872

THEORY AN1) PRACTICE OF ENGINEERING.

Book II.

To measure any Triangle , as that ofA B C. Multiply the length of B D,the perpendicular, by the base A C,and half their product will be the area :this will be evident if we observe thatin measuring the square we multiplythe two sides which form a right angletogether ; the product will be the areaof the square, as that of ABCD.

In any triangle the difference of thesquares of the two sides is equal to thedifference of the squares of the seg-ments of the base, or of the two linesor distances included between the ex-tremes of the base and the perpendi-cular. Let ACB be any triangle,having B D perpendicular to A C, thenwill the difference of the squares A B,

Fig. 1129.

Fig.

that is, AB* - CB* =

Fig. 1131.

B C be equal to the difference of the squares of AD, CD;

A D 8 — C D 8 .

For since AB* is equal to AD*+BD*,and C B 8 is equal to C D 2 + B D 8 ,therefore the difference between A B 8 and C D 8is equal to the difference between A D*4- B D 8 ,and CI) ! + BD ! ,

or equal to the difference between AD 8 and CD 8 , by taking away thecommon square B D 8 .

The rectangle of the sum and difference of the two sides of any triangle is equal to therectangle of the sum and difference of the distances between the perpendicular and the twoextremes of the base, or equal to the rectangle of the base and the difference or sum of thesegments, according as the perpendicular falls within or without the triangle.

To measure Rhomboidal Figures _Let fall a perpendicular A n

from one angle to the opposite side, and this perpendicularmultiplied by the side gives the area required; but whenthis cannot be done, prolong one side; let fall a perpen-dicular on the side of the rhombus, and proceed as before.

For example, to measure the rhombus ABCD, whoseside A B is 30 feet long ; as this figure has no right angle,it is necessary that one should be established; this is effectedby placing the cross staff on CD, in such a position that apiquet at A may be seen through one of the pinnules onone diameter, and D C through those of the other, as at E ;

*hen measure the perpendicular, 27 feet, which multiplied by 30, the length of the side A B,produces 810 feet as the superficial area of the figure ABCD.

To measure a Trapezium . — Add together the two v G

parallel sides, and multiply their sum by the length ofa perpendicular line contained between their twosides, and which is the height of the trapezium; theproduct will give the area.

To ascertain the content of the figure FGHI,where the two sides F G and IH are parallel, theshortest being 30 feet 9 inches, and the longest 60 feet:place the cross staff on I H, as at K, so that throughthe pinnules of one diameter the piquets I and H may be seen ; then through the otherview the piquet F, where KL, which is 20 feet, is the perpendicular. Add together thetwo parallel sides, which will be 90 feet 9 inches; this multiplied by 20 will produce 1815square feet, the half of which will be the area sought. N

To measure the Trapezoid M N O P, whose side M N is 4feet. Plant piquets at the four angles: draw from theangle P the diagonal PN, which will divide it into twotriangles; find their superficial content, and add themtogether for the area of the trapezoid. For example, findthe area of the triangle PNM, the base of which, P N, is 6feet 6 inches, and the perpendicular RM 2 feet; these mul-tiplied together produce 13 feet, the half of which, 6 feet 6inches, is the area of the triangle ; proceed in the samemanner for the area of the other part of the figure, and add them together.

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Fig. 1132.