Mensuration for beginners [With] Answers1883 |
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Resultat 1-5 av 39
Side 12
... square piece of ground whose side measures 18 ft . 9 in . , at 2s . 8d . per square foot . ( 15 ) Find the cost of ... links . The lengths of lines measured with a chain are generally expressed in links , or chains and links , every ...
... square piece of ground whose side measures 18 ft . 9 in . , at 2s . 8d . per square foot . ( 15 ) Find the cost of ... links . The lengths of lines measured with a chain are generally expressed in links , or chains and links , every ...
Side 13
... square links in a square chain . 32. To reduce square chains to square links . RULE . - Multiply by 10000 ; or , in the case of decimal num . bers , remove the point four places to the right . Ex . Reduce to square links , 7 sq . ch . , 7 ...
... square links in a square chain . 32. To reduce square chains to square links . RULE . - Multiply by 10000 ; or , in the case of decimal num . bers , remove the point four places to the right . Ex . Reduce to square links , 7 sq . ch . , 7 ...
Side 14
... square links . RULE . - Multiply by 100000 ; or , in the case of decimal numbers , remove the point five places to the right . Ex . Reduce to square links , 5 ac . , 5 ac . 2875 sq . lks . , 3.75 ac . 5 ac . = 500000 sq . lks . 5 ac ...
... square links . RULE . - Multiply by 100000 ; or , in the case of decimal numbers , remove the point five places to the right . Ex . Reduce to square links , 5 ac . , 5 ac . 2875 sq . lks . , 3.75 ac . 5 ac . = 500000 sq . lks . 5 ac ...
Side 15
... sq . lks . = 77 · 625 sq . ch . 40. To reduce square links and square chains to acres , roods , and poles . RULE . - Reduce them to acres and decimal parts of an acre ( Arts . 36 , 38 ) , and reduce the decimal parts to roods and poles ...
... sq . lks . = 77 · 625 sq . ch . 40. To reduce square links and square chains to acres , roods , and poles . RULE . - Reduce them to acres and decimal parts of an acre ( Arts . 36 , 38 ) , and reduce the decimal parts to roods and poles ...
Side 16
... square field whose side measures 625 links . ( 2 ) The side of a square is 23 ch . 75 lks . Find its acreage . ( 3 ) Required the area of a square field whose perimeter measures 1000 links . ( 4 ) Find the area of a square field 22 ch ...
... square field whose side measures 625 links . ( 2 ) The side of a square is 23 ch . 75 lks . Find its acreage . ( 3 ) Required the area of a square field whose perimeter measures 1000 links . ( 4 ) Find the area of a square field 22 ch ...
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Vanlige uttrykk og setninger
9 ft A B C D ABCD acres angle subtended Area of base broad centre circle circular circum Circumference of base contains cube cubic foot curved surface cwts cylinder depth diagonal diam Diameter of base equal equilateral triangle find the area Find the cost find the expense find the height find the length find the number Find the side find the volume following dimensions found by Art heptagon hexagon hypotenuse length of carpet Multiply number of cubic number of degrees papering a room parallel sides parallelopiped paving perimeter perpendicular distance perpendicular height polygon prism quotient Radius of base rectangle regular polygon Required the area rhomboid rhombus right cone right-angled triangle round RULE sector slant height solid content square chains square feet square field square links square pyramid square root square yard thick trapezium trapezoid triangular field whole surface width
Populære avsnitt
Side 18 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Side 52 - A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another.
Side 52 - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.
Side 41 - RULE. — Multiply half the sum of the two parallel sides by the perpendicular distance between them, and the product will be the area.
Side 45 - 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 103 - A SPHERE is a solid bounded by a curved surface, every part of which is equally distant from a point within, called the centre.
Side 101 - The area of the curved surface of a cone is equal to one-half the product of the slant hight by the circumference of the base (660).
Side 105 - A reservoir is 24 ft. 8 in. long, by 12 ft. 9 in. wide ; how many cubic feet of water must be drawn off to make the surface sink 1 foot?
Side 38 - RULE. from half the sum of the three sides, subtract each side separately; multiply the half sum and the three remainders together, and the square root of the product will be the area required.
Side 33 - A rhombus is that which has all its sides equal, but its angles are not right angles.