bers) the whole will be equal. 3. If from equal things we take equal things, the remainder will be equal, and the reverse in respect to unequal things. 4. The whole is greater than any of its parts. 5. Two right lines do not contain a space. 6. All the angles within a circle cannot amount to more nor less than 360 degrees, nor in a semicircle to more nor less than 180 degrees. 7. The value, or measure, of an angle is not affected or changed by the lines whereby it is formed being either lengthened or shortened. 8. Two lines standing at an angle of 90 degrees from each other will not be affected by any change of position of the entire figure in which they meet, but will still be mutually perpendicular.

After thus much preparation, we may conclude the student to be ready to proceed in the solution of problems, which we shall study to exhibit in the most simple, as well as in a progressive

manner.

PROBLEM I.

To describe an equilateral triangle upon a given line. Let A B (fig 1) be the given Tine, with an opening of your compasses equal to its length: from each end, A and B, draw the arcs C D and E F, to whose point of intersection at C draw the lines AC and B C.

PROBLEM II.

PROBLEM IV.

To raise a perpendicular on a given point in a line. Fig. 4. With a moderate opening of your compasses, and placing one of its legs a little above or below the given line, describe a circle passing through the given point A on the line B C ; then draw a line from the place where the cir cle cuts at D, so as to pass through E, the centre to F on the opposite side of the circle: the line F A will be the perpendicular required.

PROBLEM V.

From a given point to let fall a perpendicular on a given line. Fig. 5. From the given point A draw the segment B C, passing under the line DE; bisect B C in F, and draw the perpendicular A F.

THEOREM VI.

The opposite angles made by intersecting lines are equal; (fig. 6.) as is shown in this figure: 0, 0, are equal; p, p, are equal; 8, 8, are equal.

PROBLEM VII.

To describe a triangle with three given lines. Fig. 7. Let A B, B C, and C D, be the three given lines; assume either of them, say A B, for a base; then with an opening equal to B C, draw the segment from the point B of the base, and with the opening C D make a segment from C: the intersection of the two segments will determine the lengths of the two lines B C and C D, and of the an

PROBLEM III.

To bisect a given line. Fig. 3. Let A B be the given line; from each end (or nearer, if space be wanting,) with an opening of your compasses rather more than half the length of A B, describe the arcs which intersect above at C, and below at D: draw the line C D, passing through the points of intersection, and the line A B will be divided into two equal parts. Observe, this is an easy mode of erecting a perpendicular upon any given line.

PROBLEM VIII.

To imitate a given angle at a given point. Fig. 8. Let A B C be the given angle, and O the point on the line O D whereon it is to be imitated. Draw the line A C, and from O measure towards D with an opening equal to A B: then from O make a segment with an opening equal to B C, and from K make a segment with an opening equal to A C; their intersection at E will give the point through which a line from Ŏwill make an angle with O D equal to the angle A B C.

THEOREM IX.

All right lines severally parallel to any given line are mutually parallel, as shown in

fig. 9, where A B, C D, E F, and G H, being all parallel to I K, are all parallels to each other severally.

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

PROBLEM X.

To draw a parallel through a given point. Fig. 10. From the end, on any part of the given line AB, draw an oblique line to the given point C. Measure the angle made by A B C, and return another of equal measurement upon the line B C, so as to make the angle B C D equal to A B C : the line C D will be parallel to the line A B. Or, as in fig. 11, you may from any points, say CD, in the line A B, draw two semicircles of equal dimensions; the tangent E F will be parallel to A B. Or you may, according to Problem 5, draw a perpendicular from the given point to the given line, and draw another line through the given point at right angles with the perpendicular, proceeding from it to the Jine whose parallel was to be made, and which will be thus found. See fig. 12.

THEOREM XI.

Parallelograms of equal base and altitude are reciprocally equal. Fig. 13. The parallelogram No. 1, is rectangular: No. 2 is inclined, so as to hang over a space equal to the length of its own base; but the line A. B, which is perpendicular thereto, divides it into two equal parts; let the left half, A BE, be cut off, and it will, by being drawn up to the right, be found to fit into the dotted space A C D. This theorem might be exemplified in various modes; but we presume the above will suffice to prove its validity.

THEOREM XII.

Triangles of equal base and altitude are 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 under the position assumed in this proposition: we have shown this by fig. 15.

PROBLEM XIII.

