Elements of Geometry and Trigonometry |
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Page 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 AN T L B E G K 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 AN T L B E G K Inscribe within the circles two regular polygons of the same number of sides ...
Page 121
... circ . OA . For , inscribe in the circle any E regular polygon , and draw Oř 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 Oř 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 ...
Page 122
... circ . CA , there- fore circ . CA = X2CA . Multiply both terms by CA ; we have CA x circ . CA = л × CA2 , or area CA = × CA2 : hence the area of a circle is equal to the product of the square of its radius by the constant number 7 ...
... circ . CA , there- fore circ . CA = X2CA . Multiply both terms by CA ; we have CA x circ . CA = л × CA2 , or area CA = × CA2 : hence the area of a circle is equal to the product of the square of its radius by the constant number 7 ...
Page 169
... circ . CA , we are to show that the convex surface of the cylinder is equal to circ . CA × H. Inscribe in the circle any regular polygon , BDEFGA , and construct on this polygon a right G F H A C B D KI prism having its altitude equal ...
... circ . CA , we are to show that the convex surface of the cylinder is equal to circ . CA × H. Inscribe in the circle any regular polygon , BDEFGA , and construct on this polygon a right G F H A C B D KI prism having its altitude equal ...
Page 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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Common terms and phrases
adjacent adjacent angles altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone convex surface Cosine 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 number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular perpendicular let fall plane MN polyedron polygon ABCDE PROBLEM proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABCDE Scholium secant segment similar Sine Cotang slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex