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" 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 of the other part. "
The Elements of Euclid; viz. the first six books, together with the eleventh ... - Página 52
por Euclides - 1814
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Papers for the schoolmaster, Volumen3

1867
...the line between the points of section. , Illustrate this by Algebra. 2. Divide a given straight line into two parts, so that, the rectangle contained by...parts shall be equal to the square of the other part. Shew how to represent the square root of any number by a geometrical figure. What is meant by magnitudes...
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A Collection of Problems and Examples Adapted to the "Elementary Course of ...

Harvey Goodwin - 1851 - 173 páginas
...of algebraical equations, or any demonstration other than Euclid's ? 3. Divide a given straight line into two parts, so that the rectangle contained by...parts shall be equal to the square of the other part. Shew that in Euclid's figure four other lines, beside the given line, are divided in the required manner....
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THE PRINCIPLES OF THE SOLUTION OF SENATE-HOUSE 'RIDERS'

FRANCIS J. JAMESON - 1851
...AC2) + 4AE2 + 4AF2, = 4BC2 + AB2 + AC2, = 4BC2 + BC2, - 5BC8. 1849. (A). Divide a given straight line into two parts, so that the rectangle contained by...parts shall be equal to the square of the other part. (ii. 11.) (B). Shew that in Euclid's figure, four other lines, beside the given line, are divided in...
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University of Durham

University of Durham - 1851
...twice the rectangle contained by the parts. 5. To divide a given straight line into two parts, such that the rectangle contained by the whole and one...parts shall be equal to the square of the other part. 6. If two circles cut one another they cannot have the same centre. 7. Equal straight lines in a circle...
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BONNYCASTLE'S INTRODUCTION TO ALGEBRA; CONTAINING THE INDETERMINATE AND ...

JAMES RYAN - 1851
...to divide a line y of 20 inches in length, into two such parts that the rectangle of the whole arid one of the parts shall be equal to the square of the other. Ans. 10^/5 — 10, and 30— 10 ^5. 5. It is required to divide the number 60 into two such parts,...
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Scholarship examinations of 1846/47 (-1853/54).

Bengal council of educ - 1852
...angle. Is this proposition included in any more general one ? (2.) To divide a given straight line into two parts, so that the rectangle contained by...parts, shall be equal to the square of the other part. Can this be solved arithmetically ? if so, find approximately into how many parts the given line must...
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Five Years in an English University, Volumen2

Charles Astor Bristed - 1852
...every triangle are equal to two right angles. 3. Divide a given straight line into two such parts, that the rectangle contained by the whole and one...parts shall be equal to the square of the other part. 4. The angle at the centre of a circle is double of the angle at the circumference, upon the same base,...
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General Report on Public Instruction in the Bengal Presidency

1852
...general one ? 2. To divide a given straight line into two parts, so that the rectangle contained hy the whole and one of the parts, shall be equal to the square of the other part. Can this be solved arithmetically ? if so, find approximately into how many parts the given line must...
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The first two books of the Elements of Euclid, with additional figures ...

Euclides - 1852
...PROP. X. PEOB. To bisect a (jiven finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line; it is required to divide it into two equal parts. Describe " upon it an equilateral triangle ABC, and bisect b the angle ACB by the straight...
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Arithmetic, algebra, differential and integral calculus, by W. Rutherford ...

Royal Military Academy, Woolwich - 1853
...divide a line of 20 inches in length into two parts such that the rectangle contained by the whole line and one of the parts shall be equal to the square of the other part. Let x = the greater part ; then 20 — x = the less part, and hence by the question we have x8 = 20...
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