To make a parallelogram equal to a given triangle, with a given inclination or angle. Fig. 16. Let B A C be the given triangle, and E D F the given angle. On the line DF measure a base equal to B C, the base of the triangle. Take B G equal to half the altitude of the triangle for the altitude

of the parallelogram, and set it off on the line ED. Draw FH parallel to F D, and HE parallel to D F, which will complete the parallelogram E F D H, equal to the triangle B A C.

PROBLEM XIV.

To apply a parallelogram to a given right line, equal to a given triangle, in a given right line figure. Fig. 17 Let A B be the given line to which the parallelogram is to be annexed. Let C be the triangle to be commuted, and D the given angle. Make BE F G equal to C, on the angle E B G; continue A B to E; carry on F E to K, and make its parallel H A L. bounded by FH, parallel to E A: draw the diagonal H K, and G M both through the point B; then K L, and the parallelogram BMAL will be equal to the triangle C, and be situated as desired.

PROBLEM XV.

To make a parallelogram, on a given inclination, equal to a right-lined figure. Fig. 18. Let A B C D be the right-lined figure, and FK H the given angle or inclination; draw the line D B, and take its length for the altitude, F K, of the intended parallelogram, applying it to the intended base line K M: now take half the greatest diameter of the triangle D C B, and set it off from K to M, and set off half the greatest diameter of the triangle DA B, and set it off from H to M: make G H to L M parallel to F K, and F G parallel to K H. The parallelogram FK G H will be equal in area to the figure A B C D, and stand at the given inclination or angle.

PROBLEM XVI.

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 C D, and the square is com pleted.

THEOREM XVII.

The square of the hypothenuse is equal to both the squares made on the other sides of a right-angled triangle. Fig. 20. This comprehends a number of the foregoing propositions, at the same time giving a very beautiful illustration of many. Let ABC be the given right-angled triangle; on each side thereof make a square. For the sake of arithmetical proof, we have assumed three measurements for them: viz. the bypothenuse at 5, one other side at 4, and

the last at 3. Now the square of 5 is 25. The square of 4 is 16, and the square of 3 is 9 it is evident the sum of the two last sides make up the sum of the hypothenuse's square; for 9 added to 16 make 25. But the mathematical solution is equally simple and certain. The squares are lettered as follow: BDCE, FG BA, and AHGK. Draw the following lines; F C, BK, A D, A L, and A E. We have already shown, that parallelograms and triangles of equal base and altitude are respectively equal. The two sides F B, B C, are equal to the two sides A B, BD, and the angle D A B is equal FBC: the triangle ABD must therefore be equal to the angle F B C. But the parallelogram BL is double the triangle ABD. The square G B is also double the triangle FBC: consequently the parallelogram BL is equal to the square G B square H C in like manner is proved to be equal to the parallelogram C L, which completes the solution. Euclid, 47th of 1st Book.

PROBLEM XVIII.

The

To make a square equal to a given rightlined figure. Fig. 22 Let A be the given right-lined figure: commute it to a parallelogram, B D, as already shown, (prob. 15.): add the lesser side E D to B E, so as to proceed to F: bisect B F in G, and from that point describe the semicircle BHF. Continue D E to H, which will give H E for the side of a square equal in area to the parallelogram B D, and to the original given figure A.

PROBLEM XX.

To find the centre of a given circle. Fig. 23. Draw at pleasure the chord A B, bisect it in D by means of a diameter, which being bisected will give F for the centre of the circle.

PROBLEM XXI.

Fig. 24. Let A B C be the given segment: draw the line A C, and bisect it in D; draw also the perpendicular B E through D, draw B A, and on it make the angle BA E, equal to DB A; this will give the point of intersection E for the centre, whence the circle may be completed. It matters not whether the segment be more or less than a semicircle.

PROBLEM XXII.

To cut a given circumference into two equal parts. Fig. 25. Draw the line A B, bisect in C; the perpendicular D C will divide the figure into two equal and similar parts.

PROBLEM XXIII.

In a given circle to describe a triangle equiangular to a given triangle. Fig. 26. Let A B C be the circle, and D E F the triangle given. Draw the line G H, touching the circle in A: make the angle HAC equal to D E F, and G A B equal to DF E: draw B C, and the triangle B A C will be similar to the triangle D E F.

PROBLEM XXIV.

About a given circle to describe a triangle similar to a given triangle. Fig 27. Let ABC be the given circle, and D E F the given triangle: continue the line E-F both ways to G and H, and having found the centre K, of the circle, draw a radius, K B, at pleasure; then from K make the angle B KA equal to D E C, and B KC equal to D F H; the tangents L N perpendicular to K C, M N perpendicular to K B, and M L perpendicular to KA, will form the required triangle.

