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 1848-1851 |
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Page 4
... rectangle contained by the whole and one of the parts shall be equal to the square of the other part . ( II . 11. ) ( B ) . Shew that in Euclid's figure , four other lines , beside the given line , are divided in the required manner ...
... rectangle contained by the whole and one of the parts shall be equal to the square of the other part . ( II . 11. ) ( B ) . Shew that in Euclid's figure , four other lines , beside the given line , are divided in the required manner ...
Page 5
... 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 involves this ...
... 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 involves this ...
Page 7
... rectangle that can be inscribed in a circle is a square . Here ( B ) is a direct application of the fact asserted in ( A ) . Let ABCD ( fig . 9 ) be any rectangle inscribed in a circle . Join AC . By the proposition , 4ABC is a right ...
... rectangle that can be inscribed in a circle is a square . Here ( B ) is a direct application of the fact asserted in ( A ) . Let ABCD ( fig . 9 ) be any rectangle inscribed in a circle . Join AC . By the proposition , 4ABC is a right ...
Page 8
Francis James Jameson. Then rectangle ABCD = 2 triangle ABC = rectangle on base AC , and between same parallels as the triangle ABC AC.BE. Now AC , being the diameter of the circle , is constant ; therefore the area of the rectangle is ...
Francis James Jameson. Then rectangle ABCD = 2 triangle ABC = rectangle on base AC , and between same parallels as the triangle ABC AC.BE. Now AC , being the diameter of the circle , is constant ; therefore the area of the rectangle is ...
Page 10
... rectangle contained by the segments of any chord passing through a given point within a circle is constant . Let AB , CD ( fig . 13 ) be any two chords of a circle inter- secting in O. Join BC , AD . Since the angles in the same segment ...
... rectangle contained by the segments of any chord passing through a given point within a circle is constant . Let AB , CD ( fig . 13 ) be any two chords of a circle inter- secting in O. Join BC , AD . Since the angles in the same segment ...
Other editions - View all
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson No preview available - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson No preview available - 2015 |
Common terms and phrases
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
Popular passages
Page 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.
Page 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.
Page 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.
Page 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...
Page 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.