| Henry Arthur Bethell - 1910 - 443 páginas
...Fig. 144, or by reducing it to one triangle, as in Fig. 145, working on the principle that any two **triangles on the same base and between the same parallels are equal.** GUNNERY CALCULATIONS. KIG. 144. Circle. F'g- 145Let r be the radius ; then Circumference = 2*r Area,... | |
| Alexander H. McDougall - 1910 - 302 páginas
...23 sq. cm. nearly; 3. 13-7 sq. cm. nearly; 4. 28'8 sq. cm.; 5. 36 ac. ; 6. 73 ac. nearly. THEOREM 5 **Triangles on the same base and between the same parallels are equal** in area. X BT Hypothesis.- — ABC, DEC are As on the same base BC and between the same ||s AD, BC.... | |
| Joseph Harrison, George Albert Baxandall - 1913 - 677 páginas
...DK. Then DKG is a triangle having an area equal to the polygon. This solution is based on the theorem **that triangles on the same base and between the same parallels are equal** in area. 31. PROBLEM. — To construct a square equal in area to a given rectangle. Determine a mean... | |
| Bennett Hooper Brough - 1920 - 477 páginas
...line parallel to AD, meeting CD produced, at G. Join A G. The method depends upon Euclids' theorem **that triangles on the same base and between the same parallels are equal** to one another. By using a parallel ruler and a pricker the drawing in of all the constructional lines,... | |
| Ivor Grattan-Guinness, Gerard Bornet - 1997 - 236 páginas
...geometry exhibited in the form of propositions are universal ie they have universal subjects. Ex. All **triangles on the same base and between the same parallels are equal.** All right angled triangles have this property that the square of the hypothenuse is equal to the squares... | |
| 1897
...; the method is mathematically accurate, and is based upon a familiar proposition of Euclid, viz., **that triangles on the same base, and between the same parallels, are equal** (vide Euc. I. 37). Suppose it is required to reduce the figure ABCDEF — which is supposed to be plotted... | |
| Education Department - 1879
...bisects it. If the diagonal also bisects the angles, show that the parallelogram is a rhombus. 2. Show **that triangles on the same, base and between the same parallels are equal** to each other. Hence show that a trapezium is equal in area to a triangle whose vertical height is... | |
| ...opposite sides of a straight line AB; join DQ, CP: prove that CDQP is a parallelogram. 4. (a) Prove **that triangles on the same base and between the same parallels are equal** in area. (6) FGH is a triangle, K is the mid.point of GH, and P is any point on FK ; prove that the... | |
| Thomas Hadyn Ward Hill
...parallelograms on the same base and between the same parallels are equal in area. From this we have **that triangles on the same base and between the same parallels are equal** in area, and the converse; and also expressions for the areas of parallelograms, triangles, quadrilaterals... | |
| 340 páginas
...depends only on its base and altitude. [Exercises 7 (a) and 7 (6) may now be taken.] Theorem 5 a. (i) **Triangles on the same base and between the same parallels are equal** in area. (ii) Conversely, if two triangles with equal areas stand on the same side of a common base,... | |
| |