Elements of Geometry and Trigonometry Translated from the French of A.M. Legendre by David Brewster: Revised and Adapted to the Course of Mathematical Instruction in the United States |
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Page 92
... ABCDE , FGHIK , be two similar polygons . From any angle A , in the polygon ABCDF , B draw diagonals AC , AD to the other angles . From the homologous angle F , in the other polygon FGHIK , draw diagonals FH , FI to the other an- gles ...
... ABCDE , FGHIK , be two similar polygons . From any angle A , in the polygon ABCDF , B draw diagonals AC , AD to the other angles . From the homologous angle F , in the other polygon FGHIK , draw diagonals FH , FI to the other an- gles ...
Page 93
... polygon ABCDE , is to the sum of the consequents FGH + FHI + FIK , or to the polygon FGHIK , as one antecedent ABC , is to its consequent FGH , or as AB2 is to FG2 ( Prop . XXV . ) ; hence the areas of similar poly- gons are to each ...
... polygon ABCDE , is to the sum of the consequents FGH + FHI + FIK , or to the polygon FGHIK , as one antecedent ABC , is to its consequent FGH , or as AB2 is to FG2 ( Prop . XXV . ) ; hence the areas of similar poly- gons are to each ...
Page 102
... polygon . Let ABCDE be the given polygon . Draw first the diagonal CE cutting off the triangle CDE ; through the point D , draw DF parallel to CE , and meet- ing AE produced ; draw CF : the poly- gon ABCDE will be equivalent to the ...
... polygon . Let ABCDE be the given polygon . Draw first the diagonal CE cutting off the triangle CDE ; through the point D , draw DF parallel to CE , and meet- ing AE produced ; draw CF : the poly- gon ABCDE will be equivalent to the ...
Page 104
... triangle EHG , the square of HE is to the square of HG as the segment EF is to the ... polygon similar to a given Let FG be the given line , and AEDCB the given ... ABCDE , as required . For , these two polygons are composed of the same ...
... triangle EHG , the square of HE is to the square of HG as the segment EF is to the ... polygon similar to a given Let FG be the given line , and AEDCB the given ... ABCDE , as required . For , these two polygons are composed of the same ...
Page 110
... polygon may be inscribed in a circle , and circum- scribed about one . Let ABCDE & c . be a regular poly- gon : describe a circle through the three points A , B , C , the centre being O , and OP the perpendicular let fall from it , to ...
... polygon may be inscribed in a circle , and circum- scribed about one . Let ABCDE & c . be a regular poly- gon : describe a circle through the three points A , B , C , the centre being O , and OP the perpendicular let fall from it , to ...
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Common terms and phrases
adjacent altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone consequently convex surface cosine Cotang cylinder diagonal diameter dicular distance divided draw drawn equal angles equally distant equations equivalent feet figure find the area formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC logarithm measured by half number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE PROBLEM proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle Scholium secant segment side BC similar sine slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex
Popular passages
Page 241 - In every plane triangle, the sum of two sides is to their difference as the tangent of half the sum of the angles opposite those sides is to the tangent of half their difference.
Page 251 - Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51° ; then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45' ; required the height of the tower.
Page 109 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their surfaces are to each other as the squares of those sides (Book IV.
Page 91 - Two similar triangles are to each other as the squares described on their homologous sides. Let ABC, DEF, be two similar triangles, having the angle A equal to D, and The angle B=E.
Page 169 - THEOREM. 7?/6 convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude.
Page 41 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Page 155 - AK. The two solids AG, AQ, having the same base AEHD are to each other as their altitudes AB, AO ; in like manner, the two solids AQ, AK, having the same base AOLE, are to each other as their altitudes AD, AM. Hence we have the two proportions, sol.
Page 86 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Page 282 - ... 1. To find the length of an arc of 30 degrees, the diameter being 18 feet. ' Ans. 4.712364. 2. To find the length of an arc of 12° 10', or 12£°, the diameter being 20 feet.
Page 93 - ABC : FGH : : ACD : FHI. By the same mode of reasoning, we should find ACD : FHI : : ADE : FIK; and so on, if there were more triangles. And from this series of equal ratios, we conclude that the sum of the antecedents...