 | John Henry Robson - 1880 - 116 pages
...problem is the sth Proposition of Euclid's Fourth Book. 12. By Euc. I. 32, Cor. 1, it is proved that " All the Interior angles of any Rectilineal figure,..."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 . -.... | |
 | Oxford univ, local exams - 1880 - 396 pages
...explain what you mean by a straight line touching a circle, and by a straight line placed in a circle. 2. All the interior angles of any rectilineal figure,...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... | |
 | Isaac Todhunter - Euclid's Elements - 1880 - 426 pages
...together equal to two right angles. [Axiom 1. Wherefore, if a side of any triangle &c. <JE». COROLLARY 1. All the interior angles of any rectilineal figure,...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,... | |
 | Elizabethan club - 1880 - 156 pages
...of the acute angles at the other point, and similarly of the obtuse angles. 3. All the angles of a rectilineal figure, together with four right angles,...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... | |
 | Sir Norman Lockyer - Electronic journals - 1880 - 668 pages
...the flat angle we may take 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... | |
 | Elias Loomis - Conic sections - 1880 - 452 pages
...is the complement of the other. PROPOSITION XXVIII. THEOREM. All the interior angles of a polygon, together with four right angles, are equal to twice as many right angles as the figure IMS sides. Let ABCDE be any polygon ; then all its interior angles A, B, C, D, E, together with four... | |
 | 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... | |
 | 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. "... | |
 | Thomas Newton Andrews - Geometry - 1881 - 168 pages
...setting off the angles at the base found thus: — In Euclid, Book I., Prop, xxxi11., it is proved that "All the interior angles of any rectilineal figure,...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... | |
 | John Gibson - 1881 - 302 pages
...EFD of the triangle DEF is subtended by the greater side, or has the greater side opposite to it. 3. All the interior angles of any rectilineal figure,...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... | |
| |
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. 
