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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 ... - Side 52
av Euclides - 1814
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The synoptical Euclid; being the first four books of Euclid's Elements of ...

Euclides - 1853 - 146 sider
...X. PROBLEM. To bisect a given 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 (I. 1.) upon it an equilateral triangle ABC, and bisect (I. 9.) the angle ACB...
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The Elements of Euclid. Books I.-VI.; XI. 1-21 XII. 1-2. A New Text ..., Bok 1

Euclides - 1853 - 334 sider
...contained by the whole line and one of the parts shall be equal to the square of the other part. Let AB be the given straight line. It is required to divide it into two such parts, that the rectangle contained by the whole line and one of the parts shall be equal to the...
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The geometry, by T. S. Davies. Conic sections, by Stephen Fenwick

Royal Military Academy, Woolwich - 1853
...PROPOSITION X. PROB. To bisect a given 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 (1. i.) upon it an equilateral triangle ABC, and bisect (9. i.) the angle ACB...
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The first six books of the Elements of Euclid, with numerous exercises

Euclides - 1853 - 176 sider
...— PROBLEM. To bisect a given 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 (i. 1) upon ab an equilateral triangle abc, and bisect (i. 9) the angle acb by...
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Solutions of the problems and riders proposed in the Senate-house ...

Cambridge univ, exam. papers - 1854 - 284 sider
...KBD. 2. Divide a given straight line into two parts so that the rectangle contained by the whole line and one of the parts shall be equal to the square of the other part. Produce a given straight line to a point such that the rectangle contained by the whole line thus produced...
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Solutions of the Problems and Riders Proposed in the Senate-house ...

William Walton, Charles Frederick Mackenzie - 1854 - 266 sider
...KBD. 2. Divide a given straight line into two parts so that the rectangle contained by the whole line and one of the parts shall be equal to the square of the other part. Produce a given straight line to a point such that the rectangle contained by the whole line thus produced...
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Orr's Circle of the Sciences: The mathematical sciences

William Somerville Orr - 1854
...the remarks at the end of this book. PROPOSITION XI.— PROBLEM. To divide a given straight line (AB) into two parts, so that the rectangle contained by the whole, and от of the parts, shall be equal to the square of the other part. From A draw AC perpendicular and...
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Minutes of the Committee of Council on Education

Great Britain. Committee on Education - 1855
...two sides of it, the angle contained by those two sides is a right angle. 3. Divide a straight line into two parts, so that the rectangle contained by...parts shall be equal to the square of the other part. Section 3. 1. The angles in the same segment of a circle are equal. 2. If from a point without a circle...
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Liber Cantabrigiensis, an account of the aids afforded to poor students, the ...

Robert Potts - 1855
...two equal parts. Prove also that the two diameters bisect each other. 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 Solve the same problem algebraically, and give the geometrical interpretation of the result. 4. The...
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The Elements of geometry; or, The first six books, with the eleventh and ...

Euclides - 1855
...PROBLEM. To bisect a given finite straight line. ; thai 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 AB (I. 1) au equilateral triangle ABC, and bisect (I. 9) the angle ACB by...
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