| WILLIAM GEORGE SPENCER - 1877
...You recollect that plane figure that has the fewest lines possible for its boundaries. A body which **has four plane, equal, and similar surfaces, is called...another ? 138. Can you place a circle, whose radius is** If inch, so that its circumference may touch two points 4 inches asunder ? 139. How many squares may... | |
| William George Spencer - 1876 - 97 páginas
...name. 135. Make two equal and similar rhomboids, and divide one into two equal and similar trianpies **by means of one diagonal, and the other into two equal...another ? 138. Can you place a circle, whose radius is** \\ inch, so that its circumference may touch two points 4 inches asunder \ 139. How many squares may... | |
| Edward Atkins - 1877
...two equal parts. Therefore, the opposite sides, &c. QED Proposition 35. — Theorem. Parallelograms **upon the same base, and between the same parallels, are equal to one another.** Let the parallelograms ABCD, EBCF be on the same base BC, and between the same parallels AF, BC ; The... | |
| D. Tierney - 1877
...not intersect, and therefore the construction for the triangle required would fail. 2. Parallelograms **upon the same base, and between the same parallels, are equal to one another.** Shew that if two triangles have two sides of the one equal to two sides of the other, each to each,... | |
| Alfonzo Gardiner - 1878
...= _. 7. The difference of two numbers is 14, and their sum is 48 : find the numbers. 8. Prove that **triangles upon the same base, and between the same parallels, are equal to one another.** 9. What do you mean by the " complements of a parallelogram " and by " applying a parallelogram to... | |
| J T. Amner - 1878
...equal to two right angles. What ratio does the angle of a regular hexagon bear to a right angle ? 2. **Triangles upon the same base and between the same parallels are equal.** A line drawn through the middle points of the sides of a triangle is parallel to the base. 3. In any... | |
| Thomas Hunter - 1878 - 132 páginas
...same reason EFGH is equal to ABGH. Hence ABCD and EFGH are equal (Ax. 1). PROPOSITION VIII.—THEOREM. **Triangles upon the same base, and between the same parallels, are equal** ABC and ABD have the same base, AB, and be between the same parallels, AB and CD; then will these two... | |
| Moffatt and Paige - 1879
...parallelogram EFGH (Ax. 1). Therefore, parallelograms upon equal bases, etc. QED Proposition XXXVII. Theorem. **Triangles upon the same base, and between the same parallels, are equal to one another.** Let the triangles ABC, DBC be upon the same base BC and between the same parallels AD, B C. Then the... | |
| Great Britain. Civil Service Commission - 1879
...same side ; and also the two interior angles on the same side together equal to two right angles. 3. **Triangles upon the same base and between the same parallels are equal to one another.** Let ABC, ABD, be two equal triangles upon the same base AB, and on opposite sides of it ; if CD be... | |
| W J. Dickinson - 1879 - 36 páginas
...angle. 36. Parallelograms upon equal bases and between the same parallels are equal to one another. 37. **Triangles upon the same base and between the same parallels are equal to one another.** Any point P is taken in the line joining an angular point A of a triangle to the middle point of the... | |
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