 | John Love - 1720 - 290 Seiten
...gives 6 Chains 28 Links, and fomething more for the Circumference. Now, to know the Superficial Content multiply half the Circumference by half the Diameter, the Product will be the Content: Half the Circumference is 3 Chains 14 Links: half the Diameter I Chain oo Links ; which multiplied... | |
 | John Love - 1760 - 288 Seiten
...6 Chains; 28 Links, and fomething more fdr the Circumference. Now, to know the fuperficial Content, multiply half the Circumference by half the Diameter, the Product will be the Content; Half the Circumference is 3 Chains, 14 Links ; half the Diameter, I Chain, o Links : which... | |
 | John Love - 1792 - 288 Seiten
...6 chains, 28 links, and fomething more for the circumference. Now to know the fuperficial content, multiply half the circumference by half the diameter, the product will be the content ; half the circumference is 3 chains, 14 links; half the diameter, I chain, o links; which... | |
 | Caleb Alexander - 1813 - 152 Seiten
...gives б eh. 28 lin and something more for the circumferenc^ Now, to. find the superficial content, multiply half the circumference by half the diameter, the product will be the content ¡ hull" the circumference is 3 ch. 14 lin. -, half the .diameter, 1 ch. 00 lin : which, multiplied... | |
 | William Templeton (engineer.) - 1833 - 224 Seiten
...area of a triangle whose base is equal to the circumference, and perpendicular equal to the radius. 6. — The area of a circle is to the square of the...Multiply half the circumference by half the diameter, and the product will be the area. EXAMPLE 1. — Required the area of a circle, the diameter being... | |
 | Robert Simson (master of Colebrooke house acad, Islington.) - 1838 - 206 Seiten
...circle is 62.832, what is the diameter? 62.832 -f- 3.1416 = 20, or 62.832 X .31831 = 20 as before. The area of a circle is to the square of the diameter...multiply half the circumference by half the diameter and the product will be the area. Required the area of a circle the diameter being 30.5. 30.52 X'7854... | |
 | William Templeton (engineer.) - 1846 - 236 Seiten
...to the radius. 5. — The area of a circle is equal to the rectangle of its radius, and a right line equal to half its circumference. 6. — The area of...multiply half the circumference by half the diameter, and the product will be the area. EXAMPLE 1. — Required the area of a circle, the diameter being... | |
 | James Robinson (of Boston.) - 1847 - 304 Seiten
...diameter? Al*t. 2O3. To FIND THE AREA OF A CIRCLE WHEN THE DIAMETER AND CIRCUMFERENCE ARE GIVEN. RULE. Multiply half the circumference by half the diameter ; the product will be the area. 1. What is the area of a circle whose circumference is 37.6992 feet, and diameter 12 feet? 37.6992-=-2... | |
 | William Templeton (engineer.) - 1848 - 256 Seiten
...the radius. 5 . — The area of a circle is equal to the rectangle of its radius, and a right line equal to half its circumference. 6. — The area of a circle is found by squaring the diameter, and multiplying by the decimal .7854 ; or by multiplying the circumference... | |
 | Charles Haslett - 1855 - 544 Seiten
...equal to the radius. 5. The area of a circle is equal to the rectangle of its radius, and a right line equal to half its circumference. 6. The area of a circle is found by squaring the diameter, ant multiplying by the decimal '7854; or by multiplying the circum... | |
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... is double that of another, contains four times the area of the other. 4. — The area of a circle is equal to the area of a triangle whose base is equal to the circumference, and perpendicular equal to the radius. 5. — The area of a circle is equal... 
