 | Oxford univ, local exams - 1880 - 396 sider
...in a circle. 2. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. 3. If the square described on one side of a triangle be equal to the squares described on the other... | |
 | John Henry Robson - 1880 - 116 sider
...proved that " All the Interior angles of any Rectilineal figure, "together with four right angles, are equal to "twice as many right angles as the figure has " sides." If, therefore, we suppose the polygon to have n sides, All its interior angles + 4.90 .= 272.90 . -.... | |
 | Elizabethan club - 1880 - 156 sider
...similarly of the obtuse angles. 3. All the angles of a rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. A floor has to be laid with tiles in the form of regular figures all equal and similar ; show what... | |
 | William Frothingham Bradbury - 1880 - 260 sider
...minus two. Let ABCDEF be the given polygon ; the sum of all the interior angles A, B, C, D, E, F, is equal to twice as many right angles as the figure has sides minus two. For if from any vertex A, diagonals AC, AD, AE, are drawn, the polygon will be divided into... | |
 | William Mitchell Gillespie - 1880 - 540 sider
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
 | Isaac Todhunter - 1880 - 426 sider
...<JE». COROLLARY 1. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has side*. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides,... | |
 | Thomas Newton Andrews - 1881 - 168 sider
...is proved that "All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides." If we have to describe a pentagon on the base AB, we must first calculate the angles at the base. Thus... | |
 | Thomas Holloway (surveyor.) - 1881 - 132 sider
...sixty degrees. 3. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. Although further systems of proof could easily be quoted, I consider the foregoing quite sufficient... | |
 | John Gibson - 1881 - 302 sider
...opposite to it. 3. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. 4. Describe a parallelogram that shall be equal to a given triangle BCD, and have one of its angles... | |
 | Marianne Nops - 1882 - 278 sider
...13). Therefore all the pairs of angles, ie all the interior and all the exterior angles of the figure, are equal to twice as many right angles as the figure has sides. But all the interior angles + four right angles = twice as many right angles as the figure has sides.... | |
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All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. 
