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" Multiply half the circumference by half the diameter, and the product will be the area. Or, divide the product of the whole circumference and diameter -by 4, and the quotient will be the area. 2. Multiply the square of the diameter by .7854, and the product... "
An essay on mechanical geometry, explanatory of a set of models - Page 67
by Benjamin Donne - 1796
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The Tutor's Guide: Being a Complete System of Arithmetic; with Various ...

Charles Vyse - Arithmetic - 1785 - 325 pages
...its Diameter is 8000 Miles ? PROBLEM VII. To find the Area of a Circle. RULES. 1 . Multiply Half the the Circumference by Half the Diameter, and the Product will be the Area. Or, 2. Multiply the Square of the Diameter by ,78,54, and' the Product will be the Area. Or, • 3....
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The principles of architecture, Volume 1

Peter Nicholson - 1809 - 426 pages
...the greater segment. PROBLEM XVI. To find the area of a circle, the diameter being given. METHOD I. Multiply half the circumference by half the diameter, and the product will be the area. i EXAMPLE I. What u the area nf a circle lahose diameter is 28 feet, and its circumference 88 feet?...
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The Tutor's Guide: Being a Complete System of Arithmetic; with Various ...

Charles Vyse - Arithmetic - 1815 - 320 pages
...perfectly round, and its diameter be 8000 miles? PROBLEM' VIL To find the area of a circle. RULES. ' 1. Multiply half the circumference by half the diameter, and the product will he the area. Or, 2. Multiply the square of the diameter by ,7854, and the product will be the area....
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The Scholar's Arithmetic: Designed for the Use of Schools in the United States

Jacob Willetts - Arithmetic - 1822 - 191 pages
...circumference to the diameter ; Or; as 3. 1416 is to 1, so is the circumference to the diameter. III. .Multiply half the circumference by half the diameter and the product will be the area. Or multiply the square of the diameter by. 7854, and the pro* duct will be the area. Or multiply the...
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A general view of the sciences and arts, Volume 1

William Jillard Hort - 1822 - 308 pages
...opposite angle, and half the product will be the area. Problem. To find the area of a circle. Rule 1. Multiply half the circumference by half the diameter, and the product will be the area. Rule 2. Multiply the square of the diameter by 7854', and the product will be the area. ProUem. To...
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A New and Complete System of Arithmetick: Composed for the Use of the ...

Nicolas Pike - Arithmetic - 1822 - 560 pages
...of the arch line ABC, and by Art. 24, the diameter FB ; then multiply half the chord of the arch ABC by half the diameter, and the product will be the area of the sector ABCE : then find the area of the triangle AEC, wboae base AC is 172, and perpendicular height...
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A Treatise on Practical Mensuration in Eight Parts ...

Anthony Nesbit - Surveying - 1824 - 476 pages
...may consult Marrat's Mechanics. Article 768. PROBLEM XV. To find the area of a circle. RULES. _ , 1. Multiply half the circumference by half the diameter, and the product will be the area. Or, divide the product of the whole circumference and diameter by 4t9 and the quotient will be the...
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A Popular Course of Pure and Mixed Mathematics ...: With Tables of ...

Peter Nicholson - Mathematics - 1825 - 372 pages
...of the greater segment. Prob. 1 6. To find the area of a circle, the diameter being given. Method 1. Multiply half the circumference by half the diameter, and the product will be the area. Ex. l. What is the area of a circle whose diameter is 28 feet, and iU circumference 88 feet ? Method...
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The Youth's Assistant in Theoretick and Practical Arithmetic

Zadock Thompson - Arithmetic - 1826 - 164 pages
...miles, what is its diameter ? Ans. 79571 nearly. Problem VI. To find the area of a circkj RULE. — Multiply half the circumference by half the diameter, and . the product will be the area. * These three methods do not exactly agree, but the last is the most correct. The exact proportion...
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The Juvenile Arithmetick, and Scholar's Guide: Wherein Theory and Practice ...

Martin Ruter - Arithmetic - 1828 - 180 pages
...diameter to the circumference. ARTICLE V. To find the superficial contents, or area, of a circle. RULE. Multiply half the circumference by half the diameter, and the product will be the answer. Or, multiply the square of the diameter by .7854; or multiply the square of the circumference...
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