 and b2 are the length of the bases and h is the height.)\n\nedit: Ooh, I just found a formula you can use if you know the length of the sides.  Let a be the length of the longer parallel side, c be the length of the shorter parallel side, and b and d be the legs.  Then\n\nA = (a+c)/(4(a-c))*\nsqrt((a+b-c+d)(a-b-c+d)*\n(a+b-c-d)(-a+b+c+d))\n\nSo in your case a = 17, c = 9, b = d = 5.\n\nA = (17+9)/(4(17-9))*\nsqrt((17+5-9+5)(17-5-9+5)*\n(17+5-9-5)(-17+5+9+5))\n= (26/32)sqrt(18*8*8*2)\n= (26/32)(48)\n= 39\n\nSorry, I have to line break sometimes because Yahoo likes to truncate.\n\nSo either way...
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I'll leave you to do some of it.\n\nLet the length of the base be x cm, the height h cm.  \n\nThe total area of base and sides is\nx^2 + 4*h, and since we are told this is 1200 sq cm, we write\nx^2 + 4h = 1200, which allows us to express h in terms of x.  [You can do that.]\n\nNow volume\nV = (x^2)*h, and if you replace h by the expression you just found, you have V as a function of x only.  I expect you've already done the mechanics of differentiation and finding max val, so I'll leave you to finish it yourself or copy from others who will post the complete working here or you can email  h_chalker@yahoo.com.au
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Area of a trapezoid = 1/2 Height (Base1 + Base 2)\n224 sqm = 1/2 (16)(20 + x)\nmultiply both sides of the equation by 2 to remove fractions\n448 = 16 (20 + x)\n448 = 320 + 16x\ncombine\n16x = 128\nx = 8 m
389991
height times the base area. divided by three.
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It sounds to me like you are talking about a trapezoid. Go to the website I have listed under my sources and see if what you are describing looks like the trapezoid on there. If it is a trapezoid, the area formula is  Area = 1/2h(b1 + b2). h=height, b1=base 1, b2=base 2
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Area = pi times r squared\n\nDivide 199 by pi, this will be r^2\n\nDo the square root of this to get r\n\nThen double it to get d\n\nIf you use the pi button for pi, it is 15.917747 ft
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It's a trapezoid.  The base is 9 (11-2) and the top is 3 (8-5).  The height is 4 (5-1).\n\nThe area is the average of the bases x the height.\n\nSo that is (9+3)/2 x 4\n\n= 6 x 4\n= 24 sq. units.
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Answers:\n\n1) B    Area = 1/2 base times altitude\n\n2) D   s= 3t  So R = 3t + 2t = 5t  \n\n3) D  Add the 4 values together and divide by 4. Make sure \n      you know how to add fractions.\n\n4) C  \n\n5) C  Change 2 and 1/4 to improper fraction 9/4. Then  \n      divide 36 by 9/4. 36/(9/4) = 36x4/9 = 16
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A = 1/2 x (a + b) x h\n90 = 1/2 x (6.5 + 5.5) x h\n90 = 1/2 x (12) x h\n90 = 6h\n90 / 6 = h\n15 = h\n\nThe height is 15.  The 'x' should be used as a multiplier.\n\nThe formula for the area of a trapezoid is also:\n\nArea = (height * (side 1 + side 2)) / 2\n\nHope that helps! =)
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SA = 5as + 5sh\n\nwhere\na = apothem length \ns = side length\nn = #sides\nh = height\n\nDo the math.
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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
666057
*\n1/2 base X altitude + area (triangle)\n\n\n45 square centimeters
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In a 45-45-90 triangle, there is an important ratio. leg-leg-hypotenuse = 1-1-square root of two.\n\nTo find the length of the leg if the hypotenuse is given...\ndivide the hypotenuse by square root of two.\nFor example... if the hypotenuse were 5, the leg would be 5/square root of 2 = about 3.5
893599
You may want to be more specific on this one... are you trying to find the largest possible area? You can inscribe lots of different trapezoids in a circle. \n\nAnd which theta are we talking about? The angle that the base makes with the circle?\n\nTry that one again,\n\nUMRmathmajor
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 :(
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Area = πab where a and b are the semi-major and semi-minor axis length
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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