Elements of Geometry and Trigonometry from the Works of A.M. Legendre: Adapted to the Course of Mathematical Instruction in the United States |
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Page 179
... triangular prism is one whose bases are triangles ; u pentangular prism is one whose bases are pentagons , & c . 7. A PARALLELO PIPEDON is a prism whose bases are parallelograms . A Right Parallelopipedon is one whose lat- eral edges ...
... triangular prism is one whose bases are triangles ; u pentangular prism is one whose bases are pentagons , & c . 7. A PARALLELO PIPEDON is a prism whose bases are parallelograms . A Right Parallelopipedon is one whose lat- eral edges ...
Page 188
... triangular prisms . Let ABCD - H be a parallelopipedon , and let a plane be passed through the edges BF and DH then will the prisms ABD - H and BCD - H be equal in volume . For , through the vertices F and B let planes be passed ...
... triangular prisms . Let ABCD - H be a parallelopipedon , and let a plane be passed through the edges BF and DH then will the prisms ABD - H and BCD - H be equal in volume . For , through the vertices F and B let planes be passed ...
Page 189
... triangular prism ABD - II , is equal to half of the parallelopipedon AG , which has the same triedral angle A , and the same edges AB , AD , and AE . PROPOSITION VIII . THEOREM . If two parallelopipedons have a BOOK VII . 189.
... triangular prism ABD - II , is equal to half of the parallelopipedon AG , which has the same triedral angle A , and the same edges AB , AD , and AE . PROPOSITION VIII . THEOREM . If two parallelopipedons have a BOOK VII . 189.
Page 190
... triangular AEI - M and BFK - L , we have the line AE equal and parallel to BF , and EI equal to FK ; hence , the face AEI is equal to BFK . In the faces EIMH and FKLG , we have , HE = .GF , EI = FK and HEI = GFK : hence , the two faces ...
... triangular AEI - M and BFK - L , we have the line AE equal and parallel to BF , and EI equal to FK ; hence , the face AEI is equal to BFK . In the faces EIMH and FKLG , we have , HE = .GF , EI = FK and HEI = GFK : hence , the two faces ...
Page 197
... the same faces , pass the planes AH , AI , dividing the prism into triangular prisms . These prisms will all have a common altitude equal to that of the given prism . K Now , the volume of any one of the BOOK VII . 197.
... the same faces , pass the planes AH , AI , dividing the prism into triangular prisms . These prisms will all have a common altitude equal to that of the given prism . K Now , the volume of any one of the BOOK VII . 197.
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Common terms and phrases
AB² AC² adjacent angles altitude angle ACB apothem Applying logarithms base and altitude bisect centre chord circle circumference circumscribed coincide cone consequently convex surface corresponding cosec Cosine Cotang cylinder denote diagonals diameter distance divided draw drawn edges equally distant feet find the area Formula frustum given angle given straight line greater hence homologous hypothenuse included angle interior angles intersection less Let ABC log sin lower base mantissa mean proportional measured by half number of sides opposite parallel parallelogram parallelopipedon perimeter perpendicular plane MN polyedral angle polyedron prism PROPOSITION proved pyramid quadrant radii radius rectangle regular polygons right angles right-angled triangle Scholium secant segment semi-circumference side BC similar sine six right slant height sphere spherical polygon spherical triangle square Tang tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence