A key to the National arithmetic

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R.S. Davis & Co., 1857
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Page 246 - DA, we have the semi-diameter of the larger circle, 5.7735027 -(- 5 = 10.7735027. By multiplying this last number by 2, we have the diameter of the larger circle, 10.7735027 X 2 = 21.5470054. As the area of a circle may be found by multiplying the square of the diameter by .785398, therefore, by dividing the area by .785398, the quotient will be the square of the diameter. The area of the lady's field is 500 acres = 80000 square rods ; 80000 -=-.785398 = 101859.28 square of the diameter ; V 101859.28...
Page 205 - ... and 3d, 8 small squares, each having its side equal in length to the thickness of the plank. Since the box is twice as wide as it is high, and twice as long as wide, the area of an end is doubled in a side, and quadrupled in the top or bottom. Then an end, a side, and the top or bottom, contain 7 times as many large squares, strips, and small squares, as an end ; that is, 14 large squares -)- 56 18 ! } \ s 1 Ш/Ш$Ж/#Ш 'ШШШМ; 'ежшшш. ': ,f
Page 204 - FIG. 1. No. 1. — Suppose the 6 pieces of which the box is made, to be so arranged (Fig. 1), that the edges of the top, bottom, and sideswill appear at the ends, as in the shaded parts about E ; and that the edges of the top and bottom will appear at the sides, as in the shaded parts above and below S. The external area of each face of the box may be considered as divided into, 1st, a large square, as at E ; 2d, 4 strips, one on each side of the large square, having each a width equal to the thickness...
Page 244 - Because ADEF is a regular figure, and the angles ADE and AFE being equal, each being a right angle, and the sides DE and FE being also equal, because they are radii of the circle DFHG, the sides AD and AF are also equal.

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