## The First Six and the Eleventh and Twelfth Books of Euclid's Elements: With Notes and Illustrations, and an Appendix in Five Books |

### Inni boken

Side 59

Hence , from either of the triangles ABD , ACD , we find the perpendicular to be

15 ; and thence the area is found , by taking

210 . The segments of the base are more easily found by means of the 4th ...

Hence , from either of the triangles ABD , ACD , we find the perpendicular to be

15 ; and thence the area is found , by taking

**half**the product of BC and AD , to be210 . The segments of the base are more easily found by means of the 4th ...

Side 207

D A If from the greater of two unequal magnitudes , there be taken more than its

remain a magnitude less than the least of the proposed magnitudes . Let AB and

C ...

D A If from the greater of two unequal magnitudes , there be taken more than its

**half**, and from the remainder more than its**half**; and so on : there will at lengthremain a magnitude less than the least of the proposed magnitudes . Let AB and

C ...

Side 219

In the circle AC describe a square ; this square is greater than the

circle : and upon the square ABCD form a pyramid having the same vertex with

the cone ; this pyramid is greater than the

before ...

In the circle AC describe a square ; this square is greater than the

**half**of thecircle : and upon the square ABCD form a pyramid having the same vertex with

the cone ; this pyramid is greater than the

**half**of the cone ; because , as wasbefore ...

Side 241

are equal : therefore ECD is

because AD , AC are equal , AH common , and the contained angles equal , the

angles at H ( I. 4. ) are equal , and are therefore right angles . Then , in the

triangles ...

are equal : therefore ECD is

**half**of BED . Again , in the triangles AHD , AHC ,because AD , AC are equal , AH common , and the contained angles equal , the

angles at H ( I. 4. ) are equal , and are therefore right angles . Then , in the

triangles ...

Side 306

the remainder is A + B. Take the

as a + b : a - b :: tan } ( A + B ) : tan ( A— B ) . This analogy gives

difference of A and B ; and * by adding this and } ( A + B ) together , A , the greater

angle ...

the remainder is A + B. Take the

**half**of this , and then , by the third proposition ,as a + b : a - b :: tan } ( A + B ) : tan ( A— B ) . This analogy gives

**half**thedifference of A and B ; and * by adding this and } ( A + B ) together , A , the greater

angle ...

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The First Six and the Eleventh and Twelfth Books of Euclid's Elements: With ... Euclid Ingen forhåndsvisning tilgjengelig - 2016 |

### Vanlige uttrykk og setninger

ABCD altitude base bisected called centre chord circle circumference coincide common cone consequently const construction contained continual cylinder demonstrated describe diagonal diameter difference divided double draw equal equal angles extremities figure fore four fourth given given circle given point given straight line greater half Hence inscribed join less magnitudes manner means meet method multiple opposite parallel parallelepiped parallelogram pass perpendicular plane polygon prism PROB produced proof PROP proportional proposition proved pyramid radius ratio reason rectangle remaining respectively right angles Schol segments semicircle shown sides similar square straight line taken THEOR third touching triangle triangle ABC twice vertical wherefore whole

### Populære avsnitt

Side 94 - 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 53 - If a straight line be divided into any two parts, the squares of the whole line and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts at the point C : the squares of AB, BC shall be equal to twice the rectangle AB, BC, together with the square of AC.

Side 143 - 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 4 - A rhombus is that which has all its sides equal, but its angles are not right angles.

Side 57 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.

Side 138 - IF a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those produced, proportionally; and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle...

Side 43 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Side 32 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Side 40 - 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.

Side 36 - PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another...