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" Prove that parallelograms on the same base and between the same parallels are equal in area. "
Examination Papers for Science Schools and Classes - Page 23
by Great Britain. Education Department. Department of Science and Art - 1882
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Euclid Revised: Containing the Essentials of the Elements of Plane Geometry ...

Euclid - Geometry - 1890 - 442 pages
...quadrilateral is a parallelogram if its diagonals bisect each other. 39 Proposition 35. THEOREM — Parallelograms on the same base, and between the same parallels, are equal in area. Q CO (2) SP RQ SP R (.3) PS B B Let ABPQ, ABRS be on same base AB, and between same \\s AB,...
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The Harpur Euclid: An Edition of Euclid's Elements

Edward Mann Langley, W. Seys Phillips - 1890 - 538 pages
...be concurrent. Note that 'the three medians of a triangle are concurrent.' PROPOSITION 35. THEOREM. Parallelograms on the same base and between the same parallels are equal to one another. Let ||gms A BCD, EBCF be on the same base and between the same ||s AF, BC, ABCD shall...
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Rider Papers on Euclid (books I. and II.)

Rupert Deakin - Euclid's Elements - 1891 - 102 pages
...to one another, and the diameter bisects the parallelogram, ie divides it into two equal parts. 35. Parallelograms on the same base and between the same parallels are equal to one another. 36. Parallelograms on equal bases and between the same parallels are equal to one another....
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A Text-book of Geometrical Deductions

James Andrew Blaikie, William Thomson - Geometry - 1891 - 160 pages
...and opposite angles of a parallelogram are equal, and either diagonal bisects the parallelogram. 35. Parallelograms on the same base and between the same parallels are equal. 36. Parallelograms on equal bases and between the same parallels are equal. 37. Triangles on the same...
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Treatise on Thermodynamics

Peter Alexander - Thermodynamics - 1892 - 228 pages
...are parallel to OX, and the lines hX', kN, Eu, L8, Da, PJTare parallel to OF. From Euclid's theorem that parallelograms on the same base and between the same parallels are equal, we get ASCD = AKQD = AKPk = AH . Ak, = ABUL = AISL =AL . Al, = AMWD = AMRm = AM. Am, = ABX'N= An VN...
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Calendar ...

University College, Galway - 1896 - 430 pages
...+ x by a? - x + 1. 5' Simplify (*+l)H (*-!). 6. Solve the equation x + 5 x - 6 x + 3 ~ x - 9' 7. Parallelograms on the same base and between the same parallels are equal in area. 8. If a straight line is bisected and also divided unequally the sum of the squares on the...
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Report of the Council of Public Instruction of the North-West Territories of ...

Northwest Territories Council of Public Instruction - 1897 - 628 pages
...(i) with no pair parallel ; (ii) with one pair parallel ; (iii) with two pairs parallel. (J) Prove that parallelograms on the same base and between the same parallels are equal in area. I. 35. (f) Point out the uses of proposition (I) in mensuration. (d) Prove that if a square...
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Longmans' School Mensuration: With an Additional Chapter and Exercises

Alfred John Pearce - 1897 - 202 pages
...parallelogram, the length and perpendicular height being given. , Let ABCD be an oblique parallelogram. Now, parallelograms on the same base and between the same parallels are equal (Euc. I. 35). Therefore the area of the oblique parallelogram ABCD equals the area of the rectangle...
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Proceedings of the American Association for the Advancement of ..., Volume 48

American Association for the Advancement of Science - Science - 1899 - 650 pages
...geometry, and in the third part a strict treatment of equivalence. Even Euclid, in proving his I. 35, "Parallelograms on the same base, and between the same parallels, are equal to one another," does notshowthat the parallelograms can be divided into pairs of pieces admitting...
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Proceedings, Volume 48

American Association for the Advancement of Science - American periodicals - 1899 - 646 pages
...geometry, and in the third part a strict treatment of equivalence. Even Euclid, in proving his I. 35, " Parallelograms on the same base, and between the same parallels, are equal to one another," does not show that the parallelograms can be divided into pairs of pieces admitting...
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