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PART II.

Geometrical Problems.

Fig. 22.

D

PROBLEM I. To draw a line parallel to another line at any given distance; as at the point D, to make a line, parallel to the A line AB. Fig. 22.

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With the dividers take the nearest distance between thepoint D and the given line AB; with that distance set one foot of the dividers any where on the line AB, as at E, and draw the arc C; through the point D draw a line so as just to touch the top of the arc C.

A more convenient way to draw parallel lines is with a parallel rule. [The parallel rules, however, found in cases of mathematical instruments, are often inaccurate.]

Fig. 23.

PROBLEM II. To bisect a given line; or, to find the middle of it. Fig. 23.

A.

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*D Open the dividers to any convenient distance, more than half the given line AB, and with one foot in A, describe an arc above and below the line, as at C and D; with the same distance, and one foot in B, describe arcs to cross the former; lay a rule from C to D, and where the rule crosses the line, as at E, will be the middle.

Fig. 24.

H
PROBLEM III. To erect a perpendicu.
lar from the end, or any part of a given line.
Fig. 24.

G

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Open the dividers to any convenient distance, as from D to A, and with one foot on the point D, from which the perpen. dicular is to be erected, describe an arc, as AEG ; set off the same distance AD, from A to E, and from E to G; upon E and G describe two arcs to intersect each other at H; draw a line from H to D, and one line will be perpendicular to the other. Note. There are other methods of erecting a perpendicu. lar, but this is the most simple.

Fig. 25.

с PROBLEM IV. From a given point as at C, to drop a perpendicular on a given line AB. Fig 25.

A

B

I'D With one foot of the dividers in C describe an arc to cut he given line in two places, as at F and G; upon F and G describe two arcs to intersect each other below the line as at D; lay a rule from C to D and draw a line from C to the giv

Perpendiculars may be more readily raised and let fall, by a small square made of brass, ivory, or wood.

Fig. 26.

CG PROBLEM V. To make an angle at E, equal to a given angle ABC. Fig. 26.

en line.

D

H Open the dividers to any convenient distance, and with one foot in B describe the arc FG; with the same distance and one foot in E, describe an arc from H; measure the arc FG, and lay off the same distance on the arc from H to I; draw a line through I to E, and the angles will be equal.

Fig. 27. PROBLEM VI. To make an acute angle equal to a given number of degrees, suppose 36. Fig. 27.

Draw the line AB of any convenient length ; from a scale of chords take 60 degrees with the dividers, and with one foot in B describe an arc from the line AB; from the same scale take the given number of degrees, 36, and lay it on the arc from C to D; draw a line from B through D, and the angle at B will be an angle of 36 degrees.

Fig. 28.

PROBLEM VII. To make an obtuse angle, suppose of 110 degrees. Fig. 28.

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B Take a chord of 60 degrees as before, and describe an arc greater than a quadrant; set off 90 degrees from B to C, and from C to E set off the excess above 90, which is 20; draw a line from G through E, and the angle will contain 110 degrees.

[It is best, however, in making obtruse angles, to take from the scale the chord of half the angle, and set it off twice. This will save taking two separate chords.] Note. In a similar manner angles may be measured; that

is, with a chord of 60 degrees describe an arc on the angular point, and on a scale of chords measure the arc in.

tercepted by the lines forming the angle. A more convenient method of making and measuring angles is to use a protractor instead of a scale and dividers.

Fig. 29.
B

0 B

I PROBLEM VIII. To make a triangle I of three given lines, as BO, BL, LO. Fig. 29, any two

of which are greater than the third.

B4

L Draw the line BL from B to L; from B, with the length of the line BO, describe an arc as at 0; from L, with the length of the line LO, describe another arc to intersect the former; from O draw the lines OB and OL, and BOL will be Fig. 30.

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25

35 36

PROBLEM IX. To make a right an. gled triangle, the hypothenuse and angles being given. Fig. 30.

C

A Suppose the hypothenuse CA 25 rods or chains, the angle at C 35° 30' and consequently the angle at A 54° 30'. See note after the 39th Geometrical Definition. Note. When degrees and minutes are expressed, they are

distinguished from each other by a small cipher at the right hand of the degrees, and a dash at the right hand of the minutes ; thus 35° 30' is 35 degrees and 30 min. . utes.

Draw the line CB an indefinite length; at C make an an. gle 35° 30'; through where that number of Degrees cuts the arc draw the line CA 25 rods, which must be taken from some scale of equal parts; drop a perpendicular from A to B, and the triangle will be completed.

[A scale of equal parts may be found on one side of Gunter's scale, occupying half its length. It will be known by slanting lines which cross it at each end. The length of the scale, not occupied by these oblique lines, is equally divided into several larger divisions, numbered on one side, and like. wise twice as many smaller, numbered on the other. In taking distances from the scale, each of these divisions, (either the larger or the smaller, as is most convenient,] must be considered 1, 10, 100, &c. rods, chains, or other dimensions of length. If each division be called 1, it will be easy to take off the required number. But the scale is not usually long enough for this. When each division is called 10, as many divisions must be taken, as there are tens in the given number. For the excess of tens, in this case, the little scale, with the oblique lines, is used. Each side of this little scale is divided into 10 equal parts, and each of these parts is, of

course, 1. Then, to take off the hypothenuse, 25, above, we should take in the dividers 2 of the divisions of the large scale, and 5 of those of this small one. There is one of these little scales for the greater, and one for the smaller divisions of the large scale.

When each division of the large scale is called 100, each of those of the small one becomes 10, and the units are found by means of the oblique lines. These are drawn across parallel lines, running the whole length of the

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the other. The parallel lines divide the width of the scale into 10 equal parts. Since each oblique line, then, in crossing the scale passes over one division of length, it is evident, that, in passing one tenth across, (that is, to the first parallel line,) it will pass over one tenth of a division of length ; in passing two tenths across (that is, to the second parallel linc,) it will pass over two tenths of a division of length, and so on. The parallel lines are numbered at the end of the scale, To take off a distance, containing hundreds, then, as 234, we must place one foot of the dividers on the second division of the larger scale, and on the parallel line marked 4, and extend the other foot to the third oblique ine. Decimals may evidently be taken off in a similar man ner; the divisions of the larger scale being made units, and those of the smaller, tenths and hundredths.]

NOTE. The length of the two legs may be found by mea.

suring them upon the same scale of equal parts from
which the hypothenuse was taken.

Fig. 31

B PROBLEM X. To make a right angled triangle, the angles and one leg being given. Fig. 31.

33' 15

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Suppose the angle at C 33° 15', and the leg AC 285.

Draw the leg AC making it in length 285; at A erect a perpendicular an indefinite length ; at C make an angle of 3315'; through where that number of degrees cuts the arc, draw a line till it meets the perpendicular at B. NOTE. If the given line CA should not be so long as the Chord of 60°, it may be continued beyond A, for the purpose of making the angle.

Fig. 32.

PROBLEM XI. To make a right angled triangle, the hypothenuse and one leg being given. Fig. 32.

40

B

28 Suppose the hypothenuse AC 40, and the leg AB 28.

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