## 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 pages |

### From inside the book

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Page 7

... right angle , ABC is a semicircle . Draw BE

... right angle , ABC is a semicircle . Draw BE

**perpendicular**to AC , and take F the centre . * If the polygon has six sides , the lines A4 , A4 , will coincide . Then rectangle ABCD = 2 triangle ABC = rectangle on EUCLID . 7. Page 11

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**perpendicular**to two planes CAB , DAB , which intersect in AB , and from D draw DF**perpendicular**to the plane CAB , meeting it in F ; shew that the line joining the points C and F , produced if necessary , is**perpendicular**to AB . Since ... Page 12

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**perpendicular**to a plane from a given point without it . ( XI . 11. ) ( B ) . Prove that equal right lines drawn from a given point to a given plane are equally inclined to the plane . Let P be the given point ; PA , PA ' , two equal ... Page 13

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**perpendicular**upon it from the focus , meets it in the tangent at the vertex . ( B ) . If PM be the ordinate at P , and T the intersection of the tangent at P with the axis , TP.TY TM.TS. Since , by ( 4 ) , SYT ( fig . 15 ) is a right ... Page 15

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**perpendicular**to QZ ' ; and therefore QY ' = SY , QZ ' = HZ . Hence , drawing QKK ' through the centre C , QK.QK ' = QY'.QZ ' SY.HZ BC2 . Therefore the distance of Q from C is constant , and the locus of Q is a circle whose centre is C ...### 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.