A Graduated Course of Elementary Problems in Practical Plane Geometry (Paperback)


This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1870 edition. Excerpt: ...with the given point p, and through D A. and E draw lines parallel to cP, meeting cA and CD in G and h Bisect H B in K, and join p K. Then p G, (c) BY STRAIGhT LINES dRAWN fROM A GIVEN POINT WITh-IN The Triangle. Divide one side X B into the re-quired number of equal parts in D, E and F. Join c with the given point P, and join P F. Through c draw C G parallel to P F; and join P G. Join p E; and through c draw c h parallel to p E. Join Ph and Pa. Bisect c h in K, and through K draw K L parallel to p A. Join p L. Then p G, P H, P L and P C divide A B C into four equal parts. If the given point be near the centre of the triangle a second side may have to be divided similarly to the first. Similarly the triangle may be divided into proportional parts. PROBLEM VII.--To bisect a given triangle. (a) By A Straight Line PARALLEL TO ThE BASE. A B c=given triangle. In one side c B find the point e, so that the square on Ce may be equal to half the square on c B (i. e. On c B describe the semicircle Cdb; bisect Cdb in D; join Cd; and from c, with Cd as radius, cut C B in E). From E draw E F parallel to the base A B. Then Ef bisects the triangle Abc--(Euc. vi. 19 and 20.) (J) BY A STRAIGhT LINE PERPENdICULAR TO ThE BASE. Bisect the base A B of the triangle A B C in D. From C draw CE perpendicular to A B. Find the point F, so that A F may be a mean proportional between the greater segment of the base (ae) made by c E, and half the base (ad). Through f draw F G parallel to E c. Then F G bisects the triangle Abc PROBLEM VIII.--To construct a triangle equal to a given triangle. (a) GIVEN THE BASE Of ThE REQUIREd TRIANGLE. In the base Ab (or base produced) of the given triangle Abc set off Ad equal to the given base. Join Cd. Through B draw a line parallel...

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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1870 edition. Excerpt: ...with the given point p, and through D A. and E draw lines parallel to cP, meeting cA and CD in G and h Bisect H B in K, and join p K. Then p G, (c) BY STRAIGhT LINES dRAWN fROM A GIVEN POINT WITh-IN The Triangle. Divide one side X B into the re-quired number of equal parts in D, E and F. Join c with the given point P, and join P F. Through c draw C G parallel to P F; and join P G. Join p E; and through c draw c h parallel to p E. Join Ph and Pa. Bisect c h in K, and through K draw K L parallel to p A. Join p L. Then p G, P H, P L and P C divide A B C into four equal parts. If the given point be near the centre of the triangle a second side may have to be divided similarly to the first. Similarly the triangle may be divided into proportional parts. PROBLEM VII.--To bisect a given triangle. (a) By A Straight Line PARALLEL TO ThE BASE. A B c=given triangle. In one side c B find the point e, so that the square on Ce may be equal to half the square on c B (i. e. On c B describe the semicircle Cdb; bisect Cdb in D; join Cd; and from c, with Cd as radius, cut C B in E). From E draw E F parallel to the base A B. Then Ef bisects the triangle Abc--(Euc. vi. 19 and 20.) (J) BY A STRAIGhT LINE PERPENdICULAR TO ThE BASE. Bisect the base A B of the triangle A B C in D. From C draw CE perpendicular to A B. Find the point F, so that A F may be a mean proportional between the greater segment of the base (ae) made by c E, and half the base (ad). Through f draw F G parallel to E c. Then F G bisects the triangle Abc PROBLEM VIII.--To construct a triangle equal to a given triangle. (a) GIVEN THE BASE Of ThE REQUIREd TRIANGLE. In the base Ab (or base produced) of the given triangle Abc set off Ad equal to the given base. Join Cd. Through B draw a line parallel...

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Product Details

General

Imprint

Rarebooksclub.com

Country of origin

United States

Release date

2013

Availability

Supplier out of stock. If you add this item to your wish list we will let you know when it becomes available.

First published

2013

Authors

Dimensions

246 x 189 x 1mm (L x W x T)

Format

Paperback - Trade

Pages

24

ISBN-13

978-1-234-37213-2

Barcode

9781234372132

Categories

LSN

1-234-37213-4



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