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 TrigonometryB. & S. Collins; W. E. Dean, printer, 1836 - 311 sider |
Inni boken
Resultat 16-20 av 45
Side 46
... being a right angle , the other angles of the parallelogram CGKB are also right angles Wherefore CGKB is a square , and it is upon the side CB . D F E ( Cor . 2. 28. 1. ) For the same reason HF also is a square , and it is 46 ELEMENTS.
... being a right angle , the other angles of the parallelogram CGKB are also right angles Wherefore CGKB is a square , and it is upon the side CB . D F E ( Cor . 2. 28. 1. ) For the same reason HF also is a square , and it is 46 ELEMENTS.
Side 49
... parallelogram CO : therefore also BN is equal to GR ; and the four rect- angles BN , CK , GR , RN are there- fore equal to one another , and so CK + BN + GR + RN 4CK . Again , be- cause CB is equal to BD , and BD equal = A CB D GK ...
... parallelogram CO : therefore also BN is equal to GR ; and the four rect- angles BN , CK , GR , RN are there- fore equal to one another , and so CK + BN + GR + RN 4CK . Again , be- cause CB is equal to BD , and BD equal = A CB D GK ...
Side 50
... parallelogram ML ; wherefore AG is equal also to RF . Therefore the four rectangles AG , MP , PL , RF , are equal to one another , and so AG + MP + PL + RF = 4AG . And it was demonstrated , that CK + BN + GR + RN = 4CK ; wherefore ...
... parallelogram ML ; wherefore AG is equal also to RF . Therefore the four rectangles AG , MP , PL , RF , are equal to one another , and so AG + MP + PL + RF = 4AG . And it was demonstrated , that CK + BN + GR + RN = 4CK ; wherefore ...
Side 54
... parallelogram , the two diagonals bisect each other ; and the sum of their squares is equal to the sum of the squares of all the four sides of the parallelogram . Let ABCD be a parallelogram , of which the diameters are AC and BD ; the ...
... parallelogram , the two diagonals bisect each other ; and the sum of their squares is equal to the sum of the squares of all the four sides of the parallelogram . Let ABCD be a parallelogram , of which the diameters are AC and BD ; the ...
Side 90
... parallelogram , and AEB a right angle , AGB ( 28. 1. ) is likewise a right angle : In the same manner , it may be shewn that the an- gles at H , K , F are right angles ; therefore the quadrilateral figure FGHK H K is rectangular ; and ...
... parallelogram , and AEB a right angle , AGB ( 28. 1. ) is likewise a right angle : In the same manner , it may be shewn that the an- gles at H , K , F are right angles ; therefore the quadrilateral figure FGHK H K is rectangular ; and ...
Andre utgaver - Vis alle
Elements Of Geometry John Playfair,William Wallace,John Davidsons Ingen forhåndsvisning tilgjengelig - 2023 |
Elements Of Geometry John Playfair,William Wallace,John Davidsons Ingen forhåndsvisning tilgjengelig - 2023 |
Vanlige uttrykk og setninger
ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal base BC bisected centre chord circle ABC circumference cosine cylinder demonstrated diameter divided 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 triangle DEF wherefore