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take equal things, the remainder will be line, describe a circle passing through the equal, and the reverse in respect to unequal given point A on the line B C; then draw a things. 4. The whole is greater than any line from the place where the circle cuts at of its parts. 5. Two right lines do not con- D, so as to pass through E, the centre to F

6. All the angles within a on the opposite side of the circle: tire line circle canuot amount to more nor less than FA will be the perpendicular required. 360 degrecs, por in a semi-circle to more nor less than 180 degrees. 7. The value,

PROBLEM V. or measure, of an angle is not affected or changed by the lines whereby it is formed cular on a given line. Fig. 5. From the

From u given point to let fall a perpendibeing either lengthened or shortened. 8. given point A draw the segment B C, passTwo lines standing at an angle

of 90 degrees ing under the line D E; bisect B C in F, from each other will not be affected by any and draw the perpendicular A F. change of position of the entire figure in which they meet, but will still be mutually

THEOREM VI. perpendicular. After thius much preparation, we may

The opposite angles made by intersecting conclude the student to be ready to pro figure: 0, 0, are equal; P, P, are equal; 8, 8,

lines are equal ; (fig. 6.) as is shown in this ceed in the solution of problems, which we shall study to exhibit in the most simple,

are equal. as well as in a progressive manner.

PROBLEM VII.

PROBLEM I.

PROBLEM II.

PROBLEM VIII.

To describe a triangle with three given lines. To describe an equilateral triangle upon a Fig. 7. Let A B, BC, and C D, be the three given line. Let AB (fig. 1.) be the given given lines; assume either of them, say A B, line, with an opening of your compasses

for a base, then with an opening equal to equal to its length; from each end, A and B C, draw the segment from the point B of B, draw the arcs C D and EF, to whose the base, and with the opening CD make a point of intersection at C draw the lines segment from C: the intersection of the AC and BC.

two segments will determine the lengths of the two lines B C and CD, and of the angle

A BC. To divide an angle equally. Fig. 2. Let BAC be the given angle, measure off equal To imitate a given angle at a given point. distances from A to B, and from A to C; Fig. 8. Let A B C be the given angle, and then with the opening B C draw alternately the point on the line O D whereon it is from B and from C the arcs which intersect to be imitated. Draw the line AC, and at D: a line drawn from A to D will bisect from O measure towards D with an opening the angle BAC.

equal to A B: then from O make a segment

with an opening equal to BC, and from PROBLEM IU.

K make a segment with an opening equal To bisect a given line. Fig. 3. Let A B

to A C: their intersection at E will give be the given line ; from each end (or nearer,

the point through which a line from O will if space he wanting), with an opening of make an angle with O D equal to the angle your compasses rather more than half the ABC. length of A B, describe the arcs which intersect above at C, and below at D: draw

All right lines sererally parallel to any the line CD, passing through the points of giren line are mutually parallel, as shown in intersection, and the line A B will be di fig. 9, where AB, CD, EF, and GH, vided into two equal parts. Observe, this being all parallel to I K, are all parallels to is an easy mode of ere cting a perpendicular each other severally. upon any given line.

N. B. They all make equal angles with the oblique line O P.

THEOREM IX.

PROBLEM IV.

PROBLEM X.

To raise a perpend icular on a given point in u line. Fig. 4. With a moderate open- To draw a parallel through a given point. ing of your compasses, and placing one of Fig. 10. From the end, on any part of the its legs a little above or below the given given line A B, draw an oblique line to the

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PROBLEM XV.

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given point C. Measure the angle made by nexed. Let C be the triangle to be com.
ABC, and return another of equal mea- muted, and D the given angle. Make
surement upon the line B C, so as to make BEF G equal to C, on the angle E BG:
the angle B C D equal to ABC: the line continue A B to E: carry on FE to K, and
C D will be parallel to the line A B. Or, make its parallel HAL, bounded by FH,
as in fig. 11, you may from any points, say parallel to E A: draw the diagonal H K and
CD, in the line A B draw two semicircles GM both through the point B; then KL;
of equal dimensions; the tangent EF will and the parallelogram B MAL will be equal
be parallel to A B. Or you may, according to the triangle C, and be situated as de.
to Problem 5, draw a perpendicular from sired.
the given point to the given line, and draw
another line through the given point at right
angles with the perpendicular proceeding

To make u parallelogram, on u given inclifrom it to the line whose parallel was to be nation, equal to a right-lined figure. Fig. 18. made, and which will be thus found. See Let A B C D be the right-lined figure, and fig. 12.

