 | Webster Wells - Geometry - 1898 - 264 pages
...of the A of any A is equal to two rt. AJ (§ 84) 127. Cor. I. TJie sum of the angles of any polygon is equal to twice as many right angles as the polygon has sides, less four right angles. "For if R represents a rt. Z., and n the number of sides of a polygon, the... | |
 | Webster Wells - Geometry - 1899 - 424 pages
...of the A of any A is equal to two rt. A] (§ 84) 127. Cor. I. The sum of the angles of any polygon is equal to twice as many right angles as the polygon has sides, less four right angles. For if R represents a rt. Z, and n the number of sides of a polygon, the sum... | |
 | William James Milne - Geometry - 1899 - 398 pages
...of sides. To prove ABCDE and FGHJK similar. Proof. By § 166, the sum of the angles of each polygon is equal to twice as many right angles as the polygon has sides less two. Since, § 374, each polygon is equiangular, and since each contains the same number of angles... | |
 | William James Milne - Geometry, Modern - 1899 - 258 pages
...of sides. To prove ABCDE and FGHJK similar. Proof. By § 166, the sum of the angles of each polygon is equal to twice as many right angles as the polygon has sides less two. Since, § 374, each polygon is equiangular, and since each contains the same number of angles... | |
 | Webster Wells - Geometry - 1899 - 450 pages
...the sum of the A of the polygon is n — 2 times 127. Cor. I. The sum of the angles of any polygon is equal to twice as many right angles as the polygon has sides, less four right angles. For if R represents a rt. Z, and n the number of sides of a polygon, the sum... | |
 | William James Milne - Geometry - 1899 - 404 pages
...polygon of any number (n) of sides, as ABCDE. Ef To prove the sum of the angles, A, B, C, D, and E equal to twice as many right angles as the polygon has sides less two. Proof. From any vertex, as J,draw the diagonals, JCand AD. The number of triangles thus formed... | |
 | International Correspondence Schools - Mining engineering - 1900 - 728 pages
...regular polygon. If not, it is an irregular polygon. The sum of the interior angles of any polygon is equal to twice as many right angles as the polygon has sides, less four right angles. To Find the Area of Any Regular Polygon.— Square one of its sides and multiply... | |
 | Science - 1900 - 870 pages
...necessary to apply the well-known principle of geometry that the turn of UK interior angle* n_f a polygon is equal to twice as many right angles as the polygon has sides, less four right angle*. Thia applies to figures having any number of sides, without regard to whether... | |
 | Edward Brooks - Geometry, Modern - 1901 - 278 pages
...etc. Proof.— First, the poly- B_ gons are mutually equiangular. For every angle in either polygon is equal to twice as many right angles as the polygon has sides, less four right angles,, divided by the number of sides. I. 36, 2. And since the number of sides in... | |
 | Thomas Franklin Holgate - Geometry - 1901 - 462 pages
...Hence all the angles of all the triangles are together equal to two right angles for each side, or to twice as many right angles as the polygon has sides. But these angles include the interior angles of the polygon, together with four right angles about the... | |
| |
... therefore the sum of the angles of all the triangles is equal to twice as many right angles as the figure has sides. But the sum of all the angles... 
