Introduction to lineal drawing and mensurationDepository of the British and Foreign School Society, 1847 - 48 sider |
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Side 7
... examples how to proceed -his own judgment and the success or failure of the class will probably induce him to omit or repeat many of the directions given . Previous to the daily lesson , the exercise should have been well practised at ...
... examples how to proceed -his own judgment and the success or failure of the class will probably induce him to omit or repeat many of the directions given . Previous to the daily lesson , the exercise should have been well practised at ...
Side 16
... examples of an ellipse . Of an oval . Draw an ellipse 12 inches by 8 inches . form is an egg ? ( Fig . 64. ) Add the diameter , and describe them . Draw an ellipse 9 inches by 6 inches . their extremities by right lines . 9 inches by 6 ...
... examples of an ellipse . Of an oval . Draw an ellipse 12 inches by 8 inches . form is an egg ? ( Fig . 64. ) Add the diameter , and describe them . Draw an ellipse 9 inches by 6 inches . their extremities by right lines . 9 inches by 6 ...
Side 32
... ( See EUCLID , Book I. Prop . xli . ) MENSURATION . Prob . I. To find the area of a square . Rule . Multiply the length of one side by itself . The result is the required area . Example . A square table has a side of 6 APPENDIX .
... ( See EUCLID , Book I. Prop . xli . ) MENSURATION . Prob . I. To find the area of a square . Rule . Multiply the length of one side by itself . The result is the required area . Example . A square table has a side of 6 APPENDIX .
Side 33
... Example . A court yard contains 1296 feet : required the length of its side . Here 1296 feet 36 feet . Ans . Prob . III . To find the area of a parallelogram or rectangle . Rule . Multiply the length by the breadth . The result is the ...
... Example . A court yard contains 1296 feet : required the length of its side . Here 1296 feet 36 feet . Ans . Prob . III . To find the area of a parallelogram or rectangle . Rule . Multiply the length by the breadth . The result is the ...
Side 34
... Example . A marble slab 8 feet long , was found to con- tain a surface of 5760 inches : required its width . Here 5760 by 96 inches 60 inches , or 5 feet . Ans . 5 feet . Prob . V. To find the solid contents of a cube . Rule . Multiply ...
... Example . A marble slab 8 feet long , was found to con- tain a surface of 5760 inches : required its width . Here 5760 by 96 inches 60 inches , or 5 feet . Ans . 5 feet . Prob . V. To find the solid contents of a cube . Rule . Multiply ...
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Introduction to Lineal Drawing and Mensuration Henry Pickton Ingen forhåndsvisning tilgjengelig - 2009 |
Vanlige uttrykk og setninger
12 inches long 15 feet 9 inch accuracy acute angle Add diagonals Appendix architrave base black board Book Borough Road called cavetto centre chalk cima recta cornice course cultivation curved lines decagon DEFINITIONS diameter Divide door Draw a 12 Draw a rectangle Draw an ellipse Draw two curves duodecagon Enclose entablature EUCLID Example extremities facia feet 6 inches fillet find the area find the solid frieze geometrical Heptagon horizontal line Illustrate on black inch circle inch lines inch sides introduction king post lineal drawing Look round MENSURATION minutes model drawing muntin nonagon oblique line ovolo panels Parallel curve pencil perpendicular height Point polygon Prob Prop proportion pupils purlines quadrant quadrilateral figure rails rectangle 12 inches required area required the area required the solid rhombus right angle right lines roof round the room Rule schools solid contents SUCCEEDING LESSONS teacher tie beam torus trapezium Tuscan column undecagon
Populære avsnitt
Side 40 - To find the solidity of a cone. RULE. Multiply the area of the base by the perpendicular height, and ^ of the product will be the solidity.
Side 40 - To find the solidity of a cylinder. RULE. — Multiply the area of the base by the altitude, and the product will be the solidity.
Side 16 - A Segment is any part of a circle bounded by an arc and its chord. 51. A Semicircle is half the circle, or a segment cut off by a diameter. The half circumference is sometimes called the Semicircle. 52. A...
Side 16 - Hexagon, of six sides; a Heptagon, seven; an Octagon, eight; a Nonagon, nine ; a Decagon, ten ; an Undecagon, eleven ; and a Dodecagon, twelve sides.
Side 35 - Triangles on the same base, and between the same parallels, are equal.
Side 35 - Any two angles of a triangle are together less than two right angles.
Side 37 - To find the area of a trapezium. RULE. — Divide the trapezium into two triangles by a diagonal, and then find the areas of these triangles ; their sum will be the area of the trapezium.
Side 14 - ... foursided figure with all its sides equal, and all its angles right angles.
Side 38 - Ibs. 7 oz. to ounces. 2. The proportion of the diameter of a circle to its circumference is very nearly 113 to 355; find the circumference of a circle whose diameter is 1 32 feet.
Side 36 - I 8.78 .|- yards breadth of the walk. PROBLEM II. To find the area of a rhombus or rhomboid. RULE. — Multiply the length of the base by the perpendicular height. 6. The base of a rhombus being 12 feet, and its height 8 feet. Required the area. Ans. 96 feet. PROBLEM III.