FK A the given angle or inclination; draw

the line D B, and take its length for the alTHEOREM XI.

titude, FK, of the intended parallelogram,
Parallelogrums of equal buse and altitude applying it to the intended base line KM:
are reciprocully equal. Fig. 13. The paral- now take half the greatest diameter of the
lelogram No. 1 is rectangular: No. 2 is in. triangle DC B, and set it off from K to M,
clined, so as to hang over a space equal to and set off half the greatest diameter of the
the length of its own base; but the line triangle D A B, and set it off from H to M:
A B, which is perpendicular thereto, divides make GH and LM parallel to FK, and
it into two equal parts: let the left half, FG parallel to KH. The parallelogram
A B E, be cut off, and it will, by being FKG H will be equal in area to the figure
drawn up to the right, be found to fit into ABC D, and stand at the given inclination
the dotted space ACD. This theorem or angle.
might be exemplified in various modes; but
we presume the above will suffice to prove
its validity.

To describe a square on a given line. Fig. 19.
Raise a perpendicular at each end of the

line A B equal to its length; draw the line Triungles of equal base and altitude are

CD, and the square is completed. reciprocally equal. Fig. 14. As every parallelogram is divisible into two equal and similar triangles, it follows that the same rule answers for both those figures nnder the The square of the hypothenuse is equal to position assumed in this proposition: we

both the squares made on the other sides of a have shown this by fig. 15.

right-ungled triangle. Fig. 20. This com

prehends a number of the foregoing proPROBLEM XIII.

positions, at the same time giving a very To make a parallelogram equal to a giren be the given right-angled triangle; on each

beautiful illustration of many. Let A BC triangle, with a giren inclination or angle.

side thereof make a square. For the sake Fig. 16. Let B AC be the given triangle,

of arithmetical proof, we have assumed and E D F the given angle. On the line DF

three measurements for them: viz. the hy. measure a base equal to BC, the base of pothenuse at 5, one other side at 4, and the the triangle. Take B G equal to half the

last at 3. Now the square of 5 is 25. The altitude of the triangle for the altitude of square of 4 is 16, and the square of 3 is 9: the parallelogram, and set it off on the line

it is evident the sum of the two last sides ED. Draw F H parallel to E D, and HE make up the sum of the hypothenuse's parallel to DF, which will complete the

square; for 9 added to 16 make 25. But parallelogram E F D H, equal to the triangle

the mathematical solution is equally simple ВА С.

and certain. The squares are lettered as PROBLEM XIV.

follow: BDCE, F G BA, and AHG K. To apply a parallelogram to a giren right Draw the following lines : FC, BK, AD, line, equal to a giren triungle, in a given right AL, and A E. We have already shown, lined figure. Fig. 17. Let A B be the given that parallelograms and triangles of equal line to which the parallelogram is to be an- base and altitude are respectively equal.

PROBLEM XVI,

THEOREM XII.

THEOREM XVII.

PROBLEM XXIII.

The two sides F B, BC, are equal to the in C; the perpendicular DC will divide two sides A B, B D, and the angle DAB the figure into two equal and similar parts. is equal F BC: the triangle ABD must therefore be equal to the angle F B C. But the parallelogram B Lis double the triangle In a given circle to describe a triangle equiA BD. The square G B is also double angular to a given triungle. Fig. 26. Let the triangle F B C: consequently the paral. A B C be the circle, and DEF the triangle lelogram BL is equal to the square G B. given. Draw the line GH, touching the The square HC in like manner is proved circle in A: make the angle HAC equal to to be equal to the parallelogram C L, which DEF, and G A B equal to DFE: draw completes the solution. Euclid, 47th of BC, and the triangle BAC will be similar 1st Book.

to the triangle-D EF.

PROBLEM XVIII.

PROBLEM XIX.

PROBLEM XXV.

PROBLEM XXIV. To divide a line so that the rectangle con- About a given circle to describe a triangle tained under the whole line, and one segment, similar to u giren triangle. Fig. 27. Let be equal to the square of the other segment. ABC be the given circle, and DEF the Fig. 21. On the given line A B describe given triangle: continne the line E F both the square A B CD; bisect AC in E, and ways to G and H, and having found the with the distance EB extend AC to F, centre, K, of the circle, draw a radius, K B, measuring from E. Make on the excess at pleasure; then from K make the angle, FA the square FH, and continue G H to BK A equal to D E C, and B K C equal to K. The square FH will be equal to the DFH; the tangents LN perpendicular to parallelogram HD.

KC, M N perpendicular to K B, and ML perpendicular to KA, will form the re

quired triangle. To make a square equal to a given rightlined figure. Fig. 22. Let A be the given right-lined figure: commute it to a parallel

To describe a circle about a given triangle. ogram, B D, as already shown (prob. 15.): Fig. 28. In the given triangle ABC, bisect add the lesser side ED B E, so as to

any two of the angles ; the intersection of proceed to F: bisect B F in G, and from their dividing lines, B D and CD, will give that point describe the semicircle BHF. the centre D, whence a circle may be deContinue D E to H, which will give HE scribed about the triangle, with the radius for the side of a square equal in area to the

DC. parallelogram B D, and to the original

PROBLEM XXVI. given figure A.

