| Thomas Hodson - Arithmetic - 1806 - 502 pages
...circumference 85000 miles ? Anfwer 7957! miles. PROBLEM XXXVIII. TO FIND THE AREA OF A CIRCLE. RULE i. Multiply half the circumference by half the diameter, and the product is the area. RULE 2. Multiply the fquare of the diameter by .7854. RULE 3. Multiply the fquare of the circumference by .07958.... | |
| Charles Hutton - Mathematics - 1807 - 464 pages
...Ans."4>'7115. Ex. 2. To find the length of an arc of 12° 10', or 12"£, the radius being 10 feet; Ans. 2'1231. PROBLEM IX. To find the Area of a Circle^. RULE 1....MULTIPLY half the circumference by half the diameter. Or multiply the whole circumference by the whole diameter, and take •£ of the product.' RULE the... | |
| Charles Vyse - Arithmetic - 1815 - 340 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... | |
| William Jillard Hort - 1822 - 308 pages
...perpendicular let fall upon it from the opposite angle, and half the product will be the area. Problem. To find the area of a circle. Rule 1. Multiply half...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... | |
| Anthony Nesbit - Surveying - 1824 - 476 pages
...pow.-r, 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... | |
| Peter Nicholson - Mathematics - 1825 - 1046 pages
...height 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... | |
| Thomas Hornby (land surveyor.) - Surveying - 1827 - 318 pages
...8'94427x 8~ 16- 18.51805 chains, the length of 3 PROBLEM 9. To find the Area of a Circle. RULE 1st. Multiply half the Circumference by half the Diameter, and the product is the area. RULE 2. Multiply the square of the diameter by .7854 and the product is the area. EXAMPLE. Required the area... | |
| James L. Connolly (mathematician.) - Arithmetic - 1829 - 266 pages
...the proportion of Van Culen, if the diameter be I , the circumference .will be 3,1415926, &e. Then multiply half the circumference by half the diameter, and the product is the area. EXAMPLES. Problem 1. Having the diameter and circumference to find the area. RULE. Every circle is... | |
| Dudley Leavitt - Mathematics - 1830 - 154 pages
...square rods, and 750-H60=4 acres and 1 1 square rods. Ans. 4. To find the content of a circle. Rule. Multiply half the circumference by half the diameter, and the product is the content. f Exam. If the circumference of a circle is 40 rods, and the diameter 12.7 rods, what is its... | |
| Edinburgh encyclopaedia - 1830 - 856 pages
...NOTE. For another approximate value to an arc of a circle, see GEOMETRY, Prop. H. Sect. 5. Part 1. PROBLEM IX. To find- the area of a circle. RULE 1. Multiply the radius by half the circumfe» rence, the product is- the area. >f enturalion of Plane Figure!.... | |
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