Les elemens d'Euclide: expliquez d'une maniere nouvelle & très-facile. Avec l'usage de chaque proposition pour toutes les parties des mathematiquesJombert, 1720 - 360 páginas |
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Página 3
... parallelogram BCPS, PB is a diagonal. 1 '. area of ABPS = 2. area of BCPS ... (V) Adding (iv) and (v) area of AASP + ... parallelogram ABCD. Show that (i) ar (APB) + ar (PCD) = ar (ABCD) (ii) ar (APD) + ar (PBC) = ar(APB) + ar (PCD) ...
... parallelogram BCPS, PB is a diagonal. 1 '. area of ABPS = 2. area of BCPS ... (V) Adding (iv) and (v) area of AASP + ... parallelogram ABCD. Show that (i) ar (APB) + ar (PCD) = ar (ABCD) (ii) ar (APD) + ar (PBC) = ar(APB) + ar (PCD) ...
Página 10
... parallelogram . So parallelograms with a given angle are easily constructed with two applications of a construction for I.31 . Euclid 1.42 requires that the constructed parallelogram also have a prescribed area , and I.44 then imposes ...
... parallelogram . So parallelograms with a given angle are easily constructed with two applications of a construction for I.31 . Euclid 1.42 requires that the constructed parallelogram also have a prescribed area , and I.44 then imposes ...
Página 10
... parallelogram fpm . About the resultant fl1 complete the parallelogram fi11k ' , then fi1 will be the upward thrust on ƒ B , and ƒ k1 the downward thrust upon the strut fi . From the point g , in like manner , mark off gn and go equal ...
... parallelogram fpm . About the resultant fl1 complete the parallelogram fi11k ' , then fi1 will be the upward thrust on ƒ B , and ƒ k1 the downward thrust upon the strut fi . From the point g , in like manner , mark off gn and go equal ...
Página 20
... parallelogram law. Two common problems in statics involve either finding the resultant force,knowing its components,or resolving a known force into two components. We will now describe how each of these problems is solved using the ...
... parallelogram law. Two common problems in statics involve either finding the resultant force,knowing its components,or resolving a known force into two components. We will now describe how each of these problems is solved using the ...
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... parallelogram DABE is equal to the parallelogram LAKN. For the same reason, the parallelogram BGFC is equal to the parallelogram NKCM. Therefore the sum of the parallelograms DABE, BGFC is equal to the parallelogram LACM, that is, to the ...
... parallelogram DABE is equal to the parallelogram LAKN. For the same reason, the parallelogram BGFC is equal to the parallelogram NKCM. Therefore the sum of the parallelograms DABE, BGFC is equal to the parallelogram LACM, that is, to the ...
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Términos y frases comunes
ABCD ainfi ajoûte auffi égaux aura même raifon bafe bafe BC baſe c'eft c'est-à-dire CD font centre commune fection compofée confequent contient Corol côtez AB côtez AC Cylindre d'Euclide Démonftration divife eft double eft égal eft perpendiculaire égal à l'angle égal au quarré égal au rectangle équiangles eſt fe coupent feconde fegment fera égal feront auffi feront égaux fert feul foient foit folide fomme font auffi font égaux font paralleles fphere fuppofe furface Geometrie Gnomonique hauteur infcrits l'antecedent ligne AB ligne AC ligne BD lindre mefure moitié de l'arc oppofez paffe paralle parallelepipede parallelograme plufieurs polygone précedente premiere prifmes propofe Propofition proportion proportionnelles PROPOSITION puifqu'ils puifque pyramide quantité quarré de AC raifon triplée rectangle compris fous THEOREM tirez la ligne Trian Triangle ABC troifiéme USAGE
Pasajes populares
Página 354 - Cercle de fa baie, qui a pour diamètre PQ^, égal à MN , puifque le diamètre d'une Sphère infcrite dans un Cylindre , doit être égal à celui de la bafe du Cylindre, félon l'idée qu'on a des figures infcrites. Or puifque la furface de la Sphere eft ¿gale à celle du Cylindre dans lequel elle eft infcrite, & que cette^furface de CyLivre Douzième.
Página 214 - ... & le cube du premier terme eft au cube du fécond , comme le premier eft au quatrième.