 | John McGregor (teacher of mathematics.) - 1792 - 532 Seiten
...milftone, whofe circumference js 22 feet. 4"f- 7.0028. PROBLEM XTVV У he diameter and • circumference of a circle being given , to find, the area. RULE. ! Multiply half the radius into the circumference, and thc'prois the area. EXAMPLE L. Required the area of a circle whofe... | |
 | James Thompson - 1805 - 100 Seiten
...?. fquare equal in area, to a rhombus whofc area is 358 ? 1/358=1 8.9108279 Anfwer. PROBLEM III. The diameter of a circle being given to find the area. RULE. Multiply the fquare of the diameter by .785398* or .7854, and the product is the area. EXAMPLE. Required the... | |
 | Thomas Hodson - 1806 - 502 Seiten
...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... | |
 | Zadock Thompson - 1826 - 176 Seiten
...be 25000 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... | |
 | Thomas Hornby (land surveyor.) - 1827 - 318 Seiten
...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... | |
 | Martin Ruter - 1828 - 180 Seiten
...the 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... | |
 | James L. Connolly (mathematician.) - 1829 - 266 Seiten
...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 - 1830 - 154 Seiten
...25x30=750 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... | |
 | Francis Walkingame - 1832 - 224 Seiten
...length is equal to half the circumference, and half the breadth equal to half the diameter ; therefore multiply half the circumference by half the diameter, and the product is the area of the circle. Thus, if the diameter of a circle, that is, the line drawn across the circle through... | |
 | James L. Connolly (mathematician.) - 1835 - 264 Seiten
...whose length is equal to half the circumference, and the breadth equal to half the diameter: therefore multiply half the circumference by half the diameter, and the product is the area of the circle. 1. Let the diameter of a circle be 22,6 and the circumference 71, what is the area of... | |
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Rule. — Multiply half the circumference by half the diameter, and the product will be the area. 
