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 60
... circumference ; and BF is greater than CF , and CF than GF . K Also there can be drawn only two equal straight lines from the point F to the circumference , one upon each side of the shortest line FD : at the point F in the straight ...
... circumference ; and BF is greater than CF , and CF than GF . K Also there can be drawn only two equal straight lines from the point F to the circumference , one upon each side of the shortest line FD : at the point F in the straight ...
Página 170
... circumference again in E ; draw also CG perpendicular to BD : produce BC to A , bisect AC in H , and join CD . It is evident , that the arcs BD , BE are each of them one - sixth of the circumference ( Cor . 15. 4. ) , and that therefore ...
... circumference again in E ; draw also CG perpendicular to BD : produce BC to A , bisect AC in H , and join CD . It is evident , that the arcs BD , BE are each of them one - sixth of the circumference ( Cor . 15. 4. ) , and that therefore ...
Página 172
... circumference , S2AH ( AC + Q ) = 500 ( 1991.44495- ) —995722.475— , and S = ( 997.85895- ) because ( 997.85895 ) 2 is greater than 995722.475 . = But the square of the chord of the ninety - sixth part of the circumference = AB ( AC — S ) ...
... circumference , S2AH ( AC + Q ) = 500 ( 1991.44495- ) —995722.475— , and S = ( 997.85895- ) because ( 997.85895 ) 2 is greater than 995722.475 . = But the square of the chord of the ninety - sixth part of the circumference = AB ( AC — S ) ...
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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