To inscribe a circle in a given triangle.

Fig. 29. In the triangle A B C, divide the To find the centre of a given circle. Fig. 23.

angles ABC, and BCA, equally by the Draw at pleasure the chord A B, bisect it give a point whence the circle EC F may

lines BD, CD. Their junction at D, will in D by means of a diameter, which being be described, with the radius D F perpenbisected will give F for the centre of the

dicular to B C. circle.

PROBLEM XX.

PROBLEM XXVII.

PROBLEM XXI.

PROBLEM XXVIII.

To inscribe a square in a given circle. Fig. To complete a circle upon a giren segment.

30. Draw the diameter AC, and, perpen Fig. 24. Let A B C be the given segment: draw the line A C, and bisect it in D; draw

dicular thereto, the diameter BD: the also the perpendicular B E through D, draw lines A B, BC, CD, and D A, will form a

correct square. B A, and on it make the angle B A E, equal to DBA; this will give the point of intersection E for the centre, whence the circle To describe a circle around a square. may be completed. It matters not whether Fig. 30. In the square A B CD, draw the the segment be more or less than a semi- diagonals A C, B D, their intersection at E circle.

will give the centre of a circle, whose raPROBLEM XXII.

dius may be any one of the four converg. To cut a given circumference into two equal ing lines; say EA, that will enclose the parts. Fig. 25. Draw the line A B, bisect square.

PROBLEM XXIX.

tablishes a very easy mode of setting off the To describe a circle within a given square.

six sides as follows: draw the diameter Fig. 31. Divide the square into four equal AB, set one leg of your compasses at A,

and draw the segment DF, and from B parts, by the lines A C, BD, whose intersection at E, shows the centre of a circle

draw the segment CE; thus dividing the to be drawn with any one of the converg. joining them, and the figure will be com

circle into six equal portions ; draw lines ing lines, say E A, as a radius.

plete. PROBLEM XXX.

PROBLEM XXXVI. To describe a square on a given circle.' To form u quindecagon, or figure of 15 Fig. 31. Divide the circle into four equal equal sides, within a circle. An equilateral parts, (or quadrants) by the lines A C, BD; triangle being inscribed within a circle, by draw the tangents GH, FK, parallel to assuming the distance between three points AC, and GF, HK, parallel to BD; which of a hexagon, say from A to C in the last will give the required square.

figure for a side, let one point of such triangle be applied to each angle of a penta

gon in succession ; its two other points will To make an isosceles triangle, having each divide the opposite sides in three equal of the angles at the base double that at the parts, as the figure changes place within the summit. Fig. 32. Cut any given line, as pentagon. A B, into extreme and mean proportions, (as in Problem 18); then, from A, as the

To change a circle to a triangle. Fig. 35. centre, draw a circle B D E, with the open

Draw the tangent A B equal to 34 diameters ing A B, and apply the line B D within its

A D of the circle, and from the centre C circumference, equal to AC, the greater draw CB, and CA: the triangle CA B portion of A B; join CD, A B D will be

will be equal in contents to the circle A D. the isosceles triangle sought.

PROBLEM XXXI.

!

PROBLEM XXXVII,

PROBLEM XXXVIII.

PROBLEM XXXII.
To describe a regulur pentagon. Fig. 33.

To change a pentagon into a triangle. Fig. Make the isosceles triangle ACD within

36. Continue the base line A B to C, and the circle ABCDE; the base CD will from the centre D let a perpendicular fall give the fifth part of the circumference.

on A B, bisecting it in E. Measure from

B a space equal to four times E B. Through PROBLEM XXXIII.

the centre D draw D F, parallel and equal To describe a regular pentagon about a cir.

to EC; draw FC: the parallelogram cle. Fig. 33. This is done by drawing pa- contained under ECDF will equal the rallels to the lines AB, BC, CD, DE,

area of the pentagon. Or the pentagon EA; making them all tangents to the cir- may be changed to a triangle by adding to cle; on the same principle, a square, a hex: A B four times its own length, and draw. agon, &c. may be drawn around a circle, ing a line from the centre, to the produced from a similar figure inscribed within it.

termination of A B; the angle at the centre

would then be obtuse. PROBLEM XXXIV. To describe a circle around a pentagon.

