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 141
... cost at 2s . 3d . per sq . foot . By decimals . 7.25 ft By duodecimals . 3.5 ft . 3625 7 ft . 3 ' 3 6 21 9 2175 3 7 ... Cost = 2s . 3d . × 253 = £ 2 17s . 1d . 4. Required the area of a rectangular table 9 ft MENSURATION OF SURFACES . 141.
... cost at 2s . 3d . per sq . foot . By decimals . 7.25 ft By duodecimals . 3.5 ft . 3625 7 ft . 3 ' 3 6 21 9 2175 3 7 ... Cost = 2s . 3d . × 253 = £ 2 17s . 1d . 4. Required the area of a rectangular table 9 ft MENSURATION OF SURFACES . 141.
Page 142
Thomas Tate (mathematical master.) 4. Required the area of a rectangular table 9 ft . 6 in . long , and 4 ft . 3 in . broad . Ans . 40 ft . 4 ′ 6 ′′ . 5. Required the cost , in the last example , at 1s . 6d . per sq . foot . Ans . £ 3 ...
Thomas Tate (mathematical master.) 4. Required the area of a rectangular table 9 ft . 6 in . long , and 4 ft . 3 in . broad . Ans . 40 ft . 4 ′ 6 ′′ . 5. Required the cost , in the last example , at 1s . 6d . per sq . foot . Ans . £ 3 ...
Page 163
... Required the cost of a block of stone 8 ft . 4 in . long , 2 ft . 6 in . broad , and 1 ft . 3 in . deep ; at 10d . per cubic foot . Ans . £ 1 . 1s . 81d . 4. What will be the cost of a log of timber 18 ft . long , 1 ft . 8 in . broad ...
... Required the cost of a block of stone 8 ft . 4 in . long , 2 ft . 6 in . broad , and 1 ft . 3 in . deep ; at 10d . per cubic foot . Ans . £ 1 . 1s . 81d . 4. What will be the cost of a log of timber 18 ft . long , 1 ft . 8 in . broad ...
Page 164
... cost of hewing all the faces of the stone in example 3 ; at 5d . per square foot ? Ans . £ 1 8s . 7ąd . 10. A box ... Required the same as in the last example , when the length is 3 ft . , breadth 21 ft . , depth 2 ft . , and the ...
... cost of hewing all the faces of the stone in example 3 ; at 5d . per square foot ? Ans . £ 1 8s . 7ąd . 10. A box ... Required the same as in the last example , when the length is 3 ft . , breadth 21 ft . , depth 2 ft . , and the ...
Page 165
... Required the same as in the last ex- ample , when AB = 14 ft . 3 in . , and AD = AE = DE = 2 ft . Ans . 24.68 c . ft ... cost sinking at 3s . 7d . per cubic yard ? Ans . £ 1 13s . 41d . 8. The exterior diameter of a metal pipe is 6 in . , the ...
... Required the same as in the last ex- ample , when AB = 14 ft . 3 in . , and AD = AE = DE = 2 ft . Ans . 24.68 c . ft ... cost sinking at 3s . 7d . per cubic yard ? Ans . £ 1 13s . 41d . 8. The exterior diameter of a metal pipe is 6 in . , the ...
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.