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ON THE

Geometry and Measurement

OF

PLANE FIGURES.

BEING

Solutions of the Theorems, Problems and Questions,
in "Wormell's Modern Geometry."

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THOMAS MURBY,

32, BOUVERIE STREET, FLEET STREET, E.C.

183.

д

152.

EDUCATIONAL WORKS PUBLISHED BY THOMAS MURBY.

MATHEMATICAL WORKS.

By RICHARD WORMELL, M.A., B.Sc.

I.-Wormell's Modern Geometry: A New Course
of Plane Geometry, in which the Theory of the
Science and its Practical Applications are treated
simultaneously.
Crown 8vo, 256 pp. Price

3s. 6d.

The explanations and illustrations of the leading problems of Geometry here treated, and the logical demonstrations of their principles, place this work amongst the very best of the kind, and its easy and graduated method makes it at once the best that has hitherto appeared.— Evening Standard.

II.-An Elementary Course of Plane Geometry.
Third Edition. Revised and Enlarged. Price 3s.

III.-An Elementary Course of Solid Geometry.
Second Edition. Price 2s. 6d.

***These works (II. & III.) are adopted by the New Brunswick Council on Education as Text-books for use in schools throughout that province.

IV. Solutions to Exercises in Solid Geometry.
Price 2s. 6d.

V.-Solutions to Exercises in the Author's
Modern Geometry. Price 2s. 6d.

These Solutions are also adapted to the Elementary Course of Plane Geometry, (II. above,) the Exercises being the same both in that work and in the Modern Geometry.

VI.-Arithmetic for Schools and Colleges.
Revised Edition. Price 2s., with answers, 35.

VII.-Graduated Arithmetic. A Selection from the preceding work. Price 9d.

VIII.-Answers to Arithmetic for Schools and Colleges. Price Is.

THOMAS MURBY, 32, Bouverie Street, Fleet Street, London, E.C.

PART I.

THEOREMS AND PROBLEMS.

CHAPTERS I. AND II.

LINES AND ANGLES.

THEOREMS ON ANGLES.

1. The bisectors of the two adjacent angles formed when one straight line meets another are perpendicular to one another.

E

Let A CD and B C D be the adjacent angles and let C E bisect the angle A C D and CF bisect the angle BCD; then shall the angle ECF be a right

angle.

Because
and

:. The sum of

ECD is half ▲ ACD,

FCD is half

ECD and

F

B C D;

FCD is half the

sum of ACD and 4 B C D ;

But we know the sum of A CD and B C D to be

two right angles.

. 4 ECF, which is the sum of ECD and ▲ FCD, is one right angle.

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