Elements of geometry: consisting of the first four,and the sixth, books of Euclid, with the principal theorems in proportion [&c.] by J. Narrien1842 |
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Resultat 1-5 av 73
Side
... perpendicular be let fall from the centre of the circle upon such line , it will be less than a semidiameter of the circle . Thus , there may be conceived to be inscribed in a circle , a polygon whose sides are less than any lines that ...
... perpendicular be let fall from the centre of the circle upon such line , it will be less than a semidiameter of the circle . Thus , there may be conceived to be inscribed in a circle , a polygon whose sides are less than any lines that ...
Side 15
... perpendicular required ; and the demonstration is the same as that which is given in the text . COR . By the help of this problem , it may be demon- strated , that two straight lines cannot have a common seg- ment . E If it be possible ...
... perpendicular required ; and the demonstration is the same as that which is given in the text . COR . By the help of this problem , it may be demon- strated , that two straight lines cannot have a common seg- ment . E If it be possible ...
Side 16
... perpendicular to it ; therefore from the given point c a perpendicular CH has been drawn to the given straight line AB . Q. E. F. In practice ; having described the arc FG , as in the text , from F and G as centres , with the same , or ...
... perpendicular to it ; therefore from the given point c a perpendicular CH has been drawn to the given straight line AB . Q. E. F. In practice ; having described the arc FG , as in the text , from F and G as centres , with the same , or ...
Side 37
... perpendicular to AB or CD ; then if AM be made equal to half the sum of the sides AB , CD , and a rectangular parallelogram be formed with AM and AE as sides , that parallelogram shall be equiva- lent to ABDC . Join A , D ; bisect AB in ...
... perpendicular to AB or CD ; then if AM be made equal to half the sum of the sides AB , CD , and a rectangular parallelogram be formed with AM and AE as sides , that parallelogram shall be equiva- lent to ABDC . Join A , D ; bisect AB in ...
Side 54
... perpendicular falls , and the straight line inter- cepted on the exterior of the triangle between the perpendicular and the obtuse angle . Let ABC be an obtuse - angled triangle , having the obtuse angle ACB , and from the point A let ...
... perpendicular falls , and the straight line inter- cepted on the exterior of the triangle between the perpendicular and the obtuse angle . Let ABC be an obtuse - angled triangle , having the obtuse angle ACB , and from the point A let ...
Andre utgaver - Vis alle
Elements of Geometry: Consisting of the First Four, and the Sixth, Books of ... Euclides Ingen forhåndsvisning tilgjengelig - 2015 |
Elements of Geometry: Consisting of the First Four, and the Sixth, Books of ... Euclides Ingen forhåndsvisning tilgjengelig - 2018 |
Elements of Geometry: Consisting of the First Four,and the Sixth, Books of ... Euclides Ingen forhåndsvisning tilgjengelig - 2013 |
Vanlige uttrykk og setninger
ABCD AC is equal adjacent angles altitudes angle ABC angle ACB angle BAC assigned base BC bisected centre circle ABC circumference cone convex surface cylinder described diameter draw drawn duplicate ratio Edition equal angles equal or equivalent equi equilateral and equiangular Euclid exterior angle fore given line given rectilineal given straight line gnomon greater Greek homologous homologous sides inscribed join Latin Let ABC measure number of sides opposite angles parallel parallelepiped parallelogram perpendicular picket plane angles prism PROB proportional proposition pyramid Q. E. D. PROP rectangle contained rectilineal figure regular polygon remaining angle right angles segment similar solid angle sphere spherical angle square of AC straight line AC THEOR touches the circle triangle ABC triangle DEF wherefore
Populære avsnitt
Side 55 - In every triangle, the square of the side subtending either of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle.
Side 47 - CB ; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB : but HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB: therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore, if a straight line, &c.
Side 12 - UPON the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity...
Side 73 - CBED is greater than a semicircle, the angles CAD, CED are equal : therefore the whole angle BAD is, equal to the whole angle BED.
Side 8 - A New Treatise on the Use of the Globes ; or, a Philosophical View of the Earth and Heavens : comprehending an Account of the Figure, Magnitude, and Motion of the Earth: with the Natural Changes of its Surface, caused by Floods, Earthquakes, &c.
Side 142 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 11 - ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC : And it has also been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore, " the angles at the base
Side 53 - AB be the given straight line ; it is required to divide it 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.
Side 30 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sidef. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles.
Side 9 - If two triangles have two sides of the one equal to two sides of the...