## 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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**parallels**are equal to one another . ( 1. 38. ) ( B ) . Let ABC , ABD be two equal triangles upon the same base AB ...**parallels**, ACEBAD'EB , = ADEB , by the construction of the figure . Now the triangles CEB , DEB being equal and ... Page 4

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**parallels**, are by ( 4 ) equal to one another . Hence ^ FEB = ^ DEB ; that is , the less equals the greater , which is absurd . Therefore CE = ED . 1851. ( A ) . In any right - angled triangle , the square described upon the side ... Page 8

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**parallels**as the triangle ABC AC.BE. Now AC , being the diameter of the circle , is constant ; therefore the area of the rectangle is proportional to BE , and is greatest when BE is greatest , i . e . when it coincides with GF . Hence ... Page 11

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**parallel**, and one of them be at right angles to a plane , the other is at right angles to the same plane . ( XI . 8. ) ( B ) . From a point E draw EC , ED perpendicular to two planes CAB , DAB , which intersect in AB , and from D draw ... Page 12

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**parallel**to EC or DF . Then , by ( 4 ) , GH is at right angles to the plane CAB ; and therefore / AGH is a right angle . Similarly , a line GH ' drawn**parallel**to ED will lie in the plane of ECD , and will be at right angles to the ...### 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.