 | John Henry Robson - 1880 - 116 pages
...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 - Electronic journals - 1880 - 668 pages
...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... | |
 | Oxford univ, local exams - 1880 - 396 pages
...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... | |
 | Elizabethan club - 1880 - 156 pages
...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... | |
 | Isaac Todhunter - Euclid's Elements - 1880 - 426 pages
...from a point F within the figure to each of its angles. And by the preceding proposition, n all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as the figure has sides. And the same angles are equal to the interior... | |
 | William Mitchell Gillespie - Surveying - 1880 - 540 pages
...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 - Geometry - 1880 - 260 pages
...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... | |
 | Thomas Holloway (surveyor.) - 1881 - 132 pages
...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 pages
...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... | |
 | Thomas Newton Andrews - Geometry - 1881 - 168 pages
...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... | |
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... figure, together with four right angles, are equal to twice as many right angles as the figure has be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. 
