## Euclid's Elements of Geometry: From the Latin Translation of Commandine. To which is Added, A Treatise of the Nature of Arithmetic of Logarithms ; Likewise Another of the Elements of Plain and Spherical Trigonometry ; with a Preface |

### Inni boken

Resultat 1-5 av 5

Side 255

For if the

thereof . First , let it be greater than triple to the Cone , and let the Square ABCD

be described in the Circle ABCD , then the Square ABCD is greater than one half

...

For if the

**Cylinder**be not triple to the Cone , it shall be greater or less than triplethereof . First , let it be greater than triple to the Cone , and let the Square ABCD

be described in the Circle ABCD , then the Square ABCD is greater than one half

...

Side 264

, and the fame Altitude is + triple of the Cone , since it is demonstrated , that every

Cone is one third Part of a

**Cylinder**to**Cylinder**. For a**Cylinder**having the + 10 of this fame Base as a Cone, and the fame Altitude is + triple of the Cone , since it is demonstrated , that every

Cone is one third Part of a

**Cylinder**, having the same Base and equal Altitude . Side 265

are equal to each other , as also the

LN , NE , EK , is equal to the Number of PR , RB , BG : The Axis KL shall be the

same Multiple of the Axis EK , as the

...

are equal to each other , as also the

**Cylinders**PR , RB , BG ; and the Number ofLN , NE , EK , is equal to the Number of PR , RB , BG : The Axis KL shall be the

same Multiple of the Axis EK , as the

**Cylinder**PG , is of the**Cylinder**GB . For the...

Side 266

Therefore the

FM is cut by'a Plane CD , parallel to the opposite Planes , it shall be as the

Therefore the

**Cylinders**EB , CM , will be also equal . And because the**Cylinder**FM is cut by'a Plane CD , parallel to the opposite Planes , it shall be as the

**Cylinder**CM is to the**Cylinder**FD , so is the Axis LN , to the Axis KL . But the**Cylinder**... Side 267

And take PM equal to L K from MN ; and let the

Plane TYS , parallel to the opposite Planes of the Circles EFGH , RO , and

conceive ES to be a

And take PM equal to L K from MN ; and let the

**Cylinder**E O be cut thro ' P by thePlane TYS , parallel to the opposite Planes of the Circles EFGH , RO , and

conceive ES to be a

**Cylinder**, whose Base is the Circle EFGH , and Altitude PM .### Hva folk mener - Skriv en omtale

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Euclid's Elements of Geometry: From the Latin Translation of Commandine. To ... John Keill Uten tilgangsbegrensning - 1723 |

### Vanlige uttrykk og setninger

added alſo Altitude Angle ABC Baſe becauſe Center Circle Circle ABCD Circumference common Cone conſequently contained Cylinder demonſtrated deſcribed Diameter Difference Diſtance divided double draw drawn equal equal Angles equiangular Equimultiples exceeds fall fame firſt fore four fourth given greater half join leſs likewiſe Logarithm Magnitudes Manner mean Multiple Number oppoſite parallel Parallelogram perpendicular Place Plane Point Polygon Priſms produced Prop Proportion PROPOSITION proved Pyramid Radius Ratio Rectangle remaining Right Angles Right Line Right-lined Figure ſaid ſame ſame Reaſon ſay ſecond Segment Series ſhall ſhall be equal Sides ſimilar ſince Sine Solid ſome Sphere Square ſtand taken Terms THEOREM thereof theſe third thoſe thro touch Triangle Triangle ABC Unity Wherefore whole whoſe Baſe

### Populære avsnitt

Side 66 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.

Side 161 - IF two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals : the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.

Side 110 - And in like manner it may be shown that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM ; therefore the five angles GHK, HKL, KLM, LMG, MGH...

Side 88 - IN a circle, the angle in a semicircle is a right angle ; but the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.

Side 22 - ... sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB...

Side 9 - ... equal to them, of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF ; but the base CB greater than the base EF ; the angle BAC is likewise greater than the angle EDF.

Side 15 - CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which...

Side 33 - ... therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, (i.

Side 111 - If two right-angled triangles have their hypotenuses equal, and one side of the one equal to one side of the other, the triangles are congruent.