 | Nathan Daboll - Arithmetic - 1825 - 248 pages
...circumference' and the product is the area; or if the diameter is ^iven vdthout the circumference, multiply the square of the diameter by ,7854 and the product will be the urea. EXAMPLES. 1. Required Hie area of a circle whose diameter is 12 inches, and circumference 37,7... | |
 | Nathan Daboll - Arithmetic - 1825 - 256 pages
...circumference, and the product is the area; or if the diameter is gi\en without the circumference, multiply the square of the diameter by ,7854 and the product will be tlie area. EXAMPLES. 1. Required the area of a circle whose diameter is 12 inches, and circumference... | |
 | Thomas Hornby (land surveyor.) - Surveying - 1827 - 318 pages
...Circle. RULE 1st. Multiply half the Circumference by half the Diameter, and the product is the area. RULE 2. Multiply the square of the diameter by .7854 and the product is the area. EXAMPLE. Required the area of the circle whose diameter AB is 600 links. (By Prob. 7th,)... | |
 | Daniel Parker - Arithmetic - 1828 - 358 pages
...between 15 and 60 ? 60X15=900 ; and ^900=30 JÎn». PROB. V. — To find the area of a circle, Rule. — Multiply the square of the diameter by ,7854, and the product will be the area. Or, multiply the square of the circumference by ,07958, and the product will be the area. 1. What is... | |
 | Nathan Daboll - Arithmetic - 1828 - 250 pages
...circumference, and the product is the area; or if tiic diameter is given without the circumference, multiply the square of the diameter by ,7854, and the product will be the ares. EXAMPLES. 1. Required the area of a circle whose diameter is 12 inches, and circumference 37,7... | |
 | John Bonnycastle - Geometry - 1829 - 256 pages
....7854. 4. w^hat is the area of a circle whose diameter is 7, and circumference 22 ? Ans. 38$. RULE II.* Multiply the square of the diameter by .7854, and the product will be the area ; or, Multiply the square of the circumference by .07958 and the product will be the area. * Demon.... | |
 | Thomas Conkling (W.) - Arithmetic - 1831 - 302 pages
...to the diameter; or, as 355 is to 133, so is the circumference, &c. RULE 3. To find the content. — Multiply the square of the diameter by .7854, and the product will be the area; or, multiply half the diameter by half the circumference; and the product will be the area. PROBLEM... | |
 | Samuel YOUNG (of Manchester.) - 1833 - 272 pages
...the diameter. Therefore, when 1 X 3-1416 the diameter is 1 the area = =r -7854. whence 4 we have RULE 2. Multiply the square of the diameter by '7854 and the Product will be the area. RULE 3. Multiply the square of the circumference by •07958 for the area. (1) Required the area of... | |
 | Francis Walkingame - 1833 - 204 pages
...the diameter. 1 X 3'14l6 Therefore, when the diameter is 1, the area = = '7854 ; whence we have RULE 2. Multiply the square of the diameter by '7854 ; and the product will be the area. RULE 3. Multiply the square of the circumference by •07958 for the area. (1) Required the area of... | |
 | Tobias Ostrander - Measurement - 1833 - 172 pages
...diameter ? Ans. 5,2521 feet. PROBLEM in. • The diameter given, to find the area. Rule—Multiply the square of the diameter by ,7854, and the product will be the area. EXAMPLES. 1. The diameter of a circle is 16 chains—How many acres are contained within its periphery... | |
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... multiply the square of the diameter by ,7854 and the product will be- the area. 
