 | J. W. Riley - 1905 - 518 σελίδες
...height area = base xr — c S— . 2i The shape of the triangle does not affect the result, as all triangles on the same base and between the same parallels are equal in area. The area of a triangle can also be determined by the following formula : \^s(s - a)(s - b)(x... | |
 | Joseph Gregory Horner - 1909 - 560 σελίδες
...angles; and the three interior angles of every triangle are equal to two right angles. (Prop. 32.) Triangles on the same base and between the same parallels are equal to one another. (Prop. 37.) Triangles on equal bases and between the same parallels are equal to one... | |
 | Alexander H. McDougall - 1910 - 316 σελίδες
...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.... | |
 | Henry Arthur Bethell - 1910 - 576 σελίδες
...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,... | |
 | Henry John Spooner - 1911 - 196 σελίδες
...equal triangles are on the same base and between the same parallels. Euclid (I. 37) proves that all triangles on the same base and between the same parallels are equal in area. Area=l S / r \ q .o O W, 4/ -ea=l 's 9/, h 7C 90. To construct a Rectangle equal in Area to... | |
 | Thomas Aloysius O'Donahue - 1911 - 288 σελίδες
...method of casting out areas is that known as " equalising," which is based on Euclid's proposition of " triangles on the same base and between the same parallels are equal." The method consists in constructing a triangle equal to the figure, the area of which is required,... | |
 | Joseph Harrison, George Albert Baxandall - 1913 - 714 σελίδες
...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... | |
 | John William Angles - 1919 - 200 σελίδες
...lines. .-. Area of triangle = | x base x perpendicular height = \bh <s— — —— — • —SB^ Triangles on the same base and between the same parallels are equal in area. The line joining the apex of a triangle to the middle point of the opposite side is called... | |
 | Bennett Hooper Brough - 1920 - 510 σελίδες
...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,... | |
 | Newfoundland Council of Higher Education - 1922 - 204 σελίδες
...Grade.) Wednesday, June. 28th, 1922.— Morning, 9 to 12. Recognized abbreviations may be used. 1. Prove that triangles on the same base and between the same parallels are equal in area. A straight line parallel to the diagonal DB of a parallelogram A BCD meets CD and CB in E... | |
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... that triangles on the same base and between the same parallels are equal... 
