Elements of Geometry and Trigonometry |
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Page 7
... Diameter of a Circle , 116 To find the length of an Arc , 117 Area of a Circle , Area of a Sector , Area of a Segment , Area of a Circular Ring , 117 118 118 119 PAGE . Area of the Surface of a Prism , CONTENTS . vi :
... Diameter of a Circle , 116 To find the length of an Arc , 117 Area of a Circle , Area of a Sector , Area of a Segment , Area of a Circular Ring , 117 118 118 119 PAGE . Area of the Surface of a Prism , CONTENTS . vi :
Page 59
... DIAMETER is a straight line drawn through the centre and terminating in the circumference . All radii of the same circle are equal . are also equal , and each is double the radius . 4. An ARC is any part of a circumference . All diameters ...
... DIAMETER is a straight line drawn through the centre and terminating in the circumference . All radii of the same circle are equal . are also equal , and each is double the radius . 4. An ARC is any part of a circumference . All diameters ...
Page 60
... when its circumference touches all of the sides of the polygon . ооо хо POSTULATE . A circumference can be described from any point as a centre . and with any radius . PROPOSITION L. THEOREM . Any diameter divides the circle , 60 GEOMETRY .
... when its circumference touches all of the sides of the polygon . ооо хо POSTULATE . A circumference can be described from any point as a centre . and with any radius . PROPOSITION L. THEOREM . Any diameter divides the circle , 60 GEOMETRY .
Page 61
... diameter is greater than any other chord . Let AD be a chord , and AB a diameter through one extremity , as A : then will AB be greater than AD . Draw the radius CD . In the tri- angle ACD , we have AD less than the sum of AC and CD ...
... diameter is greater than any other chord . Let AD be a chord , and AB a diameter through one extremity , as A : then will AB be greater than AD . Draw the radius CD . In the tri- angle ACD , we have AD less than the sum of AC and CD ...
Page 62
... diameters AB and EF ADB be applied to the semi - circle EGF , it will coincide with it , and the semi - circumference ADB will coincide with the semi - circumference EGF . But the part AMD is equal to the part ENG , by hypothesis ...
... diameters AB and EF ADB be applied to the semi - circle EGF , it will coincide with it , and the semi - circumference ADB will coincide with the semi - circumference EGF . But the part AMD is equal to the part ENG , by hypothesis ...
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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