The Principles of the Solution of the Senate-house 'riders,' Exemplified by the Solution of Those Proposed in the Earlier Parts of the Examinations of the Years 1848-1851Macmillan & Company, 1851 - 116 σελίδες |
Αναζήτηση στο βιβλίο
Αποτελέσματα 1 - 5 από τα 50.
Σελίδα 1
... equal to twice the base . Let AB ( fig . 1 ) be the given finite straight line . With centre A and radius equal to 2AB describe a circle CDF ; and with centre B and radius equal to 2AB describe a circle CEF . Let the circles intersect ...
... equal to twice the base . Let AB ( fig . 1 ) be the given finite straight line . With centre A and radius equal to 2AB describe a circle CDF ; and with centre B and radius equal to 2AB describe a circle CEF . Let the circles intersect ...
Σελίδα 2
... equal to two sides in the other , each to each , and the third side BD is common to both , the triangles are equal , and therefore the angles , each to each . Therefore ABD = 4CDB and ADB = △ CBD ) Hence ABCD is a parallelogram . AB ...
... equal to two sides in the other , each to each , and the third side BD is common to both , the triangles are equal , and therefore the angles , each to each . Therefore ABD = 4CDB and ADB = △ CBD ) Hence ABCD is a parallelogram . AB ...
Σελίδα 3
... equal and AB = DC , therefore EB = ED . Therefore the triangles BCE , ECD , being upon equal bases and between the same parallels , * are equal to one another , Hence AECD ABCE ABEA 1851 . AAED . ( A ) . Triangles upon equal bases and ...
... equal and AB = DC , therefore EB = ED . Therefore the triangles BCE , ECD , being upon equal bases and between the same parallels , * are equal to one another , Hence AECD ABCE ABEA 1851 . AAED . ( A ) . Triangles upon equal bases and ...
Σελίδα 4
Francis James Jameson. Then triangles CEB , FEB , being on equal bases and between the same parallels , are by ( 4 ) equal to one another . Hence ^ FEB = ^ DEB ; that is , the less equals the greater , which is absurd . Therefore CE = ED ...
Francis James Jameson. Then triangles CEB , FEB , being on equal bases and between the same parallels , are by ( 4 ) equal to one another . Hence ^ FEB = ^ DEB ; that is , the less equals the greater , which is absurd . Therefore CE = ED ...
Σελίδα 5
... equal to a given rectangle . ( II.14 . ) ( B ) . Given a square , and one side of a rectangle which is equal to the square ; find the other side . If , in ( A ) , AC ( fig . 6 ) be the given rectangle , the proof of the proposition ...
... equal to a given rectangle . ( II.14 . ) ( B ) . Given a square , and one side of a rectangle which is equal to the square ; find the other side . If , in ( A ) , AC ( fig . 6 ) be the given rectangle , the proof of the proposition ...
Άλλες εκδόσεις - Προβολή όλων
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2015 |
Συχνά εμφανιζόμενοι όροι και φράσεις
AC² AN.NM Arithmetic arithmetical progression axis bisects body C₁ Cambridge centre of gravity chord CHURCHILL BABINGTON circle cloth cone Conic Sections conjugate hyperbola constant curvature curve cycloid describe diameter direction directrix distance drawn Edition ellipse equations equilibrium Fellow of St fluid focus geometrical given point Hence horizontal hyperbola inches inclined inscribed John's College joining latus-rectum least common multiple Lemma length locus meet mirror move number of seconds oscillation parabola parallel parallelogram particle perpendicular plane polygon pressure prop proportional proposition prove pullies quadrilateral quantity radius ratio rays rectangle refraction right angles sewed shew sides specific gravity spherical square straight line string surface tan² tangent triangle ABC Trinity College tube V₁ vary vertex vertical W₁ weight
Δημοφιλή αποσπάσματα
Σελίδα 4 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.
Σελίδα 6 - The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.
Σελίδα 11 - AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced if necessary ; if from any point C of AB, a perpendicular be drawn to AB meeting AP and .BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional of CE and CF.
Σελίδα 9 - IF the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another...
Σελίδα 4 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.