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 StatesA.S. Barnes and Company, 1838 - 359 σελίδες |
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Σελίδα 120
... circ . CA ; and its area , by area CA : it is then to be shown that circ . CA circ . OB :: CA : OB , and that area CA area OB :: CA2 : OB2 A. M N P T L B E G Inscribe within the circles two regular polygons of the same number of sides ...
... circ . CA ; and its area , by area CA : it is then to be shown that circ . CA circ . OB :: CA : OB , and that area CA area OB :: CA2 : OB2 A. M N P T L B E G Inscribe within the circles two regular polygons of the same number of sides ...
Σελίδα 121
... circ . OA . For , inscribe in the circle any E regular polygon , and draw Of perpendicular to one of its sides . Then the area of the polygon will be equal to OF , multiplied by the perimeter ( Prop . IX . ) . F Now , let the number of ...
... circ . OA . For , inscribe in the circle any E regular polygon , and draw Of perpendicular to one of its sides . Then the area of the polygon will be equal to OF , multiplied by the perimeter ( Prop . IX . ) . F Now , let the number of ...
Σελίδα 122
... circ . CA , there- fore circ . CA = 7x2CA . Multiply both terms by CA ; we have CAx circ . CA M B » D = л × CA2 , or area СA = л × CA2 : hence the area of a circle is equal to the product of the square of its radius by the constant ...
... circ . CA , there- fore circ . CA = 7x2CA . Multiply both terms by CA ; we have CAx circ . CA M B » D = л × CA2 , or area СA = л × CA2 : hence the area of a circle is equal to the product of the square of its radius by the constant ...
Σελίδα 169
... circ . CA , we are to show that the convex surface of the cylinder is equal to circ . CA x H. Inscribe in the circle any regular polygon , BDEFGA , and construct on this polygon a right HA G F B D E prism having its altitude equal to H ...
... circ . CA , we are to show that the convex surface of the cylinder is equal to circ . CA x H. Inscribe in the circle any regular polygon , BDEFGA , and construct on this polygon a right HA G F B D E prism having its altitude equal to H ...
Σελίδα 170
... circ . CA ( Book V. Prop . VIII . Cor . 1 . & 2. ) ; the inscribed prism then coincides with the cylinder , since their altitudes are equal , and their convex surfaces per- pendicular to the common base : hence the two solids will be ...
... circ . CA ( Book V. Prop . VIII . Cor . 1 . & 2. ) ; the inscribed prism then coincides with the cylinder , since their altitudes are equal , and their convex surfaces per- pendicular to the common base : hence the two solids will be ...
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adjacent altitude angle ACB ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone convex surface cosine Cotang cylinder diagonal diameter dicular distance divided draw drawn 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 middle point number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE PROBLEM Prop proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABCDE 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 trian triangle ABC triangular prism vertex