An Elementary Treatise on Plane and Solid Geometry |
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Page xiii
... triangular , quadrangular , pentagonal , hex- agonal , & c . prism ( 347 ) ; cylinder , its axis , right cylinder ( 348 ) ; generation of right cylinder ( 349 ) ; parallelopiped , right parallelopiped ( 350 ) ; cube , unit of solidity ...
... triangular , quadrangular , pentagonal , hex- agonal , & c . prism ( 347 ) ; cylinder , its axis , right cylinder ( 348 ) ; generation of right cylinder ( 349 ) ; parallelopiped , right parallelopiped ( 350 ) ; cube , unit of solidity ...
Page xiv
... triangular prism ( 392 ) , 125 Frustum of pyramid or cone , its convex surface and bases ( 393 ) ; generation of frustum of right cone ( 394 ) , Area of convex surface of frustum of regular pyramid or of 126 right cone ( 395–397 ) , 126 ...
... triangular prism ( 392 ) , 125 Frustum of pyramid or cone , its convex surface and bases ( 393 ) ; generation of frustum of right cone ( 394 ) , Area of convex surface of frustum of regular pyramid or of 126 right cone ( 395–397 ) , 126 ...
Page 113
... c . is equal to the altitude . 347. Definitions . A prism is triangular , quadran- gular , pentagonal , hexagonal , & c . , according as its Cylinder . Parallelopiped . base is a triangle , a 10 * CH . XVI . § 347. ] 113 SOLIDS .
... c . is equal to the altitude . 347. Definitions . A prism is triangular , quadran- gular , pentagonal , hexagonal , & c . , according as its Cylinder . Parallelopiped . base is a triangle , a 10 * CH . XVI . § 347. ] 113 SOLIDS .
Page 117
... perpendicular to the base ABCD . But any other face may as well be assumed for the base of AF as ABCD ; taking , then , the rectangle ABEH for bu Solidity of Triangular Prism , Prism , Cylinder . the CH , XVI . § 362. ] 117 SOLIDS .
... perpendicular to the base ABCD . But any other face may as well be assumed for the base of AF as ABCD ; taking , then , the rectangle ABEH for bu Solidity of Triangular Prism , Prism , Cylinder . the CH , XVI . § 362. ] 117 SOLIDS .
Page 118
... triangular prism is the product of its base by its altitude . Demonstration . Let ABC DEF ( fig . 170 ) be a triangular prism . Draw BG parallel to AC , CG parallel to AB , GH par- allel to AD , meeting the plane ADF in H. Join ÈH ; FH ...
... triangular prism is the product of its base by its altitude . Demonstration . Let ABC DEF ( fig . 170 ) be a triangular prism . Draw BG parallel to AC , CG parallel to AB , GH par- allel to AD , meeting the plane ADF in H. Join ÈH ; FH ...
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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°.