Elements of Geometry and Trigonometry |
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Page 17
... rectangles and rhomboids . 1st . A RECTANGLE is a parallelogram whose angles are all right angles . A SQUARE is an equilateral rectangle . 2d . A RHOMBOID is a parallelogram whose angles are all oblique . A RHOMBUS is an equilateral ...
... rectangles and rhomboids . 1st . A RECTANGLE is a parallelogram whose angles are all right angles . A SQUARE is an equilateral rectangle . 2d . A RHOMBOID is a parallelogram whose angles are all oblique . A RHOMBUS is an equilateral ...
Page 95
... 7 ) ; which was to be proved . Cor . Triangles having equal bases and equal altitudes are equal , for they are halves of equal parallelograms . PROPOSITION III . THEOREM . Rectangles having equal altitudes , BOOK IV . 95.
... 7 ) ; which was to be proved . Cor . Triangles having equal bases and equal altitudes are equal , for they are halves of equal parallelograms . PROPOSITION III . THEOREM . Rectangles having equal altitudes , BOOK IV . 95.
Page 96
... rectangles whose altitudes AD and HK are equal , and whose bases AB and HE are commensurable : then will the areas of the rectangles be proportional to their bases . D K Suppose that AB is to HE , as 7 is to 4. Conceive AB to be divided ...
... rectangles whose altitudes AD and HK are equal , and whose bases AB and HE are commensurable : then will the areas of the rectangles be proportional to their bases . D K Suppose that AB is to HE , as 7 is to 4. Conceive AB to be divided ...
Page 97
... rectangles be proportional to their bases . For , place the rectangle HEFK upon the rectangle ABCD , so that it shall take the position AEFD . Then , if the rectangles are not pro- portional to their bases , let us sup- pose that ABCD ...
... rectangles be proportional to their bases . For , place the rectangle HEFK upon the rectangle ABCD , so that it shall take the position AEFD . Then , if the rectangles are not pro- portional to their bases , let us sup- pose that ABCD ...
Page 98
... rectangles : then will ABCD be to AEGF , as AB × AD is to AE × AF For , place the rectangles so that the angles DAB and ... rectangle AEGF will bo the superficial unit , and we shall have , ABCD : 1 :: AB × AD : 1 ; ABCD AB × AD : hence ...
... rectangles : then will ABCD be to AEGF , as AB × AD is to AE × AF For , place the rectangles so that the angles DAB and ... rectangle AEGF will bo the superficial unit , and we shall have , ABCD : 1 :: AB × AD : 1 ; ABCD AB × AD : hence ...
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Common terms and phrases
ABCD AC² adjacent angles altitude angle ACB apothem Applying logarithms base and altitude bisect centre chord circle circumference circumscribed coincide cone consequently convex surface corresponding cosec cosine Cotang cylinder denote diagonals diameter distance divided draw drawn edges equally distant feet find the area Formula frustum given angle given line given point greater hence homologous hypothenuse included angle interior angles intersection less Let ABC log sin logarithm lower base mantissa mean proportional measured by half middle point number of sides opposite parallel parallelogram parallelopipedon perimeter perpendicular plane MN polyedral angle polyedron prism PROBLEM PROPOSITION proved pyramid quadrant radii radius rectangle regular polygons right angles right-angled triangle Scholium secant segment side BC similar sine slant height sphere spherical polygon spherical triangle square Tang tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence