BOOK XIV.

APPLICATIONS OF ALGEBRA TO GEOMETRY.

692. WHEN it is proposed to solve a geometrical problem by aid of Algebra, draw a figure which shall represent the several parts or conditions of the problem, both known and required.

Represent the known parts by the first letters of the alphabet, and the required parts by the last letters.

Then, observing the geometrical relations that the parts of the figure have to each other, make as many independent equations as there are unknown quantities introduced, and the solution of these equations will determine the unknown quantities or required parts.

To form these equations, however, no definite rules can be given; but the best aids may be derived from experience, and a thorough knowledge of geometrical principles. It should be the aim of the learner to effect the simplest solution possible of each problem.

PROBLEM I.

693. In a right-angled triangle, having given the hy pothenuse, and the sum of the other two sides, to deter mine these sides.

Let ABC be the triangle, right-angled at B. Put A C

=

C

a, the sum AB

If A C 5, and the sum AB+BC=7, y = 4 or 3,

=

and x = 3 or 4.

PROBLEM II.

694. Having given the base and perpendicular of a triangle, to find the side of an inscribed square.

Let ABC be the triangle,

and H E F G the

inscribed

that is, the side of the inscribed square is equal to the product of the base by the altitude, divided by their sum.

PROBLEM III.

695. Having given the lengths of two straight lines drawn from the acute angles of a right-angled triangle to the middle of the opposite sides, to determine those sides.

Let ABC be the given triangle,

and AD, BE the given lines.

Α

which values of x and y are half the base and perpendiculars of the triangle.

PROBLEM IV.

696. In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a point within to the three sides, to determine these sides.

Let A B C be the equilateral triangle, and D E, DF, DG the given perpendiculars from the point D. Draw DA, DB, DC to the vertices of the three angles, and let fall the perpendicular, CH, on the base, AB. = DG b,

=

C,

A

C

DE

HG

B

Put DE = a, DF and AH or BH, half the side of the equilateral triangle, = x. Then AC or BC= 2x, and CHAC-AH2 = √ 4 x2 — x2 = √ 3x2= x √3. Now, since the area of a triangle is equal to the product of half its base by its altitude (Prop. VI. Bk. IV.),

The triangle ACB=ABXCH=xxx√3 = x2√3.

ABD=AB×DG= x×c

BCD BC×DE= x×a

=

ACD-AC-DF=xx b

=Cx.

=ax.

=bx.

But the three triangles A BD, BCD, ACD are together

equal to the triangle A C B.

Hence, 2√3 = ax + bx + c x = x (a+b+c),

697. Cor. Since the perpendicular, CH, is equal to x3, it is equal to a+b+c; that is, the whole perpendicular of an equilateral triangle is equal to the sum of all the perpendiculars let fall from any point in the triangle to each of its sides.

PROBLEM V.

698. To determine the radii of three equal circles described within and tangent to a given circle, and also tangent to each other.

Let AF be the radius of the given circle, and BE the radius of one of the equal circles described within it. Put AF=a, and BE =x; then each side of the equilateral triangle, BCD, formed by joining the centers of the required circles, will be represented by 2x, and its altitude, CE, by 4x2- x2, or x √3.

B

The triangles BCE, A BE are similar, since the angles BCE and ABE are equal, each being half as great as one of the angles of the equilateral triangle, and the angle BEC is common.

699. In a right-angled triangle, having given the base, and the sum of the perpendicular and hypothenuse, to find these two sides.

PROBLEM VII.

700. In a rectangle, having given the diagonal and perimeter, to find the sides.