| Benjamin Peirce - Geometry - 1847 - 204 pages
...surface generated by the circular segment ABCD is a zone. Hence the area of the surface of a sphere is **the product of its diameter by the circumference of a great circle.** And, the area of a zone is the product of its altitude by the circumference of a great circle. 474.... | |
| Jeremiah Day - Logarithms - 1848 - 354 pages
...surface of a hemisphere is equal to *the product of its radius into the circumference of its base; and **the surface of a sphere is equal to the product of its diameter** into its circumference. Cor. 1. From this demonstration it follows, that the surface of any segment... | |
| Charles Davies - Trigonometry - 1849 - 372 pages
...described by ED + DC, is equal to (HI +1P) x circ. ON, or equal to HP x circ., ON. PROPOSITION X. THEOREM. **The surface of a sphere is equal to the product of...diameter by the circumference of a great circle. Let** ABCDE be a semicircle. Inscribe in it any regular semi-polygon, and from the centre O draw OF perpendicular... | |
| Elias Loomis - Conic sections - 1849 - 252 pages
...base of the frustum, its upper base, and a mean proportional between them. PROPOSITION VII. THEOREM. **The surface of a sphere is equal to the product of its diameter** b\j the circumference of a great circle. Let ABDF be the semicircle by the revolution of which the... | |
| Charles Davies - Geometry - 1850 - 218 pages
...by ED+DC, is equal to (HI+IP)x circumference OJV, or equal to HPx circumference ON. THEOREM XXIII. **The surface of a sphere is equal to the product of...diameter by the circumference of a great circle. Let** ABODE be a semicircle. Inscribe in it any regular semi-polygon, and from the centre 0 draw OF perpendicular... | |
| Charles Davies - Geometry - 1850 - 238 pages
...by ED+DC, is equal to (HI+IP)x circumference ON, or equal to UP x circumference ON. THEOREM XXIII. **The surface of a sphere is equal to the product of...diameter by the circumference of a great circle. Let** ABCDE be a semicircle. Inscribe in it any regular semi-polygon, and from the centre O draw OF perpendicular... | |
| Jeremiah Day - Geometry - 1851 - 418 pages
...surface of a hemisphere is equal to the product of its radius into the circumference of its base ; and **the surface of a sphere is equal to the product of its diameter** into its circumference. Cor. 1. From this demonstration it follows, that the surface of any segment... | |
| Adrien Marie Legendre - Geometry - 1852 - 436 pages
...described by ED-\_DC, is equal to (HI '+ IP) x circ. ON, or equal to HP x circ. ON. PROPOSITION X. THEOEEM. **The surface of a sphere is equal to the product of...diameter by the circumference of a great circle. Let** ABCDE be a semicircle. Inscribe in it a regular semi-polygon, and from the centre 0 draw OF perpendicular... | |
| Charles Davies - Geometry - 1886 - 340 pages
...(HI+IP)x circumference ON, or equal to HPx circumference ONTHEOREM XXIIIThe surface of a sphere 1s **equal to the product of its diameter by the circumference of a great** circleLet ABCDE be a semicircle- Inscribe in it any regular semi-polygon, and from the centre O draw... | |
| Horatio Nelson Robinson - History - 1853 - 334 pages
...the application of the preceding rules. From theorem 16, book VII, Geometry, we learn that The convex **surface of a sphere is equal to the product of its diameter** into its circumference. The surface of a segment is equal to the circumference of the sphere, multiplied... | |
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