## Elements of Geometry and Trigonometry: With an Easy and Concise System of Land SurveyingJames I. Cutler and Company, James I. Cutler & Company, printers, 1829 - 115 pages |

### From inside the book

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**three Sides**of a Triangle be given , its Area may be thus obtained : -**From half the sum of the three Sides**,**subtract each Side**sever- ally , then extract the Square Root of the continued**Product**of the**half sum**and three remainder for the ...### Common terms and phrases

20 Rods 40 Dist Area in Rods calculate the Area Chains and Links Circle Circumferentor Column of Half Column of Meridian Column of South contained Angle Course and Distance Decimal Dep Lat describe an Arch Diagonal Dividers draw a Line draw AC Draw the Line East EXAMPLE Field Book find the Angles find the Area given Line Half Meridian Distances Hypothenuse Instrument Intersection Lat Dep Latitude and Departure length Line AB Line BC Line CD Line of Chords measured Meridian Line multiplied Needle North Areas Northings or Southings number of Acres number of Degrees Parallel Lines Parallelogram Perpendicular Height PROBLEM Protract Random Line remainder Rhombus Right Angled Triangle Right Line Roods second Departure Column South Areas Square Rods Square Root Stake Station Stationary Distance Stationary Lines Subtract Survey Take the Course taken Trapezium Trapezoid Triangle ABC upper Meridian Distance upper numbers Westing

### Popular passages

Page 2 - District Clerk's Office. BE IT REMEMBERED, that on the tenth day of August, AD 1829, in the fifty-fourth year of the Independence of the United States of America, JP Dabney, of the said district, has deposited in this office the title of a book, the right whereof he claims as author, in the words following, to wit...

Page 8 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds.

Page 31 - From half the sum of the three sides subtract each side severally. Multiply the half sum and the three remainders...

Page 15 - The hypothenuse of a right-angled triangle may be found by having the other two sides ; thus, the square root of the sum of the squares of the base and perpendicular, will be the hypothenuse. Cor. 2. Having the hypothenuse and one side given to find the other; the square root of the difference of the squares of the hypothenuse and given side will be the required side.

Page 2 - States entitled an act for the encouragement of learning hy securing the copies of maps, charts and books to the author., and proprietors of such copies during the times therein mentioned, and also to an act entitled an act supplementary to an act, entitled an act for the encouragement of learning by securing the copies of maps, charts and books to the authors and proprietors of such copies during the times therein mentioned and extending the benefits thereof to the arts of designing, engraving and...

Page 30 - Base ; then multiply the Base by half the Perpendicular, or the Perpendicular by half the Base ; the Product will be the Area. Or, multiply the whole Base by the whole Perpendicular, and half the Product will be the Area.

Page 15 - KCML, the sum of the two parallelograms or square BCMH ; therefore the sum of the squares on AB and AC is equal to the square on EC. QED Cor. 1; The hypothenuse of a right-angled triangle may be found by...

Page 7 - Fig. 4. 12. The radius of a circle is a line drawn from the centre to the circumference ; as CB. Fig. 4. Therefore all radii of the same circle are equal. 13. The diameter of a circle is a right line drawn from one side of the circumference to the other, passing through the centre ; and it divides the circle into two equal parts, called semicircles ; as AB or DE.

Page 8 - The chord of an arc of 60 degrees is equal in length to the radius of the circle of which the arc is a part. 17. The segment of a circle is a part of a circle cut off by a chord.

Page 14 - C, with right lines, and bisect these lines; the point D, where the bisecting lines cross each other, will be the centre of the circle required. Therefore, place one...