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 Elements of Plane and Spherical TrigonometryE. Duyckinck, and George Long, 1824 - 333 sider |
Inni boken
Resultat 1-5 av 79
Side 29
... perpendicular to a given straight line , of unlimited length , from a given point without it . Let AB be a given straight line , which may be length both ways , and let C be a point without it . draw a straight line perpendi- cular to ...
... perpendicular to a given straight line , of unlimited length , from a given point without it . Let AB be a given straight line , which may be length both ways , and let C be a point without it . draw a straight line perpendi- cular to ...
Side 62
... perpendicular falls , and the straight . line intercepted betweeen the perpendicular and the obtuse angle . Let ABC be an obtuse angled triangle , having the obtuse angle ACB , and from the point A let AD be drawn ( 12. 1. ) perpendicular ...
... perpendicular falls , and the straight . line intercepted betweeen the perpendicular and the obtuse angle . Let ABC be an obtuse angled triangle , having the obtuse angle ACB , and from the point A let AD be drawn ( 12. 1. ) perpendicular ...
Side 63
... perpendicular and the acute angle at B ; and it is manifest that ( 47. 1. ) AB + BC2 = AC2 + 2BC2 = AC2 + 2BC.BC . Therefore in every triangle , & c . Q. E. D. A 1 B PROP . XIV . PROB . To describe a square that shall be equal to a ...
... perpendicular and the acute angle at B ; and it is manifest that ( 47. 1. ) AB + BC2 = AC2 + 2BC2 = AC2 + 2BC.BC . Therefore in every triangle , & c . Q. E. D. A 1 B PROP . XIV . PROB . To describe a square that shall be equal to a ...
Side 64
... perpendicular to BC , and because BEA is a right angle , AB ( 47. 1. ) BE + AE2 and = AC2 = CE2 + AE2 ; wherefore AB2 + AC BE + CE2 + 2AE2 . But be- cause the line BC is cut equally in D , and unequally in E , BE2 + CE2 = ( 9.2 . ) 2BD ...
... perpendicular to BC , and because BEA is a right angle , AB ( 47. 1. ) BE + AE2 and = AC2 = CE2 + AE2 ; wherefore AB2 + AC BE + CE2 + 2AE2 . But be- cause the line BC is cut equally in D , and unequally in E , BE2 + CE2 = ( 9.2 . ) 2BD ...
Side 66
... perpendicular falls , is said to be farther from the centre B. An arch of a circle is any part of the circumference . V. A segment of a circle is the figure con- tained by a straight line , and the arch which it cuts off . 1 VI . An ...
... perpendicular falls , is said to be farther from the centre B. An arch of a circle is any part of the circumference . V. A segment of a circle is the figure con- tained by a straight line , and the arch which it cuts off . 1 VI . An ...
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Vanlige uttrykk og setninger
ABC is equal ABCD altitude angle ABC angle ACB angle BAC angle EDF arch AC base BC bisected centre circle ABC circumference cosine cylinder demonstrated diameter draw equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater hypotenuse inscribed join less Let ABC Let the straight line BC magnitudes meet multiple opposite angle parallel parallelepipeds parallelogram perpendicular polygon prism PROB produced proportionals proposition Q. E. D. COR Q. E. D. PROP radius ratio rectangle contained rectilineal figure remaining angle segment semicircle shewn side BC sine solid angle solid parallelepipeds spherical angle spherical triangle straight line AC THEOR third touches the circle triangle ABC triangle DEF wherefore