 | Nicolas Pike - 1808 - 470 páginas
...190986 63662 95493 103-76906= diameter. This proportion is the most accurate. ART. 14. TofadtheAremof a Circle. RULE. — Multiply half the diameter by half the circumference and the produdt is the area. If For further proof of thi> matter: If a cylindrical pint, beer meafure, wbofe... | |
 | Nathan Daboll - 1815 - 250 páginas
...: 22 : : 35 : 110 ,9ns. — and inversely as 22 : 7 : : 110 : 35, the diameter, &e. ART. 5. To fiud the area of a Circle. RULE. Multiply half the diameter by half the circumference^ and the product is the area ; or if the diameter is given without the circumference, multiply the square of... | |
 | Nathan Daboll - 1817 - 252 páginas
...Aa7 : 22 : : 35 : 110 .4ns. — and inversely ••1322 : 7 : : 110 : 35, the diameter, &c. ART. 5. To find the area of, a Circle. • RULE. * Multiply...half the diameter by half the circumference, and the product is the area; or if the diameter is given without the circumference, multiply the square of... | |
 | Nathan Daboll - 1818 - 246 páginas
...— As 7 : 22 : : 35 : l10^ns. — and inversely as. 22 : 7 : : 110 : 35, the diameter, &c. ART. 5. To find the area of a Circle. ' RULE. Multiply half the diameter by half the circumference, and the product is the area ; or if the diameter is given • •without the circumference, multiply the square... | |
 | Nicolas Pike - 1822 - 562 páginas
...326 326 • 160986 (53662 95493 103-76906 = diameter. This proportion it the most accurate. ART. 14. To find the Area of a Circle, RULE. Multiply half the diameter by half the circumference and the product is the area. If the diameter be given, find the circumference by Art. 12. If the circumference... | |
 | Nicolas Pike - 1832 - 538 páginas
...ameter. This 160 proportion i« 154 1338 the most accu1065 rate. 60 44 2730 — 2485 16 245 ART. 14. To find the Area of a Circle. RULE. Multiply half the diameter by half the circumference and (he product is the area. If the diameter be given, find the circumference by Art. 12. If the circumference... | |
 | Nathan Daboll - 1828 - 268 páginas
...As 7 i 22 : : 3o : i10.fe. — and inversely as 22 : 7 : : 110 : 35, the diameter, &c,.':, ART. 5. To find the area of a Circle. RULE. Multiply half the diameter by half the circumferencel and the product is the a^ea ; or if the diameter is tiiven without the ciicuinference,... | |
 | Daniel Parker - 1828 - 358 páginas
...50,2/556 : Then 50,2656x113-^355=16+ Ans. Or 1 : ,31831 : : 50,2656 : Then 50,2656 X, 31831=16+ Dia. 3. To find the area of a circle Rule. — Multiply half the diameter by half of the circumference, and the product will be the area : or, multiply the square of the diameter by... | |
 | William Kinne - 1829 - 246 páginas
...diameter of a circle whose circumference is 110 rods? As 22 : 7 : : 110 : 35 rods the diam. ABSPROBLEM 11. To find the area of a circle. RULE. — Multiply half the diameter by half the circumference, and the product is 1(19 area ; or, if the diameter alone is given, multiply the square of the diameter by ,785398... | |
 | Nathan Daboll - 1829 - 268 páginas
...— As 7 : 22 : : 35 : 110 Ana. — and inversely as yz : 7 : : 110 : 35, the diameter, &c. ART. 5. To find the area of a Circle. RULE. Multiply half the diameter by half the circumference, und the product is the area ; or if the diameter is given •without the circumference, multiply the... | |
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To find the area of a Circle. ' RULE. Multiply half the diameter by half the circumference, and the product is the area... 
