| Edinburgh encyclopaedia - 1830 - 856 pages
...is ™* 10, and its altitude CD is 6. Hence, by the rule. The area = fx 10 x 6 ss 40. PROBLEM XIII. To find the area of an ellipse. RULE. Multiply the product of the two axes by the number .7854 for the area. Let a denote the transverse axis AB, and fi the con- Fig.... | |
| Francis Walkingame - 1833 - 204 pages
...The two axes of an ellipse are 60 and 45 yards respectively : what is the circumference ? Problem 13. To find the area of an Ellipse. RULE. Multiply the product of the axes by '7854, for the area. (1 ) Required the area of an ellipse whose axes are 35 and 25. (2) What... | |
| Thomas Tate (mathematical master.) - 1848 - 284 pages
...and no— 10-5 ft; and their distance apart, AC = CE = &c. == -5 ft. Ans. 19-625 sq. ft. 17. PROBLEM. To find the area of an ellipse. RULE. Multiply the product of the two diameters by •7854, and the result will be the area. Let ABA' B' be an ellipse whose major diameter... | |
| James Elliot - 1851 - 152 pages
...What is the convex surface of the remainder of the sphere ? Ans. 42 sq. ft. 69 — in. PROBLEM XXX. To find the Area of an Ellipse. RULE. Multiply the product of the axes by •7854 — , or the product of the semi-axes by 3•1416-. FORMULA. Area = ab x n. Or, Area... | |
| Daniel Leach - Arithmetic - 1853 - 622 pages
...of thread and following it round with a pencil, keeping it extended in the form of a triangle. 373. To find the area of an ellipse, — RULE. Multiply...feet, 9 inches ? MENSURATION OF SOLIDS. SECTION XLI. k 374. A SPHERE, or globe, is a solid, bounded by a curve surface, every point of which is equally... | |
| James Stewart Eaton - Arithmetic - 1857 - 376 pages
...inscribed regular hexagon and triangle ? - Ans. Hexagon 14. > Triangle 24.2487112, j ' 483. PKOB. 15. — To find the area of an ellipse, RULE. — Multiply the product of the semi-axes by 3.141592. NOTE. — The proof of this rule is found in Conic Sections ; however, it is... | |
| Elias Loomis - Trigonometry - 1859 - 218 pages
...is the area of the ring included between their circumferences ? Ans., 18849.55 PROBLEM XIII. (103.) To find the area of an ellipse. RULE. Multiply the product of the semi-axes by 3.14159. For demonstration, see Geometry, Ellipse, Prop. 21. Ex. 1. What is the area of... | |
| Elias Loomis - Logarithms - 1859 - 372 pages
...is the area of the ring included between their circumferences ? Ans., 18849.55 PROBLEM XIII. (103.) To find the area of an ellipse. RULE. Multiply the product of the semi-axes by 3.14159. For demonstration, see Geometry, Ellipse, Prop. 21. Ex. 1. "What is the area... | |
| Thomas Baker (C.E.) - 1865 - 174 pages
...the heights of its arcs 4 and 20 feet ; required the area. . Ans. 57'867 square yards. PROBLEM XII. To find the area of an ellipse. RULE. — Multiply the product of the semiaxes TP, CP by 3' 14 16 for the area. C m FORMULA. A = ab TT, in which a and b are the semiaxes.... | |
| Lorenzo Fairbanks - 1875 - 472 pages
...inscribed square, when the diameter of the circle ""1. AC* =- = AB1. , — THE ELLIPSE. PROBLEM I. 721. To find the area of an ellipse. RULE. — Multiply the product of the two diameters by .7854. EXAMPLES. 1. Required the area of an ellipse of which the diameters are 6 and... | |
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