| Henry Sherwin - 1772 - 116 sider
...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 - 270 sider
...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 - 100 sider
...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... | |
| 1845
...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,... | |
| ...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
...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 - 322 sider
...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 - 194 sider
...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 - 178 sider
...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 - 186 sider
...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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