A graduated course of problems in practical plane and solid geometry |
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Side 139
From the foregoing notes the truth of the following theorem will be readily seen :(
E . ) “ If a parallelogram and a triangle be upon the same base and between the
same parallels , the parallelogram shall be double of the triangle " ( in area ) .
From the foregoing notes the truth of the following theorem will be readily seen :(
E . ) “ If a parallelogram and a triangle be upon the same base and between the
same parallels , the parallelogram shall be double of the triangle " ( in area ) .
Side 140
To construct a parallelogram on a given base AB equal in area to a given
parallelogram CDEF . 1 . Find AG the fourth proportional less to AB , CD , CF ( Pr
. 141 ) . 2 . At A , raise the perpendicular AH equal to AG , and complete the ...
To construct a parallelogram on a given base AB equal in area to a given
parallelogram CDEF . 1 . Find AG the fourth proportional less to AB , CD , CF ( Pr
. 141 ) . 2 . At A , raise the perpendicular AH equal to AG , and complete the ...
Side 144
To construct a parallelogram equal to any given triangle ABC both in area and
perimeter . 1 . Biscci BC in D ( Pr . 1 ) . 2 . Produce BA beyond A , and make A E
equal to AC , and bisect BE in F . 3 . Find the altitude AG of the given triangle ( Pr
...
To construct a parallelogram equal to any given triangle ABC both in area and
perimeter . 1 . Biscci BC in D ( Pr . 1 ) . 2 . Produce BA beyond A , and make A E
equal to AC , and bisect BE in F . 3 . Find the altitude AG of the given triangle ( Pr
...
Side 145
Then the parallelogram BDLK is equal both in area and perimeter to the given
triangle ABC . Problem 165 . To divide any parallelogram ABCD into two parts ,
proportionate in area to a given divided line EF , from a point o in one side . M 1 .
Then the parallelogram BDLK is equal both in area and perimeter to the given
triangle ABC . Problem 165 . To divide any parallelogram ABCD into two parts ,
proportionate in area to a given divided line EF , from a point o in one side . M 1 .
Side 165
Then BHK will be the required triangle , and it is onethird of the given triangle
ABC . ( B ) To construct a parallelogram one - third of a given parallelogram
ABCD . 1 . As in case A , find EG the mean proportional ( Pr . 140 ) . 2 . On AB ,
mark off ...
Then BHK will be the required triangle , and it is onethird of the given triangle
ABC . ( B ) To construct a parallelogram one - third of a given parallelogram
ABCD . 1 . As in case A , find EG the mean proportional ( Pr . 140 ) . 2 . On AB ,
mark off ...
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altitude arc cutting Atlas axis base Bisect the angle called centre circumference cloth complete cone construct contained curve cutting cylinder describe a circle describe an arc describe arcs diagonal diameter distance divide draw a line draw lines Draw the line edge elevation ellipse equal equal in area equilateral triangle face four given circle given line given point given square ABCD given straight line given triangle ABC half height hexagon horizontal plane inches inclined inscribe intersection isosceles triangle Join length Maps mark meeting Method NOTE obtain parallel parallelogram pass pentagon perpendicular Philips plane of projection polygon prism Problem produced projection projectors pyramid radii radius rectangle rectilineal figure regular represent required circle respectively right angles scale semicircle sides similar solid straight line Take touching traces trapezium vertical plane
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Side 193 - A cone is a solid figure described by the revolution of a right-angled triangle about one of the sides containing the right angle, which side remains fixed. If the fixed side be equal to the other side containing the right angle, the cone is called a right-angled cone ; if it be less than the other side, an obtuse-angled ; and if greater, an acute-angled cone. XIX. The axis of a cone is the fixed straight line about which the triangle revolves.
Side 123 - A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less.