. 3-1416 X 20 + 3-1416 x 11 x 192= 925*2012 square inches, and divided by 144 B \^:== 6-425 feet nearly. PROBLEM find the solid content of the frustwn of a cone. Rule. To the product of the diameters of the two ends add tliesum of their squares; multiply this sum by the perpendicularheight and by -2618 ; the product is the solid content. Example 1. Required the solid content of the frustum in ProblemIV., whose perpendicular EF = 18 inches. 20 X 11 = 220, and 220 + 20^ + 11 x 18 x -2618 = 3491-8884cubic inches, and divided by 172
. 3-1416 X 20 + 3-1416 x 11 x 192= 925*2012 square inches, and divided by 144 B \^:== 6-425 feet nearly. PROBLEM find the solid content of the frustwn of a cone. Rule. To the product of the diameters of the two ends add tliesum of their squares; multiply this sum by the perpendicularheight and by -2618 ; the product is the solid content. Example 1. Required the solid content of the frustum in ProblemIV., whose perpendicular EF = 18 inches. 20 X 11 = 220, and 220 + 20^ + 11 x 18 x -2618 = 3491-8884cubic inches, and divided by 1728 = 2 0208 cubic feet nearly. 88 Example 2. Required the content, in imperial gallons, of theinverted frustum of a cone A B C D, whose inner dimensions are 3^feet deep, 18 inches diameter at bottom, and 22 inches diameter attop. 22 X 18 = 396, and 396 + 22-^ + Ib^ X 4213238*7024X -2618 =: —————— = 47 •745 galls, nearly. 277*274 ^ -^ Or, 13238-7024 x 0*00360654 = 47-75 gal-lons nearly, as before. PROBLEM yi. To find the solid content of the frustum of a To the sum of the areas of the two ends add the squareroot of their product; multiply this sum by theperpendicular height, and -J- of the product is thesolid content. Example. Required the solid content of thefrustum of a pyramid A B C D, whose perpen-dicular height = 24 inches, the area of the base= 144 inches, and area of the top end = 64. 144 + 64 r= 208, and Vl44 x 64 = 96 ; then208 + 96 X 24 144 X 64=: 2432 cubic inches, and -^ 1728
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Keywords: ., bookcentury1800, bookdecade1850, booksubjectenginee, bookyear1856
