 | Henry Sherwin - 1772 - 454 pàgines
...thereof. Whence we have the following Rule to Meafure any Circle -wbofe Diameter is given. Multip'y the Square of the Diameter by ,7854 •and the Product is nearly the Area thereof. ' If С be the Circumference of any Circle, then fince 3.1416 = с • с с с1 4 í> : i : : С :... | |
 | Thomas Hornby (land surveyor.) - 1827 - 318 pàgines
...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,)... | |
 | William Templeton - 1841 - 112 pàgines
...balance. THE AREA OF THE VALVE. To find the area, or number of square inches in a valve. RULE. — Multiply the square of the diameter by .7854 and the product is the area, — thus, 2.5" = 6.25 X .7854 = 4.9087 the area. THE SMOKE BOX. The shell of the smoke box... | |
 | Nathan Scholfield - 1845 - 894 pàgines
...the circle re quired. Or thus : 31.4159 x^ =78.5398 as before. RULE II. — When the diameter only is given. Multiply the square of the diameter by .7854, and the product will be the area of the circle. For the ratio of the square described on the diameter of a circle,... | |
 | 458 pàgines
...7539.8160 125.6636 = 125 ft. 8 in. nearly = arc. PROBLEM XII. To find the area of a circle. 1. When the diameter is given, multiply the square of the diameter by .7854, and the product will be the area. 2. When the circumference and diameter both are given, multiply half the circumference... | |
 | William Templeton (engineer.) - 1848 - 118 pàgines
...balance. THE AREA OF THE VALVE. To find the area, or number of square inches in a ralve. RULE. — Multiply the square of the diameter by .7854 and the product is the area, — thus, 2.5" = 6.25 X .7854 = 4.9087 the area. THE SMOKE BOX. The shell of the smoke box... | |
 | Charles Haynes Haswell - 1858 - 350 pàgines
...7, so is the circumference to the diameter. To ascertain the Area of a Circle (Fig. 22). RULE. — Multiply the square of the diameter by .7854, and the product is the area. Or, multiply the square of the circumference by .07958. Or, multiply half the circumference... | |
 | Hoy D. Orton - 1866 - 202 pàgines
...the diameter by half the circumference, and the product is the area ; or, which is the same thing, multiply the square of the diameter by .7854, and the product is the area. To find the solidity of a sphere or globe. RULE. — Multiply tJie cube of the diameter by... | |
 | Leroy J. Blinn - 1866 - 216 pàgines
...product is the Circumference. 2. Multiply the circumference by 81831, the product is the diameter. 3. Multiply the square of the diameter by .7854, and the product is the area. 4. Multiply the square root of the area by 1.12837, the product ia the diameter. 5. Multiply... | |
 | Hoy D. Orton - 1871 - 202 pàgines
...the diameter by half the< circumference, and the product is the area; ort. which is /he same thing, multiply the square of the? diameter by .7854, and the product is the area. To find the solidity of a sphere or globe. RULE. — Multiply the cube of the diameter ojf.... | |
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J of the square of their difference, then multiply by the hight, and divide as in the last rule. Having the diameter of a circle given, to find the area. RULE. — Multiply half the diameter by half the circumference, and the product is the area ; or,... 
