Product Code Database
This historic book may have numerous typos and missing text
Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1879 edition. Excerpt: ... MATHEMATICAL DEMONSTRATION OF SQUARING THE CIRCLE. RULES i. Eleven-fourteenths (PS) of the diameter equals onefourth (i) of the circumference. 2. To find the area of the circle. Multiply the diameter by the radius, and divide the product by 7; the quotient multiplied by ii gives the area of the circle. EXAMPLE: Diameter 28X i4=392-=-7=56x ii=6i6, area of circle. 3. To find the side of a square the area of which shall be equal to the area of the circle. Divide the diameter by i4, multiply the quotient by n, add to the product one-tenth (jij) of the diameter, and annex the first figure in the quotient. EXAMPLE: Diameter 28--i4=2. Product 2xii=22. one-fourth (j) of the circum. 2.82; one-tenth (j1,) of the diameter Side of square ) with quotient annexed..equal in area to 24.82 area of circle. ) Proof: 28X22=6i6; the square root of which is 24.8i93. Add the quotient when it consists of two or more figures. EXAMPLE: Diameter 224-4-i4=i6 Product i6x u= i76 One-fourth () of the circum. 22.4 One-tenth d''T) of the diam. Side of square ) i6 Quotient added. equal in area to area of circle. ) i98.56 Proof: 224X176=39424 the square root of which is 198.5547 4. Eleven-fourteenths (?) of the area of the circle equals the area of a square whose sides are equal to the circumference. 5. Seven-elevenths (?) of the area of the circle equals the area of an inscribed square. 6. One-fourth () of the circumference multiplied by nine (9), the product divided by ten (10), equals the side of an inscribed square, nearly. 7. To find the diameter when the circumference is given, Multiply by seven (7) and divide by 22. 8. To find the diameter when the area of the circle is given. Divide by fourteen (14), and multiply the quotient by eleven (1 1); the square...
|
