| Civil service commission - 1883
...stated as a corollary to that proposition. g. — (1) 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. (2) If А В be so divided in C, and D be the middle... | |
| Joseph Hughes - 1883
...rectangle contained by the parts. Euclid II., 4. 3. To divide a given straight line into two parts, so **that the rectangle contained by the whole, and one of the parts,** shall be equal to the square on the other part. Euclid П., u. I. Prove that (a+i + c) Algebra. - (30... | |
| SHEPPARD & ST. JOHN - 1884
...angles by the straight line joining their centres. 3. Divide a given straight line into two parts so **that the rectangle contained by the whole and one of the parts** shall be equal to the square of the other part. 4. The angles in the same segment of a circle are equal... | |
| Cambridge univ, exam. papers - 1884
...together equal to the square on the whole line. 5. 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. Prove that the rectangle contained by the two parts is... | |
| F. B. Stevens - 1884 - 175 páginas
...side of the secant line is equal to two right angles. 3. To divide a 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. 4. To divide a given line into two parts, such that the... | |
| Stewart W. and co - 1884
...parts shall be equal to the square of the other part. It is required to divide AB into two parts, so **that the rectangle contained by the whole and one of the parts** shall be equal to the square of the other part. Upon AB describe the square ACDB ; T • bisect AC... | |
| Oxford univ, local exams - 1885
...of its angles equal to a given rectilineal angle. 8. 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. WEDNESDAY, DECEMBER 12, from 9.30 AM to 12. 4 (I). Algebra.... | |
| GEORGE BRUCE HALSTED - 1885
...AG2 = GF* = GH2 = AH2 + A(?, by 243 ; 112 PROBLEM II. 313. To divide a given sect into two parts so **that the rectangle contained by the whole and one of the parts** shall be equal to the square on the other part. Jf X GIVEN, the sect AB. REQUIRED, to find a point... | |
| 1885
...equal to one another, and that the diagonal bisects it. 6. Divide a straight line into two parts, so **that the rectangle contained by the whole and one of the parts** shall be equal to the square of the other part. Female Candidates for Class I. will omit tit* 1st ami... | |
| George Bruce Halsted - 1886 - 366 páginas
...AG* = GF* = GH* = AH* + AG*, by 243 J 112 PROBLEM II. 313. To divide a given sect into two parts so **that the rectangle contained by the whole and one of the parts** shall be equal to the square on the other part. Jf X GIVEN, the sect AB. REQUIRED, to find a point... | |
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