Mensuration for beginners [With] Answers1883 |
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Resultat 1-5 av 35
Side 4
... divide the perimeter by 4 , and the quotient will give the length of the side , from which the area may be found . ] ( 13 ) Required the area of a plot of land whose perimeter is 18.4 yds . ( 14 ) If 176 yds . of fencing are required to ...
... divide the perimeter by 4 , and the quotient will give the length of the side , from which the area may be found . ] ( 13 ) Required the area of a plot of land whose perimeter is 18.4 yds . ( 14 ) If 176 yds . of fencing are required to ...
Side 7
... dividing by 9 . If the side is reduced to inches , the area will be expressed in square inches , which may be reduced to square feet by dividing by 144 , and then to squaro yards ( if required ) by further dividing by 9 . [ The lengths ...
... dividing by 9 . If the side is reduced to inches , the area will be expressed in square inches , which may be reduced to square feet by dividing by 144 , and then to squaro yards ( if required ) by further dividing by 9 . [ The lengths ...
Side 10
... . , hence 7'6 " 84 + 6 = 90 sq . in . 26. To express square inches as primes and seconds . RULE . - Divide the number of inches by 12 , the quotient will be the number of primes , and the remainder 10 MENSURATION FOR BEGINNERS .
... . , hence 7'6 " 84 + 6 = 90 sq . in . 26. To express square inches as primes and seconds . RULE . - Divide the number of inches by 12 , the quotient will be the number of primes , and the remainder 10 MENSURATION FOR BEGINNERS .
Side 12
... Divide by 100 ; or , in the case of decimal numbers , remove the point two places to the left . Ex . Reduce to chains , 500 lks . , 525 lks . , 162.5 ch . 500 lks.5 ch . 525 lks.5 ch . 25 lks . 162.5 lks . = 1.625 ch . Ex . IX . ( 1 ) ...
... Divide by 100 ; or , in the case of decimal numbers , remove the point two places to the left . Ex . Reduce to chains , 500 lks . , 525 lks . , 162.5 ch . 500 lks.5 ch . 525 lks.5 ch . 25 lks . 162.5 lks . = 1.625 ch . Ex . IX . ( 1 ) ...
Side 13
... Divide by 10000 ; or , in the case of decimal numbers , remove the point four places to the left . Ex . Reduce to square chains 40000 sq . lks . , 40250 sq . lks . , 9232.5 sq . lks . 40000 sq . lks . = 4 sq . ch . = 40250 sq . lks . 4 ...
... Divide by 10000 ; or , in the case of decimal numbers , remove the point four places to the left . Ex . Reduce to square chains 40000 sq . lks . , 40250 sq . lks . , 9232.5 sq . lks . 40000 sq . lks . = 4 sq . ch . = 40250 sq . lks . 4 ...
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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.