 | Charles Hutton - 1807 - 464 σελίδες
...17'4,20-6, 14% IQ'5,20'1, {J4'4; what is the area? Ans. 1550'64. PROBLEM 1 , To. find the Area of an Ellisis or Oval. MULTIPLY the longest diameter, or axis, by the shortest ; then multiply the product by the decimal -7854, fot the area. As appears from cor. 2, theor. 3, of the Ellipse, in the Conic Sections.... | |
 | Charles Hutton - 1811 - 494 σελίδες
...equidistant places 174, 20'6, 14% 16-5, 20'1, i!4-4; what is the area ? Ans. 1550-64. PROBLEM XIV. To find the Area of an Ellipsis or Oval. MULTIPLY...axis, by the shortest ; then multiply the product by the decimal '7854, for the area. As appears from cor. 2, theor. 3, of the Ellipse, in the Conic Sections.... | |
 | Charles Hutton - 1816 - 610 σελίδες
...what is tbc area ? Ans. 155O-64. • PROBLEM XIV. To find the Area of an Eltisit or Oval. "Mui-TiPtY the longest diameter, or axis, by the shortest ; then multiply the product by the decimal -785* for ihe area. As appears from cor. 2, theor. 3, of the Ellipse, m the Conic Sections.... | |
 | Charles Hutton - 1822 - 616 σελίδες
...-. 16-5, SO-I, 24-4; what is the area ? Ana. 1650-64. PROBLEM XIV. To find the Area of an Ell i sis or Oval. MULTIPLY the longest diameter, or axis, by the shortest ; then multiply the product by the decimal -7854, for the area. As appears from cor. 2, theor. 3, of the Ellipse, in the Conic Sections.... | |
 | Charles Hutton - 1831 - 632 σελίδες
...equidistant places 17-4, 20-6, 14-2, 16-5, 20-1, 21-4 ; what is the area ? Ans. 1550-64. rHOBLEH XIV. Tn find the Area of an Ellipsis or Oval. MULTIPLY the...axis, by the shortest ; then multiply the product by the decimal -7854, for the area. As appears from cor. 2, Iheor. 3, of the Ellipse, in the Conic Sections.... | |
 | Charles Hutton - 1831 - 660 σελίδες
...equidistant places 17-4, 20-6, 14-2, 16-5, 20-1, 24-4 ; what is the area ? Ans. 1550-64. PROBLEM XIV. Toßnd the Area of an Ellipsis or OvaL MULTIPLY the longest...axis, by the shortest ; then multiply the product by the decimal -7854, for the area. As appears from cor. 2, theor. 3, of the Ellipse, in the Conic Sections.... | |
 | Frederick Augustus Griffiths - 1839 - 348 σελίδες
...trapezoids, and add them all together for the whole area. To find the area of an ellisis or oval.—Multiply the longest diameter or axis by the shortest, then multiply the product by -7854 for the area. To find the area of a parabola, or its segment.—Multiply the base by the perpendicular height and... | |
 | Frederick Augustus Griffiths - 1840 - 436 σελίδες
...diameter being 50 inches. 50 X 50 X '7854 „_, , {square inches. __ yoi'io j 2 C The Area required. To find the Area of an Ellipsis, or Oval, Multiply...diameters are 25 inches, and 18 inches. 25 X 18 X -7854 == 353-43 S. Inches. Area required. To find the Area of a Parabola, or its Segment. Multiply the base... | |
 | Frederick Augustus Griffiths - 1854 - 406 σελίδες
...-7854 for the area. Example.— Required the area of the ellipse, whose diameters are 25 inches, and 18 inches. 25 x 18 x -7854 = 353-43 square inches. Area...required. To find the area of a parabola, or its segment. height, and take twobase is 20x12 = 240 | of 240 = 1 60 square feet. Area required. MENSURATION OF... | |
 | Frederick Augustus Griffiths - 1868 - 540 σελίδες
...for the area. Example. — Required the area of the ellipse, whose diameters are 25 inches, and 18 inches. 25 X 18 X -7854 = 353-43 square inches. Area required. To ßnd the area of a parabola, or its segment. Multiply the base by the perpendicular height, and take... | |
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MULTIPLY the longest diameter, or axis, by the shortest ; then multiply the product by the decimal -7854, for the area. As appears from cor. 2, theor. 3, of the Ellipse, in the Conic Sections. Ex. 1. Required the area of an ellipse whose two axes are... 