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 M N 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 M N T L B E G 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 D Now , let the number ...
... 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 D Now , let the number ...
Page 122
... circ . CA , there- fore circ . CA = x2CA . terms by CA ; we have = л × CA2 , or area CA : Multiply both CA × circ . CA A D M B × CA2 : hence the area of a circle is equal to the product of the square of its radius by the constant number ...
... circ . CA , there- fore circ . CA = x2CA . terms by CA ; we have = л × CA2 , or area CA : Multiply both CA × circ . CA A D M B × CA2 : hence the area of a circle is equal to the product of the square of its radius by the constant number ...
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 H A C F B D KEI prism having its altitude equal to ...
... 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 H A C F B D KEI prism having its altitude equal to ...
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 altitude angle ACB angle BAC 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 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 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 19 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Page 232 - ... the logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator.
Page 11 - A right-angled triangle is one which has a right angle. The side opposite the right angle is called the hypothenuse.
Page 168 - The radius of a sphere is a straight line drawn from the centre to any point of the surface ; the diameter or axis is a line passing through this centre, and terminated on both sides by the surface.
Page 31 - Hence, the interior angles plus four right angles, is equal to twice as many right angles as the polygon...
Page 18 - America, but know that we are alive, that two and two make four, and that the sum of any two sides of a triangle is greater than the third side.
Page 20 - In an isosceles triangle the angles opposite the equal sides are equal.
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 159 - S-ahc be the smaller : and suppose Aa to be the altitude of a prism, which having ABC for its base, is equal to their difference. Divide the altitude AT into equal parts Ax, xy, yz, &c. each less than Aa, and let k be one of those parts ; through the points of division...
Page 64 - To inscribe a circle in a given triangle. Let ABC be the given triangle. Bisect the angles A and B by the lines AO and BO, meeting at the point 0.