APPLICATION OF ALGEBRA

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GEOMETRY.

WHEN it is proposed to resolve a geometrical problem

algebraically, or by algebra, it is proper, in the first place, to draw a figure that shall represent the several parts or conditions of the problem, and to suppose that figure to be the true one. Then, having considered attentively the nature of the problem, the figure is next to be prepared for a solution, if necessary, by producing or drawing such lines in it as appear most conducive to that end. This done, the usual symbols or letters, for known and unknown quantities, are employed to denote the several parts of the figure, both the known and unknown parts, or as many of them as necessary; as also such unknown line or lines as may be easiest found, whether required or not. Then proceed to the operation, by observing the relations that the several parts of the figure have to each other; from which, and the proper theorems in the foregoing elements of geometry, make out as many equations independent of each other, as there are unknown quantities employed in them: the resolution of which equations, in the same manner as in arithmetical problems, will determine the unknown quantities, and resolve the problem proposed.

As no general rule can be given for drawing the lines, and selecting the fittest quantities to, substitute for, so as always to bring out the most simple conclusions, because different problems require different modes of solution; the best way to gain experience, is to try the solution of the same problem in different ways, and then apply that which succeeds best, to other cases of the same kind, when they afterwards occur. The following particular directions, however, may be of

some use.

1st, In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or so as to form similar triangles. And if an angle be given, it will be proper to let the perpendicular be opposite to that angle, and to fall from one end of a given line, if possible.

2d, In selecting the quantities proper to substitute for, those are to be chosen, whether required or not, which lie nearest the known or given parts of the figure, and by means of which the next adjacent parts may be expressed by addition and subtraction only, without using surds.

3d, When two lines or quantities are alike related to other parts of the figure or problem, the best way is, not to make use of either of them separately, but to substitute for their sum, or difference, or rectangle, or the sum of their alternate quotients, or for some line or lines, in the figure, to which they have both the same relation,

4th, When the area, or the perimeter, of a figure, is given, or such parts of it as have only a remote relation to the parts required: it is sometimes of use to assume another figure similar to the proposed one, having one side equal to unity, or some other known quantity. For, hence the other parts of the figure may be found, by the known proportions of the like sides, or parts, and so an equation be obtained. For examples, take the following problems.

PROBLEM I.

In a Right-angled Triangle, having given the Base (3), and the Sum of the Hypothenuse and Perpendicular (9); to find both

these two Sides.

LET ABC represent the proposed triangle, right-angled at B. Put the base AB = 3b, and the sum ACBC of the hypothenuse and perpendicular 9 s; also, let r denote the hypothenuse AC, and y the perpendicular BC.

B

52 = 25y + y2 = y2 + b2, 2 - 25y = b2,

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s2 - b2 = 2sy,

-

25

b2

y=4=BC,

N. B. In this solution, and the following ones, the notation is made by using as many unknown letters, r and y, as

there

there are unknown sides of the triangle, a separate letter for each; in preference to using only one unknown letter for one side, and expressing the other unknown side in terms of that letter and the given sum or difference of the sides; though this latter way would render the solution shorter and sooner; because the former way gives occasion for more and better practice in reducing equations, which is the very end and reason for which these problems are given at all.

PROBLEM II.

In a Right-angled Triangle, having given the Hypothenuse (5) ; and the Sum of the Base and Perpendicular (7); to find both these two Sides.

LET ABC represent the proposed triangle, right-angled at B. Put the given hypothenuse AC = 5 = a, and the sum ABBC of the base and perpendicular 7s; also let a denote the base AB, and y the perpendicular BC.

4 and 3, the values of x and y.

PROBLEM III.

In a Rectangle, having given the Diagonal (10), and the Perimeter, or Sum of all the Four Sides (28); to find each of the Sides severally.

LET ABCD be the proposed rectangle; and put the diagonal AC = 10 = d, and half the perimeter AB + BC or AD + DC = 4 = a; also put one side AB = T, and the other side BC= y. Hence, by right-angled triangles, And by the question

x2 + y2
= ď2
x + y = a

Then by transposing y in the 2d, gives
This value substituted in the 1st, gives a

a

-

y 2ay+2y2 = d2 Transposing

Having given the Base and Perpendicular of any Triangle; ta find the Side of a Square Inscribed in the same.

LET ABC represent the given triangle, and EFGH its inscribed square. Put the base AB b, the perpendicular CD = a, and the side of the square GF or GH = DI; then will cr = CD

DI =

Then, because the like lines in the

similar triangles ABC, GFC, are propor

F

tional (by theor. 84, Geom.), AB: CD :: GE: CI, that is, bax: a - x. Hence, by multiplying extremes and means, ab-bx = ax, and transposing br, gives ab = ax

br; then dividing by a + b, gives x =

ab

= GF

a.+ b or GH the side of the inscribed square: which therefore is of the same magnitude, whatever the species or the angles of the triangles may be.

PROBLEM V.

In an Equilateral Triangle, having given the lengths of the three Perpendiculars, drawn from a certain Point within, on the three Sides; to determine the Sides.

F

LET ABC represent the equilateral triangle, and DE, DF, DG, the given perpendiculars from the point D. Draw the lines DA, DB, DC, to the three angular points; and let fall the perpendicular CH on the base AB. Put the three given perpendiculars DE = 4, DF = b, DG = c, and put = AH or BH, half the side of the equilateral triangle. Then is AC or BC= 2x, and by right-angled triangles the perpendicular CHAC2 — AH2 =√4x2x22 √3x2=x√3. x2 = √/3x2 = x√/ 3.

HG

Now,

Now, since the area or space of a rectangle, is expressed by the product of the base and height (cor. 2, th. 81 Geom.), and that a triangle is equal to half a rectangle of equal base and height (cor. 1, th. 26), it follows that,

the whole triangle ABC is AB X CHX √/3=x2√3,

=

But the three last triangles make up, or are equal to, the whole former, or great triangle;

that is,

3 ax + bx + cx; hence, dividing by r, gives x √3 = a + b + c, and dividing by √3, gives a+b+c half the side of the triangle sought.

√3

Also, since the whole perpendicular CH is = x/3, it is therefore a + b + c. That is, the whole perpendicular CH, is just equal to the sum of all the three smaller perpendiculars DE DF + DG taken together, wherever the point D is situated.

PROBLEM VI.

IN a Right-angled Triangle, having given the Base (3), and the Difference between the Hypothenuse and Perpendicular (1); to find both these two Sides.

PROBLEM VII.

IN a Right-angled Triangle, having given the Hypothenuse (5), and the Difference between the Base and Perpendicular (1); to determine both these two Sides.

PROBLEM VIII.

HAVING given the Area, or Measure of the Space, of a Rectangle, inscribed in a given Triangle; to determine the Sides of the Rectangle.

PROBLEM IX.

IN a Triangle, having given the Ratio of the two Sides, together with both the Segments of the Base, made by a Perpendicular from the Vertical Angle; to determine the Sides of the Triangle.

PROBLEM X.

IN a Triangle, having given the Base, the Sum of the other two Sides, and the Length of a Line drawn from the

Vertical