PROBLEM XXV.

To describe a circle about a given triangle. Fig. 28. In the given triangle ABC, bisect any two of the angles; the intersection of their dividing lines, B D and C D, will give the centre D, whence a circle may be described about the triangle, with the radius D C.

PROBLEM XXVI.

To inscribe a circle in a given triangle. Fig. 29. In the triangle A B C, divide the angles A B C, and B ̊C A, equally by the lines B D, C D. Their junction at D will give a point whence the circle E C F may be described, with the radius D F per

and equal to EC; draw FC: the parallelogram contained under ECDF will equal the area of the pentagon. Or the pentagon may be changed to a triangle by adding to AB four times its own length, and drawing a line from the centre, to the produced termination of AB; the angle at the centre would then be obtuse.

PROBLEM XXXIX."

To draw a spiral line from a given point. Fig. 37. Draw the line AB through the given point C, and from C draw the semicircle DE, and then shift to D for a centre, and make the semicircle AE in the opposite side of the line: shift again from D to C for a centre, and draw the semicircle FG; and then continue to change the centres alternately, for any number of folds you may require; the centre C serving for all above, the centre D for all below, the line AB.

With respect to the application of geo metry to its pristine intent, namely, the measurement of land, we must refer our readers to SURVEYING; under which head it will be found practically exemplified. We trust sufficient has been here said to show the utility and purposes of this important science, and to prove serviceable to such persons as may not have occasion for deep research, or for extensive detail. GEORGIC, a poetical composition upon the subject of husbandry, containing rules therein, put into a pleasing dress, and set off with all the beauties and embellishments of poetry.

GEORGINA, in botany, a genus of the Syngenesia Superflua class and order. Receptacle chaffy; no down; calyx double; the outer many-leaved; inner oneleaved, eight-parted. There are three species.

GERANIUM, in botany, crane's bill, a genus of the Monadelphia Decandria class and order. Natural order of Gruinales. Gerania, Jussieu. Essential character: calyx five-leaved; corolla five-petalled, regular; nectary five honied glands, fastened to the base of the longer filaments; fruit five-grained, beaked; beaks simple, naked, neither spiral nor bearded. There are thirty-two species. There are five species indigenous to the United States. The root of one of these, G. maculatum, or spotted crane's bill, is an astringent, and the decoction of it, made with milk, is useful in cholera infantum.

GERARDIA, in botany, so called in honour of John Gerarde, our old English botanist, a genus of the Didynamia Angiespermia class and order. Natural order

of Personatæ. Scrophulariæ, Jussieu. Essential character: calyx five-cleft; corolla two-lipped, lower lip three-parted, the lobes emarginate, the middle segments two parted; capsule two-celled, gaping. There are ten species. GERMINATION. When a seed is placed in a situation favourable to vegetation, it very soon changes its appear ance; the radicle is converted into a root, and sinks into the earth; the plumula rises above the earth, and becomes the trunk or stem. When these changes take place, the seed is said to germinate; the process itself has been called germination, which does not depend upon the seed alone; something external must affect it. Seeds do not germinate equally and indifferently in all places and seasons; they require moisture and a certain degree of heat, and every species of plant seems to have a degree of heat peculiar to itself, at which its seeds begin to germinate; air also is necessary to the germination of seeds; it is for want of air, that seeds which are buried at a very great depth in the earth either thrive but indifferently, or do not rise at all. They frequently preserve, however, their germinating virtues for many years within the bowels of the earth; and it is not unusual, upon a piece of ground being newly dug to a considerable depth, to observe it soon after covered with several plants, which had not been seen there in the memory of man. Were this precaution frequently repeated, it would perhaps be the means of recovering certain species of plants which are regarded as lost; or which, perhaps, never coming to the knowledge of botanists, might hence appear the result of a new creation. Light is supposed to be injurious to the process, which af fords a reason for covering the seeds with the soil in which they are to grow, and for carrying on the business of malting in darkened apartments, malting being nothing more than germination, conducted with a particular view.

GEROPOGON, in botany, a genus of the Syngenesia Polygamia Equalis class and order. Natural order of Compositæ Semiflosculose, or compound flowers, with semi-florets or ligulate florets only. Cichoraceæ, Jussieu. Essential character: calyx simple; receptacle with bristle-shaped chaffs; seeds of the disk with a feathered down of the ray, with five awns. There are three species.

GESNERIA, in botany, so named in honour of Conrad Gesner, of Zurich, the famous botanist and natural historian, a genus of the Didynamia Angiospermia,