Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids: to which are Added, Edlemnts of Plane and Spherical TrigonometryW.E. Dean, 1836 - 311 páginas |
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Página 197
... prism BNM is to the parallelopiped CD as the triangle AEM to the parallelogram LG . For by the last Cor . the prism BNM is to the prism DPG as the triangle AME to the triangle CGF , and therefore the prism BNM is to twice the prism DPG ...
... prism BNM is to the parallelopiped CD as the triangle AEM to the parallelogram LG . For by the last Cor . the prism BNM is to the prism DPG as the triangle AME to the triangle CGF , and therefore the prism BNM is to twice the prism DPG ...
Página 204
... prisms that stand upon the equal sections are equal ( 1. Cor . 8 3. Sup . ) , that is , the prism which stands on the base BCD , and which is between the planes BCD and NQL , is equal to the prism which stands on the base FGH , and ...
... prisms that stand upon the equal sections are equal ( 1. Cor . 8 3. Sup . ) , that is , the prism which stands on the base BCD , and which is between the planes BCD and NQL , is equal to the prism which stands on the base FGH , and ...
Página 205
... prism having a triangular base may be divided into tnree pyramids that have triangular bases , and that are equal to another . Let there be a prism of which the base is the triangle ABC , and let DEF be the triangle opposite the base ...
... prism having a triangular base may be divided into tnree pyramids that have triangular bases , and that are equal to another . Let there be a prism of which the base is the triangle ABC , and let DEF be the triangle opposite the base ...
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Términos y frases comunes
ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal arc AC base BC bisected centre chord circle ABC circumference cosine cylinder demonstrated diameter divided draw Prob equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROP proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle solid parallelopiped spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC wherefore