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... multiply the square of the diameter by ,7854 and the product will be- the area.
The complete measurer: or, The whole art of measuring, containing the ... - Page 49
by Thomas Keith - 1817

## The Tutor's Guide: Being a Complete System of Arithmetic; with Various ...

Charles Vyse - Arithmetic - 1785 - 325 pages
...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. Multiply the Square of the Circumference by ,079574, and the Product will be the Area. Or,...

## A Treatise on Mensuration, Both in Theory and Practice

Charles Hutton - Measurement - 1788 - 728 pages
...the area CAE of the whole circle or feftor accordingly. ^_. £. D. * RULE II. « Multiply the fquare of the diameter by '7854, and the product will be the area* EXAMPLES. i, What is the area of a circle whofe diameter is I, and circumference 3'i4i6? Vl4-l6 XI...

## The Tutor's Guide: Being a Complete System of Arithmetic; with Various ...

Charles Vyse - Arithmetic - 1806 - 342 pages
...RULES. 1. Multiply Half the Circumference by Half the Diameter, and tin- Product will 1& the Area. Or, 2. Multiply the Square of the Diameter by ,7854, and the Product will be the Area. Or, 3. Multiply the Square of the Circumference by ,079574, and the Product will be the Area. Or, 4....

## Daboll's Schoolmaster's Assistant: Improved and Enlarged. Being a Plain ...

Nathan Daboll - Arithmetic - 1815 - 240 pages
...circumference^ and the product is the area ; or if the diameter is given without the circumference, multiply the square of the diameter by ,7854 and the product will be the area. EXAMPLE8. 1. Required the area of a circle whose diameter 'is 12 inches, and circumference 37,7 inches....

## Daboll's Schoolmaster's Assistant: Improved and Enlarged. Being a Plain ...

Nathan Daboll - Arithmetic - 1818 - 240 pages
...circumference, and the product is the area ; or if the diameter is given • •without the circumference, multiply the square of the diameter by ,7854 and the product will be- the area. EXAMPLES. 1. Required the area of a circle whose diameter is 12 inches, and circumference 37,7 inches....

## Hawney's Complete Measurer: Or, The Whole Art of Measuring: Being a Plain ...

William Hawney - Geometry - 1820 - 312 pages
...circumference by half the diameter : Or take -J of the product of the whole circumference and diameter,.and it will give the area. 2. Multiply the square of the...area. 3. As 452 is to 355, so is the square of the diameter to the area. 4. As 14 is to 11, so is the square of the diameter to the area. 5. Multiply...

## A general view of the sciences and arts, Volume 1

William Jillard Hort - 1822 - 308 pages
...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 find the area of an ellipse. Multiply the product of the two axes, by the number 7854. for...

## A New and Complete System of Arithmetick: Composed for the Use of the ...

Nicolas Pike - Arithmetic - 1822 - 560 pages
...1 5. The Diameter being given to find the Area of a Circle without finding the Circumference. RULE. Multiply the square of the diameter by -7854,* and the product will be the area of the circle, whose diameter was given. EXAMP. The diameter of a circle being 12, to find the area...

## Dictionary of the Mathematical and Physical Sciences, According to the ...

James Mitchell - Mathematics - 1823 - 576 pages
...Diameter being given. 1. Multiply the diameter by 3-14159, and the product will be the circumference. 2. Multiply the square of the diameter by -7854, and the product •will be the area. Or general, if we put diameter = D, circumference = C, area — A, and 3-141S9 = 1', we have the following...

## A Treatise on Practical Mensuration in Eight Parts ...

Anthony Nesbit - Surveying - 1824 - 476 pages
...area. Or, divide the product of the whole circumference and diameter by 4t9 and the quotient will be the area. 2. Multiply the square of the diameter by .7854, and tlie product will be the area. 3. Multiply the square of the circumference by .07958, and the product...