Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids ; to which are Added, Elements of Plane and Spherical TrigonometryW.E. Dean, 1836 - 311 sider |
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Resultat 1-3 av 35
Side 125
... proportional to A , B , C. Take two straight lines DE , DF , containing any angle EDF ; and upon these make DG equal ... proportional HF is found . PROP . XIII . PROB . To find a mean proportional between two given straight lines . Let ...
... proportional to A , B , C. Take two straight lines DE , DF , containing any angle EDF ; and upon these make DG equal ... proportional HF is found . PROP . XIII . PROB . To find a mean proportional between two given straight lines . Let ...
Side 126
... proportional : And parallelograms which have one angle of the one equal to one angle of the other , and their sides about the equal angles reciprocally proportional , are equal to one another . F E D B Let AB , BC be equal parallel- A ...
... proportional : And parallelograms which have one angle of the one equal to one angle of the other , and their sides about the equal angles reciprocally proportional , are equal to one another . F E D B Let AB , BC be equal parallel- A ...
Side 167
... proportional to the first two , will be equal to the square of the number proportional to the third ; that is , if A.C = B2 , mxr = n × n , vr = n2 . For by this Prop . A.C : B2 :: mxr : n2 ; but A.C = B2 , therefore mxr = n2 . Nearly ...
... proportional to the first two , will be equal to the square of the number proportional to the third ; that is , if A.C = B2 , mxr = n × n , vr = n2 . For by this Prop . A.C : B2 :: mxr : n2 ; but A.C = B2 , therefore mxr = n2 . Nearly ...
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Vanlige uttrykk og setninger
ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal arc AC base BC bisected centre chord circle ABC circumference cosine cylinder demonstrated diameter divided draw Prob equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROP proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle solid parallelopiped spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC wherefore