An Elementary Treatise on Plane and Solid Geometry |
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Page vi
... extremities of a line ( 42 ) , CHAPTER VII . Sides and Angles of Polygons , . Plane figure , polygon , and its perimeter ( 43 ) ; triangle , quadrilateral , pentagon , hexagon ( 44 ) ; equilateral , isosce- les , and scalene triangle ...
... extremities of a line ( 42 ) , CHAPTER VII . Sides and Angles of Polygons , . Plane figure , polygon , and its perimeter ( 43 ) ; triangle , quadrilateral , pentagon , hexagon ( 44 ) ; equilateral , isosce- les , and scalene triangle ...
Page 3
... , the limits of surfaces are lines , and the extremities of lines are points . The Point , then , on account of its simplicity , deserves our first consideration . The Position of a Point ; its Direction and Distance.
... , the limits of surfaces are lines , and the extremities of lines are points . The Point , then , on account of its simplicity , deserves our first consideration . The Position of a Point ; its Direction and Distance.
Page 12
... extremities of the line AB , is greater than that of two other lines DA and DB , similarly drawn , but included by them . Demonstration . Produce DA to E. We have , by art 17 , and AC + CE > AD + DE , DE + BE > DB . Oblique Lines ...
... extremities of the line AB , is greater than that of two other lines DA and DB , similarly drawn , but included by them . Demonstration . Produce DA to E. We have , by art 17 , and AC + CE > AD + DE , DE + BE > DB . Oblique Lines ...
Page 13
... extremities of the line AB . 2. Any point without the perpendicular , as F , is at unequal distances from the same extremities A and B. Demonstration . 1. The distances EA and EB are equal , since they are oblique lines drawn at equal ...
... extremities of the line AB . 2. Any point without the perpendicular , as F , is at unequal distances from the same extremities A and B. Demonstration . 1. The distances EA and EB are equal , since they are oblique lines drawn at equal ...
Page 16
... extremities are the same points . The triangles will therefore coincide , and must be equal . 52. Corollary . Hence , when two sides and the included angle of one triangle are respectively equal to those of another , the other side and ...
... extremities are the same points . The triangles will therefore coincide , and must be equal . 52. Corollary . Hence , when two sides and the included angle of one triangle are respectively equal to those of another , the other side and ...
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Common terms and phrases
ABCD &c AC AC adjacent angles altitude angle ABC angle BAC arc BC base and altitude bisect CD fig centre chord circumference convex surface Corollary cylinder DEF fig Definitions Demonstration denote diameter divided Draw equal arcs equal distances equilateral equivalent four right angles frustum given angle given circle given line given polygon given sides given square greater half the product Hence homologous sides hypothenuse infinitely small inscribed circle isoperimetrical isosceles Join AC Let ABCD line AB fig lines drawn mean proportional number of sides oblique lines parallel lines parallelogram parallelopiped perimeter perpendicular plane angles plane MN polygon ABCD prism Problem radii radius ratio rectangles regular polygon respectively equal right triangles Scholium secant sector segment side AC similar polygons similar triangles solid angle Solution sphere spherical polygon spherical triangle tangent Theorem triangles ABC triangular prism vertex vertices whence
Popular passages
Page 148 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Page 90 - To construct a parallelogram equivalent to a given square, and having the difference of its base and altitude equal to a given line.
Page 24 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Page 40 - One side and two angles of a triangle being given, to construct the triangle. Solution.
Page 79 - Construct, by § 145, a right triangle, of which the hypothenuse BC (fig. 79) is equal to the side of the greater square, and the leg AB is equal to the side of the less square ; and AC is the side of the required square.
Page 142 - THEOREM. If two triangles on the same sphere, or on equal spheres, are mutually equiangular, they will also be mutually equilateral. Let A and B be the two given triangles; P and Q their polar triangles. Since the angles are equal in the triangles A and B, the sides will be equal in. their polar triangles P and Q (Prop.
Page 6 - The preface and commentary to the Antigone are even more creditable to Mr. Woolsey's ability than those to the Alcestis. The sketch of the poem, in the preface, is written with clearness and brevity. The difficulties in this play, that call for a commentator's explanation, are far more numerous than in the Alcestis.
Page 70 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
Page 5 - A surface is that which has length and breadth, without thickness. 6. A plane is a surface, in which any two points being taken, the straight line joining those points lies wholly in that surface.
Page 137 - Each side of a spherical triangle is less than the sum of 'the other two sides. 48. The sum of the sides of a spherical polygon is less than 360°.