| John Potter - Mathematics - 1753 - 568 pages
...of the Square whofe Side is 15.062. The fecond Example may be tried in the fame Manner. Article 67. **The Area of a Circle being given, to find the Diameter. Rule.** Multiply the given Area by 1.2732, and the Product is the Square of the Diameter; then extracting the... | |
| Thomas Keith - Arithmetic - 1822 - 354 pages
...the given superficies consists of yards, or feet, &c. Prop. 5. Given the area or surface of a circle **to find the diameter. Rule.- Divide the area by '7854, and extract the square root of the quotient.** Prop. 6. The base and perpendicular of a right angled triangle being given, to find the hypothenuse.... | |
| Beriah Stevens - Arithmetic - 1822 - 436 pages
...whose circumference is 44 1 44 X44 X ,07958, = 1 54,06608, answer. PROBLEM 16lh. The area of a circle **given, to find the diameter. RULE. Divide the area by ,7854 and extract the** equare root of the quotient ; the root ivill be the diameter sought. . PROBLEM 17th. The area given,... | |
| Nicolas Pike - Arithmetic - 1822 - 562 pages
...?puare. the square. Or',a.< 13-545: 12:: 12: 10C31 Or, 37-7H-3-5 16=10 C3t. •-the side. • ART. 18. **The Area of a Circle being given, to find the Diameter. RULE.** Multiply the given area by 1-2732, and the product will be the square of the diameter ; then, extracting... | |
| Oliver Welch - Arithmetic - 1826 - 242 pages
...circle whose circumference is 44 rods ? 44X44X07968=154.06688, Ans. PROBLEM VI. The area of a circle **given to find the diameter. RULE. — Divide the area....7854 and extract the square root of the quotient** ; the root is the diameter sought. (See application of square roo.t, case 2.) PROBLEM VII. The area... | |
| William Kinne - 1829 - 246 pages
...whose circumference is 30 rods? 30 x30x, 07958=7 1,62200 rods. Ans. PROBLEM 1V. The area of a circle **given to find the diameter. RULE. — Divide the area by ,7854 — and** the square root ef the quotient is the required diameter. EXAMPLES. 1. Required the diameter of a circle... | |
| William Ruger - Arithmetic - 1832 - 282 pages
...the diameter of a circle whose circumference is 94,248 rods ? As 355 : 113 : : 94,248 : Ans. 30 rods. **The area of a circle being given, to find the diameter....,7854 and extract the square root of the quotient;** the root will be the diameter sought. Or, as 355 : 452 ::so is the area : to the square of the diameter;... | |
| George Alfred - Arithmetic - 1834 - 336 pages
...circle be 20, what is the diameter of another 4 times less than the given one ? Ans. 10. • PROBLEM 5. **The area of a circle being given to find the diameter. RULE. Divide the area** of the given circle by .7854, and the square root of the quotient will be the diameter required. NOTE.... | |
| Arithmetic - 1841 - 200 pages
...circle whose circumference IB 44 rods? 44X44X-07953=154-0668S., Ans. PROBLEM VI. The area of a circle **given to find the diameter. RULE. — (!). Divide...-7854 and extract the square root of the quotient;** the root is the diameter sought. (See application of sqtare root, case.2.) PROBLEM VII. The area given... | |
| William Ruger - Arithmetic - 1841 - 268 pages
...113 :: 94,248 : Ans. 30 rods. The area of a circle being giten, to -find the diameter. RULE.-^Divide **the area by ,7854, and extract the square -root of the quotient** ; the root will be the diameter sought. Or, as 355 : 452 : : so is the area : to the square of the... | |
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