Principles of geometry, mensuration, trigonometry, land-surveying, and levelling, with their application to the solution of practical problems in estimation, surveying and railway engineering |
From inside the book
Results 1-5 of 44
Page 8
... diameter , having two saw slits AC and DB cut at right angles to each other . This circular piece of wood is placed upon a staff which has a pointed end to enable the surveyor to push it T D into the ground . In order to show the use of ...
... diameter , having two saw slits AC and DB cut at right angles to each other . This circular piece of wood is placed upon a staff which has a pointed end to enable the surveyor to push it T D into the ground . In order to show the use of ...
Page 10
... diameter . Now what is the length of the diameter as compared with the radius ? P. The diameter will be double the radius ; because the diameter BF is made up of the radii AB and A F. Problems . 1. The height of a wall BC is 20 ft ...
... diameter . Now what is the length of the diameter as compared with the radius ? P. The diameter will be double the radius ; because the diameter BF is made up of the radii AB and A F. Problems . 1. The height of a wall BC is 20 ft ...
Page 12
... diameter divides the circle into two equal parts . For if BCDF be folded over upon BF , the curved line BCDF will exactly cover the curved line BLGF , because every point in these curved lines is at the same distance from the centre A ...
... diameter divides the circle into two equal parts . For if BCDF be folded over upon BF , the curved line BCDF will exactly cover the curved line BLGF , because every point in these curved lines is at the same distance from the centre A ...
Page 55
... the circle at the point A , then the DAQ , cut off by the A chord AD , is equal to the B , at the cir cumference of the circle , standing on the arc AD cut off . Q Ꭰ B For since AB is a diameter , [ QAB = D 4 PRINCIPLES OF GEOMETRY . 55.
... the circle at the point A , then the DAQ , cut off by the A chord AD , is equal to the B , at the cir cumference of the circle , standing on the arc AD cut off . Q Ꭰ B For since AB is a diameter , [ QAB = D 4 PRINCIPLES OF GEOMETRY . 55.
Page 56
Thomas Tate (mathematical master.) For since AB is a diameter , [ QAB = a right angle ; but < DAB + B = a right angle ; .. LQAB = LDAB + LB ; from these equals take away DAB ; :: LDAQ = / B. Applications of the preceding theorems . 1 ...
Thomas Tate (mathematical master.) For since AB is a diameter , [ QAB = a right angle ; but < DAB + B = a right angle ; .. LQAB = LDAB + LB ; from these equals take away DAB ; :: LDAQ = / B. Applications of the preceding theorems . 1 ...
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.