 | Cambridge univ, exam. papers - 1856 - 200 pages
...Prove that all the internal angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides; and that all the external angles are together equal to four right angles. In what sense are these propositions... | |
 | Frederick Walter Simms - Leveling - 1856 - 258 pages
...together all the internal angles, marked by dotted segments of circles; and subtract their sum from twice as many right angles as the figure has sides, less four, for the angle db e. Example. — Let the angles denoted by the dotted segments at the different letters... | |
 | Euclides - 1856 - 168 pages
...with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. XVI. If two triangles have two sides of the one equal to two sides of the other, each to each, and... | |
 | Adrien Marie Legendre - Geometry - 1857 - 444 pages
...are triangles in the figure ; that is, as many times as there are sides, less two. But this product is equal to twice as many right angles as the figure has sides, less four right angles. Cor. 1. The sum of the interior angles in a quadrilateral is equal to two right angles multiplied by... | |
 | William Mitchell Gillespie - Surveying - 1857 - 538 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. " Calculate... | |
 | W. Davis Haskoll - Civil engineering - 1858 - 422 pages
...and in an irregular polygon they may be all unequal. The interior angles of a polygon are together equal to twice as many right angles as the figure has sides, less four. On this is based the theory of the traverse, of which further explanation will be given in another... | |
 | Elias Loomis - Conic sections - 1858 - 256 pages
...that is, together with four right angles (Prop. V., Cor. 2). Therefore the angles of the polygon are equal to twice as many right angles as the figure has sides, wanting four right angles. Cor. 1. The sum of the angles of a quadrilateral is four right angles ;... | |
 | Horatio Nelson Robinson - Geometry - 1860 - 472 pages
...triangles is equal to two right angles, (Th. 11) ; and the sum of the angles of all the triangles must be equal to twice as many right angles as the figure has sides. But the sum of these angles contains the sum of four right angles about the point p ; taking these... | |
 | Sir Norman Lockyer - Electronic journals - 1880 - 668 pages
...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 - 394 pages
...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... | |
| |
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. 
