## The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The first six books, together with the eleventh and twelfth |

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Resultat 1-5 av 5

Side 262

Book XII . greater than the half of the circle ABCD + . upon the square ABCD bop

erect a prism of the fame altitude with the

of the

Book XII . greater than the half of the circle ABCD + . upon the square ABCD bop

erect a prism of the fame altitude with the

**cylinder**; this prism is greater than halfof the

**cylinder**; because if a square be described about the circle , and a prism ... Side 272

... and are therefore all equal to one anoB.15 Def.1 . ther . therefore the line GH is

the circumference of a circle 6 of which the center is the point K. therefore the

plane GH divides the

same ...

... and are therefore all equal to one anoB.15 Def.1 . ther . therefore the line GH is

the circumference of a circle 6 of which the center is the point K. therefore the

plane GH divides the

**cylinder**AD into the**cylinders**AH , GD ; for they are thesame ...

Side 273

KF and

KM , the

KF and

**cylinder**GD , any equimultiples whatever , viz . the axis Book XII . KM and**cylinder**GQ , and it has been demonstrated if the axis KL be greater than the axisKM , the

**cylinder**PG is greater than the**cylinder**GQ ; and if equal , equal , and ... Side 274

See N. THE " HE bases and altitudes of equal cones and

reciprocally proportional ; and if the bases and altitudes be reciprocally

proportional , the cones and

the circles ABCD ...

See N. THE " HE bases and altitudes of equal cones and

**cylinders**arereciprocally proportional ; and if the bases and altitudes be reciprocally

proportional , the cones and

**cylinders**are equal to one another . b , A. 5 . 3 Letthe circles ABCD ...

Side 275

But let the bases and altitudes of the

proportional , viz . the base ABCD to the base EFGH , as the altitude MN to the

altitude KL . the

be equal ...

But let the bases and altitudes of the

**cylinders**AX , EO , be reciprocallyproportional , viz . the base ABCD to the base EFGH , as the altitude MN to the

altitude KL . the

**cylinder**AX is equal to the**cylinder**EO . First , let the base ABCDbe equal ...

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added alſo altitude angle ABC angle BAC baſe becauſe Book Book XI caſe circle circle ABCD circumference common cone cylinder Definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides firſt folid fore four fourth given angle given in poſition given in ſpecies given magnitude given ratio given ſtraight line greater Greek half join leſs likewiſe magnitude manner meet muſt oppoſite parallel parallelogram perpendicular plane priſm produced PROP proportionals Propoſition pyramid radius reaſon rectangle rectangle contained remaining right angles ſame ſecond ſegment ſhall ſhewn ſides ſimilar ſolid ſphere ſquare ſquare of AC taken THEOR theſe third thro triangle ABC wherefore whole

### Populære avsnitt

Side 157 - D ; wherefore the remaining angle at C is equal to the remaining angle at F ; Therefore the triangle ABC is equiangular to the triangle DEF. Next, let each of the angles at C, F be not less than a right angle ; the triangle ABC is also, in this case, equiangular to the triangle DEF.

Side 24 - ... be equal, each to each ; and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, ECA equal to the angles DEF, EFD, viz.

Side 45 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.

Side 73 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...

Side 163 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.

Side 81 - ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal (4.

Side 323 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.

Side 80 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.

Side 36 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.