what is the area of 9" dia ?
470650
diameter = 9 inches\n\nRadius = 4.5 inches\n\npi * r^2 = 3.14159 * 4.5^2 = 63.61725...
15658
The volume of a cone is:    1/3Bh\nThat's 1/3 (B times h)\n  B is the area of the base.\n  h is the vertical height of the cone when you  measure it from the point straight down.\nSo the answer is 1/3 times (9x10).\nOr, 1/3 times 90.\nFinal answer is 90/3 = 30 feet cubed.\nThis is assuming the area of the base is 10 (or 9 - - you did not specify. But the answer is the same either way. Did you give a Base AREA? or a Base DIAMETER?)
327427
Each side would have length sqrt(3), so the area would be 3 m^2.
460116
It depends what it looks like.\nWithout seeing the actual picture I cannot find the area.
210010
8-9 inches at the least.no.
501756
length x width = area\n3 x 6 = 18
814915
Area = ½ ab SinC\n Area = ½(55)(29) Sin 61º\n Area = ½(1595) Sin 61º\n Area = 797∙5 Sin 61º\n Area = 797∙5(0∙874 619 707)\n Area = 697∙509 2165\n Area ≈ 697∙51 square units.
27117
19 feet 9 inches but there is talk of lenghtening it.
865884
A = lb = 98 in² (l = length and b = width, both in inches)\nSo b = 98/l\nWorking area W = (l - 8)(b - 4)\n= lb - 4l - 8b + 32\n= 98 - 4l - 8b + 32\n= 130 - 4l - 8b\n= 130 - 4l - 784/l\n\ndW/dl = - 4 + 784/l²\n= 0 for stationary points\nSo 4 = 784/l²\nie 4l² = 784\nThus l² = 196\nl = √196 (ignoring negative result)\n= 14 inches\nd²W/dl² = - 2 * 784/l³ < 0 when l = 14 so area is minimised when l = 14\n\nWhen l = 14 b = 7 (lb = 98)\nSo the dimensions to minimise paper used is 14 inches X 7 inches
1175861
Area of Semicircle = (PI * R * R)/2\n\nAns. = 2.262 sq. cm
224714
Perimeter for a square is 4*side\nSo side of square = 36/4=9 inches\n\nAnd area of square is side*side\nthus area is 9*9=81 sq inches !!!
145637
2 to 4 inches flased and 7 to 9 erect
720499
7.3m (width) by 2.4m (height)
220777
use the following formula .. \n\ntotatal surface area = \n[(H/D)*4(pie)(0.5D*0.5D)] + [ (H-t)/(D-2t)*4(pie)*(0.5D-t)(0.5D-t)] + [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]\n\nthis is considering D = tank diametere is from outer surface and not the cavity diameter \n\n\nouter surface area of cover = (H/D)*4*(pie)(0.5D*0.5D)\ninner surface area of cover = [(H-t)/(D-2t)*4*(pie)*(0.5D-t)(0.5D-t)] \n\nthe base line are = [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]
147303
2 times pi times the radius\nor\npi times the diameter\n\npi = 3.14\n\nThe first person to answer gave you the formula for the area, not circumference.
907964
let l be the length and w be the width\n lw=111\n\nbut it has a margin so Area= (l-11cm)(w-4)\ndifferntiate\nbut we have 3 unknowns so we use the first ea lw=111\nl=111/w\n\nA=(111/w-11)(w-4)\n\ndA/dw= (111/w-11)+(w-4)(-111/w^2)=0\n\n111/w-444/w^2= 111/w-11\ncancel 111/w\n-444/w^2=-11\ncrossmultiply\nw^2=-444/-11\nw=6.4 cm\nl=111/6.4=17.3
996399
The amount of sand you need is,\n\nVolume V = A x h\n\nwhere A = pi * r^2 is the circular area\nand h is the required height.\n\nIn your case, r = 18"/2 = 9"\n\nSo, for 1" height, you will need,\n\nV = 22/7 * 9^2 * 1 = 254.57 cu in of sand.