Principles of geometry, mensuration, trigonometry, land-surveying, and levelling, with their application to the solution of practical problems in estimation, surveying and railway engineering |
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Page iii
... INSCRIBED , AS A TRIBUTE OF ESTEEM , FOR THE IMPROVEMENTS THEY HAVE EFFECTED IN OUR METHODS OF EDUCATION , BY THEIR HUMBLE AND MOST OBEDIENT SERVANT , THE AUTHOR . PREFACE . THE present work is intended to supply Teachers.
... INSCRIBED , AS A TRIBUTE OF ESTEEM , FOR THE IMPROVEMENTS THEY HAVE EFFECTED IN OUR METHODS OF EDUCATION , BY THEIR HUMBLE AND MOST OBEDIENT SERVANT , THE AUTHOR . PREFACE . THE present work is intended to supply Teachers.
Page 62
... inscribe a square in a given circle . * If the angles are incommensurable . Let the less angle DQK be placed upon the greater , so that LACE LDQK , and arc AE = arc DK ; then if the theorem above demonstrated is not true , let LACB LDQK ...
... inscribe a square in a given circle . * If the angles are incommensurable . Let the less angle DQK be placed upon the greater , so that LACE LDQK , and arc AE = arc DK ; then if the theorem above demonstrated is not true , let LACB LDQK ...
Page 63
... inscribed , by bisecting the arcs AD , DC , & c . , and joining these points of division . Exercises . 1. Let r the radius of the circle ; required the side of the inscribed square . From the right - angled triangle AEB , we have , AB2 ...
... inscribed , by bisecting the arcs AD , DC , & c . , and joining these points of division . Exercises . 1. Let r the radius of the circle ; required the side of the inscribed square . From the right - angled triangle AEB , we have , AB2 ...
Page 64
... inscribed hexagon is equal to the radius of the circle . Cor . 1. The chord of 60 ° is equal to the radius . Cor . 2. By joining the points A , C , and E , we should ob- viously form the inscribed equilateral triangle . Cor . 3. Regular ...
... inscribed hexagon is equal to the radius of the circle . Cor . 1. The chord of 60 ° is equal to the radius . Cor . 2. By joining the points A , C , and E , we should ob- viously form the inscribed equilateral triangle . Cor . 3. Regular ...
Page 65
... inscribed circle . Ans . 2 . 3. The base of a right - angled triangle is 18 , and the hy- potenuse is 30 ; required the radius of the inscribed circle . Ans . 6 . 70. THEOREM . A circle may be described about , or in- scribed within ...
... inscribed circle . Ans . 2 . 3. The base of a right - angled triangle is 18 , and the hy- potenuse is 30 ; required the radius of the inscribed circle . Ans . 6 . 70. THEOREM . A circle may be described about , or in- scribed within ...
Common terms and phrases
ABCD base bisect breadth centre chain-line chains chord circum circumference cone construct convex surface cosine cubic cubic foot curve cutting cylinder describe diagonal diameter dicular distance divided draw equilateral triangle feet field-book figure find the area foot formula frustum given point hence hypotenuse inches inscribed isosceles triangle join labour length let fall logarithm measured miles Multiply number of degrees parallel parallelogram perpen perpendicular height plane polygon prism PROBLEM protractor pyramid radius rectangle regular polygon Required the area Required the cost Required the solidity right angles right-angled triangle scale of equal sector similar triangles sine slant height sphere square station straight line tangent theodolite THEOREM thickness trapezium trapezoid triangle ABC Trigonometry vernier
Popular passages
Page 143 - From half the sum of the three sides, subtract each side severally; multiply the half sum, and the three remainders together, and the square root of the product will be the area required. Example. — Required the area of a triangle, whose sides are 50, 40, and 30 feet. 50 + 40+30 ; — 60, half the sum of the three sides.
Page 66 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
Page 170 - RULE. To the sum of the areas of the two ends add four times the area of a section parallel to and equally distant from both ends ; multiply this sum by the perpendicular height, and J of the product will be the solidity.
Page 167 - To find t/ie solidity of a cone, or pyramid. — RULE. — Multiply the area of the base by the perpendicular height, and one-third of the product will be the content To find the solidity of the frustum of a cone.
Page 37 - Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area.
Page 146 - To find the area of a trapezoid. RULE. Multiply half the sum of the two parallel sides "by the perpendicular distance between them : the product will be the area.
Page 164 - To find the solidity of a cylinder. RULE. — Multiply the area of the base by the altitude, and the product will be the solidity.
Page 169 - To twice the length of the base add the length of the edge ; multiply the sum by the breadth of the base, and by one-sixth of the height.
Page 110 - I label the two new points e and /." FIG. 2 With the help of this figure he then proceeds to the usual proof of the theorem that the area of a parallelogram is equal to the product of the base by the altitude, establishing the equality of certain lines and angles and the congruence of the pair of triangles.