| John Henry Robson - 1880
...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
...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 - 400 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
...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
...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 - Geometry - 1880 - 443 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 - 240 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 - 524 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 - 160 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
...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... | |
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