 | 1856 - 422 páginas
...the figure has sides, wanting two. It is plain, therefore, that it will then contain only tti'i'i* as many right angles as the figure has sides, wanting four right angles. From this corollary another may be drawn regarding regular polygons. Since all the sides and angles... | |
 | William Mitchell Gillespie - 1856 - 478 páginas
...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. "... | |
 | Cambridge univ, exam. papers - 1856 - 200 páginas
...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... | |
 | Euclides - 1856 - 168 páginas
...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... | |
 | William Mitchell Gillespie - 1857 - 538 páginas
...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. "... | |
 | Adrien Marie Legendre, Charles Davies - 1857 - 442 páginas
...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... | |
 | Elias Loomis - 1858 - 256 páginas
...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 ; of a pentagon, six right angles... | |
 | W. Davis Haskoll - 1858 - 422 páginas
...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... | |
 | Isaac Sharpless - 1879 - 282 páginas
...But ACD+ACB = 2R; BAC+ABC+ACB = 2R. Corollary 1.—All the interior angles of a polygon are together equal to' twice as many right angles as the figure has sides, minus four right angles. Let ABODE be a polygon, and let n represent the number of its sides. Draw... | |
 | W J. Dickinson - 1879 - 44 páginas
...produced to meet, the angles formed by these lines, together with eight right angles, are together equal to twice as many right angles as the figure has sides. Same proposition. ABC is a triangle right-angled at A, and the angle B is double of the angle C. Show... | |
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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. 
