Geometrical Analysis: Or the Construction and Solution of Various Geometrical Problems from Analysis, by Geometry, Algebra, and the Differential Calculus; Also, the Geometrical Construction of Algebraic Equations, and a Mode of Constructing Curves of the Higher Order by Means of PointsJ. B. Lippincott & Company, 1872 - 279 σελίδες |
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Αποτελέσματα 1 - 5 από τα 25.
Σελίδα 23
... ABCD to be the required square , having the difference between the diagonal AC and the side AB equal to a given quantity ; then produce the side AB until AE AC , and BE will be equal to that given quantity . Join CE . Then in the ...
... ABCD to be the required square , having the difference between the diagonal AC and the side AB equal to a given quantity ; then produce the side AB until AE AC , and BE will be equal to that given quantity . Join CE . Then in the ...
Σελίδα 26
... ABCD represent the required parallelogram . Produce the side AB until the produced part BE = BC ; then the side . AE = AB + BC = half the given perimeter , and is known . Hence , in the triangle ABC we have the angle B , the opposite ...
... ABCD represent the required parallelogram . Produce the side AB until the produced part BE = BC ; then the side . AE = AB + BC = half the given perimeter , and is known . Hence , in the triangle ABC we have the angle B , the opposite ...
Σελίδα 28
... ABCD the side of the square required . Limits . - 1 . With the same base AB , and the same perpendicular height CD , the point D may be taken anywhere in the line AB , and the side of the inscribed square will always be the same length ...
... ABCD the side of the square required . Limits . - 1 . With the same base AB , and the same perpendicular height CD , the point D may be taken anywhere in the line AB , and the side of the inscribed square will always be the same length ...
Σελίδα 33
... ABCD represent the required trapezoid , having B and C right angles . Parallel to BD draw CG , meeting AB , produced , in G. Then ACG is a right angle , being AFB , and CGBD . Hence , by similar triangles GBC and GCA , GB : GC :: GC ...
... ABCD represent the required trapezoid , having B and C right angles . Parallel to BD draw CG , meeting AB , produced , in G. Then ACG is a right angle , being AFB , and CGBD . Hence , by similar triangles GBC and GCA , GB : GC :: GC ...
Σελίδα 57
... ABCD be the required rhombus , whose diagonals intersect each other at E. Since in a rhombus the sides are all equal , each side is one - fourth of the given perimeter . Also the diagonals cross each other at right angles . Now , take ...
... ABCD be the required rhombus , whose diagonals intersect each other at E. Since in a rhombus the sides are all equal , each side is one - fourth of the given perimeter . Also the diagonals cross each other at right angles . Now , take ...
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AB² ABCD abscissa AC² AD² Analysis angle ABC angle BAC axis base BC² Calculation.-1 CD² centre chord circular segment common tangent construction Construction.-Draw curve describe a circle determine the triangle diagonal diameter differential draw a line ellipse equal equation erect the perpendicular fall the perpendicular find angle given angle given circle given line given point given radius Hence hyperbola hypothenuse inscribed intersect join latus rectum let fall meet parabola parallel to BC parallelogram plane triangle Prob produced quadratrix radii represent the required required triangle rhombus right angle right-angled triangle Scholium segment semicircle side BC similar triangles square tangent Theorem trapezium triangle ABC triangle are given vertical angle Whence
Δημοφιλή αποσπάσματα
Σελίδα 212 - If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, and the part of it produced...
Σελίδα 212 - PROP. X. THEOR. IF a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.
Σελίδα 141 - In a triangle, having given the base, the sum of the other two sides, and the length of a line drawn from the vertical angle to the middle of the base, to find the sides of the triangle.
Σελίδα 10 - ... algebra, or the modern analysis, from the mechanical facility of its operations, has contributed, especially on the Continent, to vitiate the taste and destroy the proper relish for the strictness and purity so conspicuous in the ancient method of demonstration. The study of geometrical analysis appears admirably fitted to improve the intellect, by training it to habits of precision, arrangement, and close application. If, the taste thus acquired be not allowed to obtain undue ascendancy, it...
Σελίδα 103 - Describe a circle which shall pass through a given point and touch a given straight line and a given circle.
Σελίδα 100 - Describe a circle which shall pass through two given points, and have its centre in a given line.
Σελίδα 27 - To trisect a right angle. 37. To divide a triangle into two parts by a line drawn parallel to a side, so that these parts shall be to each other as two given straight lines. 38. To divide a triangle into two parts by a line drawn perpendicular to the base, so that these parts shall...
Σελίδα 142 - In any plane triangle, as the base is to the sum of the other two sides, so is the difference of those sides to the difference of the segments of the base made by a perpendicular let fall from the vertical angle.
Σελίδα 32 - The lengths of three lines drawn from the three angles of a plane triangle to the middle of the opposite sides, being 18, 24, and 30, respectively ; it is required to find the sides.
Σελίδα 151 - PROBLEM XVII. IN a Right-angled Triangle, having given the Perimeter or Sum of all the Sides, and the Perpendicular let fall from the Right Angle on the Hypothenuse ; to determine the Triangle, that is, its Sides. PROBLEM XVIII.