and then we should have to construct Va2 + m2 + n3. Now, when there are several positive squares contained under the radical sign, as √a2+m2 +n2 +p+ &c. we may make

√a2+m2=h,

E

√h2+n2 = i, √i3 +p2 = k, and so on; and, as each of the quantities is determined by the preceding, the last will give the value of √a2+m2+n2 +p2+ &c. In order to construct these quantities in the most simple manner, each hypothenuse is to be regarded successively as a side; having, for example, taken AB = a, and raised the perpendicular AC =e, we may join BC, which will be h; then at the point C if we raise upon BC the perpendicular CD = n; and having drawn c BD, which will be i, at the extremity D, we may raise upon BD the perpendicular DE = p, and BE will be k, and equal to √a2+m2+n2+p2.

If some of the squares are negative, we may combine the method just given with that for constructing a ba

D

8th. Lastly, if the quantity to be constructed be of this form

multiplying by ✓d+e, will change it into a

√(b+c) (d+e); d+e

then, by finding a mean proportional between b + c and d +e,

and calling it m, we have

a m

d+ e'

which is easily constructed.

The construction often becomes much more simple by setting out always from the same principles; but these simplifications are derived from certain considerations which are peculiar to each question, and consequently can be made known only as the occasion presents itself. We will merely remark, in concluding, that although the construction of the radical quantities, which we have been considering, reduces itself to finding fourth proportionals, mean proportionals, and constructing right-angled triangles, still we can arrive at constructions more or less simple or elegant by the method employed for finding these mean proportionals; we shall now, therefore, introduce two other methods of finding a mean proportional between two given lines.

The first consists in describing upon the greater AB (see 3d diagram to Art. 5) of two given lines a semicircle ACB,

and, having taken a part AD equal to the less, raising a perpendicular DC and drawing the chord AC, which will be a mean proportional between AB and AD; for, by drawing CB, the triangle ACB is right angled, (Prop. XIX, Cor. 2, B. III El. Geom.) and consequently AC is a mean proportional between the hypothenuse AB and the segment AD. (Prop. XVII, Cor. 5, B. IV, El. Geom.)

The second method consists in drawing a line AB, equal to the greater given line, and having taken a part AC equal to the less, describing upon the remainder BC a semicircle CDB, to which if we draw the tangent AD; this tangent is a mean proportional between AB and AC. (Prop. XXVII, B. IV, El. Geom.)

B

C

It is evident, therefore, that rational quantities may always be constructed by means of straight lines, and radical quantities of the second degree may be constructed by means of the circle and straight line united.

As to radical quantities of higher degrees, their construction depends upon the combination of different curved lines. We will now proceed to the consideration of questions, the solution of which depends either upon rational quantities or radical quantities of the second degree.

Geometrical questions, and modes of forming equations therefrom, and their solutions.

6. The precepts usually given in algebra for putting questions into equations, are equally applicable to questions in geometry. Here, also, the thing sought is to be represented by some symbol; and the equation is to be constructed in such manner, as to express the relations of the quantity represented by such symbol, in quantities that are known, or in those whose values are attainable; and the reasoning is to be conducted by the aid of this symbol, and of those which represent the other quantities, algebraically, as if the whole were known, and we were proceeding to verify it; this method of proceeding is called analysis.

Although in expressing geometrical questions by algebraic equations, we have more resources and more facilities according as we are acquainted with a greater number of the properties of lines, surfaces, &c., still, as algebra itself furnishes the means of discovering these properties, the number of propositions really necessary is very limited. The two proposi tions that similar triangles have their homologous sides proportional; and, that, in a right angled triangle the square of the

hypothenuse is equivalent to the sum of the squares of the two other sides, are the fundamental propositions and the basis of the application of algebra to geometry.

But there are many ways of making use of these propositions, according to the nature of the question, and there is always a discretion to be exercised in the choice of the means, and manner of applying them; and this discertion can only be acquired by practice.

When a geometrical question is to be resolved algebraically, it will be necessary to construct a figure that shall represent the several parts or conditions of the problem under consideration, and, if possible, get such expressions for the unknown quantities in terms of those that are known, as may easily be determined, according to the known properties of the figure. But if it so happens, that the required quantity can have no expression which will render it available under the present construction, we may, frequently, by drawing lines having certain relations to the known parts of the figure and also to the unknown, so connect the known to those that are required, as to get available expressions for their values. Having proposed a figure as above, we may, by means of the proper geometrical theorems, proceed to make out as many independent equations as there are unknown quantities; and the resolutions of these will give the solution.

PROBLEM I.

The base BC, and the sum of the hypothenuse AB and perpendicular AC, of a right angled triangle being given, to determine the triangle.

=

Let BC b, and AC = x, and if AB + AC be represented by s, then will the hypothenuse AB be represented by s―x.

Therefore, by the properties of the right angled triangle (Prop. XXIV, B. IV, El. Geom.) we have AC + BC = AB'

omitting a which is common to both sides of the equation, and transposing the other numbers we have,

which is the value of the perpendicular AC; where s and b may be any numbers whatever, provided s be greater than b. (Prop. X, B. II, El. Geom.)

If this quantity is to be constructed, it may be resolved (Art. 2, Ex. 4,) into the form

and constructed as follows. From any point C, draw an indefinite line, CM, and perpendicular thereto another indefinite line, CN,

=

Set off on CM, CB = 2s, and take also CE = one of the factors of the numerator, as s+b, and take CF the other factor, viz: s-b, draw BF and also EH parallel thereto, and CH will be the value of a required.

In like manner, if the base and the difference of the hypothenuse Mand perpendicular be given, we

shall have by putting d for the difference and the other letters b and x as before; d+x for the hypothenuse.

To desribe a square in a given triangle.

(By a given triangle is understood a triangle in which the construction is known, viz: one whose sides, angles, altitude, &c., are known.)

It will be perceived that this question resolves itself into the determination of some point, G, in the altitude EF, through which a line AB, drawn parallel to HI, shall be equal to GF; we may, therefore, determine in algebraic expression, for AB, and also for GF, and put them equal to each other and we shall have a solution.

Let us therefore designate the known altitude EF by a, the known base HI by b, and the unknown line, GF by z; then will EG = a― x.

Since AB is parallel to HI, we shall have

EF: EG:: FI: GB :: EI: EB :: HI: AB consequently, EF: EG:: HI : AB

x::b: AB,
bx

a

GF = x,

Or,

ab

ab - bx

therefore,

= x

a

In order to construct this quantity, it is necessary to find a fourth proportional to a+b, b, and a (Art. 2,) which may be done as follows:

From F to O apply a line FO = a+b, that is, = EF + HI, and join EO; then, having taken FM = HI = b, draw MG parallel to EO, which, by its meeting with EF, gives the determination of GF, or the value of x; for the similar triangles EFO, GFM, give FO: FM:: FE: FG

Given the base BC, and the angles B and C of the triangle ABC, to determine the altitude AD.

(Angles are made to enter into an algebraic expression by the aid of lines employed in trigonometry, viz, sines, tangents, &c. Thus when it is said that an angle is given, it may be understood that the value of its sine or tangent is given.)

If we designate BC by a, and AB by y, we shall have CD: AD: radius: tan. ACD, (Trigonometry) or if we designate the radius by r, and the tangent of the angle C by t, we have, CD; y::r: t