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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Results 1-5 of 42
Page 5
... feet ; and so on . If the measuring unit is not exactly contained in the pro- posed line , the length of the remaining portion must be expressed in certain fractional parts of the unit . To construct a scale of equal parts for comparing ...
... feet ; and so on . If the measuring unit is not exactly contained in the pro- posed line , the length of the remaining portion must be expressed in certain fractional parts of the unit . To construct a scale of equal parts for comparing ...
Page 15
... feet in the height of the tower = 335 ft . nearly . 4. To find the height of a steeple BC , whose base B is in- accessible . The surveyor measured a base line AD = he then took the CDB = 52 ° , and the Required the height C B of the ...
... feet in the height of the tower = 335 ft . nearly . 4. To find the height of a steeple BC , whose base B is in- accessible . The surveyor measured a base line AD = he then took the CDB = 52 ° , and the Required the height C B of the ...
Page 16
... feet in the height , as required , = 64 ft . 5. To find the height of a tower CB standing upon a hill AB . The sur- veyor measured in the same straight line BD = 50 ft . , and DA = = 75 ft .; he then found the CDB = 41 ° , ite and 24 ...
... feet in the height , as required , = 64 ft . 5. To find the height of a tower CB standing upon a hill AB . The sur- veyor measured in the same straight line BD = 50 ft . , and DA = = 75 ft .; he then found the CDB = 41 ° , ite and 24 ...
Page 32
... feet in a square yard . ( 4. ) Show that there are 144 square inches in a square foot . ( 5. ) Show that the square described upon the whole line is four times the square described upon one half the line . ( 6. ) Show that a rectangle 1 ...
... feet in a square yard . ( 4. ) Show that there are 144 square inches in a square foot . ( 5. ) Show that the square described upon the whole line is four times the square described upon one half the line . ( 6. ) Show that a rectangle 1 ...
Page 34
... feet . 2. Required the area of a parallelogram whose base is 5.6 ft . , and perpendicular 3-2 ft . Ans . 17.92 sq . ft . 3. Required the side of a square , which shall contain the same surface as a parallelogram whose base is 16 yds ...
... feet . 2. Required the area of a parallelogram whose base is 5.6 ft . , and perpendicular 3-2 ft . Ans . 17.92 sq . ft . 3. Required the side of a square , which shall contain the same surface as a parallelogram whose base is 16 yds ...
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.