 | William Ruger - Arithmetic - 1836 - 274 pages
...inch? Ans. £237 10s. id. To find the area of a circle, the circumference and diam'eter being given. RULE. — Multiply half the circumference by half the diameter, and the product will be the area. Or multiply the whole circumference by the whole diameter, and one fourth of the... | |
 | Thomas William Silloway - Carpentry - 1858 - 236 pages
...when it is less than a semicircle. PROBLEM X. TO FIND THE AREA OF A CIRCLE, THE DIAMETER BEING GIVEN. RULE. — Multiply half the circumference by half the diameter, and the product will be the area. PROBLEM XI. TO FIND THE AREA OF A SECTOR OF A CIRCLE. RULE. — Multiply the radius,... | |
 | Francis Walkingame - 1865 - 222 pages
...circumference will be 3-14159265, &c., of which 3-1416 is sufficient in most cases. Then the rule teaches to multiply half the circumference by half the diameter, and the product is the area: that is, multiply 1-5708 by -5 (viz., half 3-1410 by half 1), and the product is -7854, which is the... | |
 | A. C. Smeaton - Building - 1867 - 314 pages
...PROBLEM VIII. To find the Area of a Circle. 1. When the diameter and circumference are both given. RULE. Multiply half the circumference by half the diameter, and the product will be the area. 2. When the diameter is given. RULE. Multiply the square of the diameter by .7854,... | |
 | Frederick Thomas Hodgson - Carpentry - 1883 - 168 pages
...the diameter. To Find the Area of a Circle. 1. When the diameter and circumference are both given. Rule. — Multiply half the circumference by half the diameter, and the product will be the area. 2. When the diameter is given. Rule. — Multiply the square of the diameter by 7854,... | |
 | Stoddard A. Felter - Arithmetic - 1868 - 368 pages
...feet; what is the diameter? • Operation.— 25.1328-f-3.1416=8 ft., diameter. Prob. VIII. — The diameter of a circle being given, to find the area. Rule. — Multiply the square of the diameter by .7854. ILLUSTRATION. — The diameter of a circle is 9 inches ; what... | |
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Rule. — Multiply half the circumference by half the diameter, and the product will be the area. 
