moved from the perpendicular SO, and
consequently equal (Book VI. Prop. V.).
One of them SA=AB; hence the four
faces of the pyramid S-ABC, are trian- A
gles, equal to the given triangle ABC.
And the solid angles of this pyramid
are all equal, because each of them is
formed by three equal plane angles:
hence this pyramid is a regular tetrae-
dron.

Construction of the Hexaedron.

Let ABCD be a given square. On the base ABCD, construct a right prism whose altitude AE shall be equal to the side AB. The faces of this prism will evidently be equal squares; and its solid angles all equal, each being formed with three right angles: hence this prism is a regular hexaedron or cube.

A

The following propositions can be easily proved.

1. Any regular polyedron may be divided into as many regular pyramids as the polyedron has faces; the common vertex of these pyramids will be the centre of the polyedron; and at the same time, that of the inscribed and of the circumscribed sphere.

2. The solidity of a regular polyedron is equal to its surface multiplied by a third part of the radius of the inscribed sphere.

3. Two regular polyedrons of the same name, are two similar solids, and their homologous dimensions are proportional; hence the radii of the inscribed or the circumscribed spheres are to each other as the sides of the polyedrons.

B

4. If a regular polyedron is inscribed in a sphere, the planes drawn from the centre, through the different edges, will divide the surface of the sphere into as many spherical polygons, all equal and similar, as the polyedron has faces.

APPLICATION OF ALGEBRA.

TO THE SOLUTION OF

GEOMETRICAL PROBLEMS.

A problem is a question which requires a solution. A geometrical problem is one, in which certain parts of a geometrical figure are given or known, from which it is required to determine certain other parts.

When it is proposed to solve a geometrical problem by means of Algebra, the given parts are represented by the first letters of the alphabet, and the required parts by the final letters, and the relations which subsist between the known and unknown parts furnish the equations of the problem. The solution of these equations, when so formed, gives the solution of the problem.

No general rule can be given for forming the equations. The equations must be independent of each other, and their number equal to that of the unknown quantities introduced (Alg. Art. 103.). Experience, and a careful examination of all the conditions, whether explicit or implicit (Alg. Art. 94,) will serve as guides in stating the questions; to which may be added the following particular directions.

1st. Draw a figure which shall represent all the given parts, and all the required parts. Then draw such other lines as will establish the most simple relations between them. If an angle is given, it is generally best to let fall a perpendicular that shall lie opposite to it; and this perpendicular, if possible, should be drawn from the extremity of a given side.

2d. When two lines or quantities are connected in the same way with other parts of the figure or problem, it is in general, not best to use either of them separately; but to use their sum, their difference, their product, their quotient, or perhaps another line of the figure with which they are alike connected.

3d. When the area, or perimeter of a figure, is given, it is sometimes best to assume another figure similar to the proposed, having one of its sides equal to unity, or some other known quantity. A comparison of the two figures will often give a required part. We will add the following problems.*

*The following problems are selected from Hutton's Application of Algebra to Geometry, and the examples in Mensuration from his treatise on that subject.

In a right angled triangle BAC, having given the base BA, and the sum of the hypothenuse and perpendicular, it is required to find the hypothenuse and perpendicular.

Put BA=c=3, BC=x, AC=y and the sum of the hypothenuse and perpendicular equal to S=9

Then,

x+y=s=9.

and x2=y2+c2 (Bk. IV. Prop. XI.)

hence,

Therefore

From 1st equ: x=s—y

x2=s2—Qsy+y2

and By subtracting, 0=s-2sy-c2 2sy=s2―c2

or

PROBLEM I.

or

Hence,

or

and

From first equation

or

y= 2s

=4=AC

+4-9 or x=5=BC.

Put BC=a=5, BA=x, AC=y of the base and perpendicular=s=7

Then

or

Hence

PROBLEM II.

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

and the sum

x+y=s=7
x2+ y2=a2

B

x=s-y

x2=s2-2sy + y2

y2=a2—s2+2sy—y2
Qy2—2sy=a2-

y2—sy= 2

By completing the square y3—sy+1s2—1a2—\s2

с

y={s±√¿a2—÷s2=4 or 3
x= {s = √ {a2 — } s2=3 or 4

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

Let ABCD be the proposed rectangle. D

or

Put AC=d=10, the perimeter=2a=28,
AB+BC=a=14: also put AB=x and BC=y.
x2+ y2=d2

Then,

and

x+y=a

From which equations we obtain,

and

PROBLEM III.

or

Let ABC be the triangle and HEFG the inscribed square. Put AB=b, CD=a, and HE or GH-x: then CI-a-x.

Hence,

or

y=}a± √÷d2—4a2=8 or 6,
x={a= √}d2—}a2=6 or 8.

We have by similar triangles

PROBLEM IV.

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

x=

AB: CD:: GF: CI

b: a:: x: a-x

ab-bx-ax

ab

= the side of the inscribed square;

a+b

which, therefore, depends only on the base and altitude of the triangle.

PROBLEM V.

A

A

S*

B

H DE B

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

Let ABC be the equilateral triangle; DG, DE and DF the given perpendiculars let fall from D on the sides. Draw DA, DB, DC to the vertices of the angles, and let fall the perpendicular CH on the base. Let DG-a, DE=b, and DF-c: put one of the equal sides AB

A.

=2; hence AH=x, and CH=√AC2—AH2=√4x2_x2

=

= √3x2=x√3.

AB × CH=x× ≈ √3=x2 √3=triangle ACB

AB×DG=x × a =αx =triangle ADB
BCXDE=xx b =bx triangle BCD

=Cx

=triangle ACD

or

Now since the area of a triangle is equal to half its base into the altitude, (Bk. IV. Prop. VI.)

therefore,

F

ACX DF=xxc

But the three last triangles make up, and are consequently equal to, the first; hence,

x2√3=ax + bx + cx=x(a+b+c);

ᎠᎬ

PROBLEM VI.

H G B

REMARK. Since the perpendicular CH is equal to x√3, it is consequently equal to a+b+c: that is, the perpendicular let fall from either angle of an equilateral triangle on the opposite side, is equal to the sum of the three perpendiculars let fall from any point within the triangle on the sides respectively.

PROBLEM VII.

In a right angled triangle, having given the base and the difference between the hypothenuse and perpendicular, to find the sides.

In a right angled triangle, having given the hypothenuse and the difference between the base and perpendicular, to determine the triangle.