PROBLEM XXXIX. Fig. 33. Bisect any two angles of a penta- To druw a spiral line from a given point. gon, and take their point of intersection, G, Fig. 57. Draw the line A B through the as a centre, using either of the converging given point C, and from C draw the senuilines, DG, or E G, for a radius. Where a cir- circle D Е, then shift to D for a centre, and cle is to be described within a pentagon, make the semi-circle A E in the opposite you must bisect any two of the faces, and side of the line : shift again from D to C raise perpendiculars at those points, which for a centre, and draw the semi-cirele FG; will meet in the centre either of the con- and then continue to change the centres verging lines serving for a radius.

alternately, for any number of folds you PROBLEM xxxv.

may require ; the centre C serving for all

above, the centre D for all below, the To inscribe a regular heragon within a cir- line A B. cle. Fig. 34. The radius of a circle being With respect to the application of geome. equal to one-sixth of its circumference, es. try to its pristine iplent, namely, the mea

surement of land, we must refer our readers ferently, or do not rise at all. They freto SURVEYING ; under which head it will quently preserve, however, their germinatbe found practically exemplified. We trusting virtues for many years within the bowsufficient has been here said to show the els of the earth; and it is not unusual, upon utility and purposes of this important a piece of ground being newly dug to a science, and to prove serviceable to such considerable depth, to observe it soon after persons as may not have occasion for deep covered with several plants which had not research, or for extensive detail.

been seen there in the memory of man. GEORGIC, a poetical composition upon Were this precaution frequently repeated, the subject of husbandry, containing rules it would perhaps be the means of recovertherein, put into a pleasing dress, and set ing certain species of plants which are reoff with all the beauties and embellishments garded as lost; or which, perhaps, never of poetry.

coming to the knowledge of botanists, GEORGINA, in botany, a genus of might hence appear the result of a new the Syngenesia Superflua class and order. creation. Light is supposed to be injurious Receptacle chaffy; no down; calyx dou. to the process which affords a reason for ble; the outer many-leaved ; inner one- covering seeds with the soil in which they leaved, eight parted. There are three spe. are to grow, and for carrying on the busicies.

ness of malting in darkened apartments; GERANIUM, in botany, crane's bill, malting being nothing more than germinaa genus of the Monadelphia Decandria tion, conducted with a particular view. class and order. Natural order of Grui- GEROPOGON, in botany, a genus of nales. Gerania, Jussieu. Essential cha

the Syngenesia Polygamia Æqualis class racter: calyx five-leaved; corolla tive-pe- and order. Natural order of Compositæ talled, regular; nectary five honied glands,

Semiflosculosæ, or compound flowers, with fastened to the base of the longer filaments ; semi-forets or ligulate forets only. Cicho. fruit five-grained, beaked; beaks simple, raceæ, Jussieu. Essential character : ca. naked, neither spiral nor bearded. There lyx simple ; receptacle with bristle-shaped are thirty-two species.

chaffs ; seeds of the disk, with a feathered GERARDIA, in botany, so called in down of the ray, with five awns. There honour of John Gerarde, our old English are three species. botanist, a genus of the Didynamia Angi

GESNERIA, in botany, so named in ospermia class and order. Natural order honour of Conrad Gesner, of Zurich, the of Personatæ. Scrophulariæ, Jussieu. Es famous botanist and natural historian, a sential character : calyx five-cleft; corolla genus of the Didynamia Angiospermia two-lipped, lower lip three-parted, the class and order. Natural order of Persolobes emarginate, the middle segments two- natæ. Campanulaceæ, Jussieu. Essential parted; capsule two-celled, gaping. There character: calyx five-cleft, sitting on the are ten species.

germ; corolla incarved and recnrved; capGERMINATION. When a seed is placed sule inferior, two-celled. There are twelve in a situation favourable to vegetation, it species. very soon changes its appearance; the ra- GETHYLLIS, in botany, a genus of the dicle is converted into a root, and sinks into

Hexandria Monogynia class and order. the earth ; the plumula rises above the Natural order of Spathaceæ, Narcissi, Jusearth, and becomes the trunk or stem, sieu. Essential character: calyx none; When these changes take place, the seed corolla six-parted; berry club-shaped, radiis said to germinate ; the process itself has cle, one-celled. There are four species. been called germination, which does not GEUM, in botany, English avens, or depend upon the seed alone; something ex- herb bennet, a genus of the Icosandria Poternal must affect it. Seeds do not germi- lygynia class and order. Natural order of nate equally and indifferently in all places Senticosæ. Rosaceæ, Jussieu. Essential and seasons, they require moisture and a character: calyx ten-cleft; petals five; certain degree of heat, and every species seeds with a kneed awn. There are nine of plant seems to have a degree of heat species, natives of Europe and North Amepeculiar to itself, at which its seeds begin rica. to germinate; air also is necessary to the GHINIA, in botany, so named in megermination of seeds; it is for want of air mory of Lucas Ghini, a famous physician that seeds which are buried at a very great and botanist of Bologna, a genus of the depth in the earth, either thrive but indif. Diandria Monogynia class and order Na

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