 | Oxford univ, local exams - 1880
...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
...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 . -.... | |
 | Sir Norman Lockyer - 1880
...Theorem XXVI. of the syllabus, that the interior angles of any polygon, together with four right angles, are equal to twice as many right angles as the figure has sides. In the new notation we would say that the sum of the interior angles of the polygon is equal to a number... | |
 | William Mitchell Gillespie - 1880 - 524 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. "... | |
 | William Frothingham Bradbury - 1880 - 240 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... | |
 | Isaac Todhunter - 1880 - 400 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 - 160 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
...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
...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
...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. 
