PROBLEM IV.

To construct a mean proportional between two given lines.

Let it be required to construct a mean proportional between P and Q. On an indefinite line AC lay off AE = P and ECQ. On AC

C

D

E

as a diameter construct a semi-circumference and from E draw a perpendicular to AC, meeting the semi-circumference at D; then is ED the required mean proportional. For, from P. 18, Cor., we have

or

AE ED ED: CE,

P: ED :: ED : Q.

PROBLEM V.

F

E

D

To divide a given line so that the greater part shall be a mean proportional between the whole line and the less part. Let CA be the given line. At A draw AD perpendicular to AC and lay off AD equal to half of AC; from D as a centre with DA as a radius draw the circumference AEF, also draw the C secant CDE; the circumference will be tangent to CA at A and its diameter FE will be equal to CA. From C as a centre with CF as a radius draw an arc cutting CA at G; then will G be the required point of division. For from P. 20, Cor., we have

CE CA: CA: CF.

Whence, by division,

G

A

CE CA: CA :: CA - CF: CF;

(1)

but CECA CF, or CG, CA - CF CA - CG = GA,

and CF

=

CG;

substituting these in (1), we have,

CG: CA :: GA: CG,

CA: CG:: CG: GA.

whence, by inversion,

The segment CG is the greater, because CA > CG.

Scho. The line CA is said to be divided in extreme and mean ratio at the point G.

PROBLEM VI.

To draw a line through a given point within a given angle so that the parts between that point and the sides of the angle shall be equal.

Let P be the given point and ACD

the given angle.

Draw PQ parallel to AC cutting CD at Q; lay off QR equal to CQ and draw RPA; then will RA be the required line. Note. Let the student show that RP = PA.

PROBLEM VII.

A

To construct a polygon on a given line as a side that shall be similar to a given polygon, the given side being homologous with any side of the given polygon.

IM

Let P denote the given line; let ACDE be a given polygon, and AC the side with which P is to be homologous. Draw KL parallel to AC and equal to P; draw AK and CL and prolong them till they meet at some point 0; draw OD and OE; then draw LM parallel to CD, MN parallel to DE, and join N and K.

E

A

Note. Let the student show: 1°. that NK is parallel to EA; and 2°. that the polygon KLMN is similar to the polygon ACDE.

BOOK IV.

MEASURE AND COMPARISON

OF AREAS.

Definitions and Explanations.

78. The unit of measure of a surface is a square described on the linear unit as a side; thus, if the linear unit is a foot, the unit of surface is a square foot.

A unit of surface is sometimes called a superficial unit.

79. The area of a surface is an expression for the surface in terms of a superficial unit.

80. The product of two lines is the number of times the linear unit is contained in one line, multiplied by the number of times the same unit is contained in the other line; the unit of the product is a superficial unit.

The operation of comparing two surfaces is purely numerical. In what follows, we shall regard two surfaces as equal when they have the same area; if they can be made to coincide throughout they are said to be equal in all their parts.

81. The altitude of a triangle is the perpendicular from the vertex of any angle of the triangle to the opposite side, or to the opposite side prolonged; the vertex considered is called the vertex of the triangle, and the opposite side is called the base of the triangle.

82. The altitude of a parallelogram, or of a trape- zoid, is the perpendicular distance between two parallel

sides; the sides considered are called bases, one being the lower base and the other the upper base.

In a parallelogram either pair of parallel sides may be taken as bases.

If two rectangles have equal altitudes they are to each other as their bases.

Let the altitudes CD and LM of the rectangles CE and LN be equal; then are the rectangles proportional to AC and KL.

且耳

C

1o. Let the bases be to each other as two whole numbers, say as 3 is to 4.

Divide AC into three equal parts and KL into four equal parts, then at the points of division draw lines perpendicular to the corresponding bases. These lines will divide the rectangle CE into three partial rectangles, and the rectangle LN into four partial rectangles, all of which are equal to each other (P. 26, Cor. 4, B. 1); hence

hence, from proportions (1) and (2), (P. 4, B. 3), we have

applied to KL and suppose that it is contained in KL, n times, with a remainder less than that part; then will

At each point of division on AC and KL, let lines be drawn respectively perpendicular to the bases. These will divide the rectangle CE into m equal rectangles, and the rectangle LN into n corresponding and equal rectangles, with a remainder less than one of these rectangles; hence,

But m is any whole number; hence, the true values of the ratios given by equations (1) and (2) are equal to each other, (P. 1, Cor., B. 3); that is,

KLMN KL

= or ACDE: KLMN :: AC: KL,

ACDE AC'

which was to be proved.

The area of a rectangle is equal to the product of its base and altitude.

K

E

A

M

D

Let AD be a rectangle and AL the F assumed superficial unit, that is, a square each of whose sides is equal to a linear unit; let the square AL be so placed that AK shall be on the prolongation of CA, and AM on the prolongation of EA. Prolong DE and LK till they meet at F, forming the rectangle AF. Now, if we take AC and AK as bases, the rectangles AD and AF will have the same altitude AE, and we shall have, (P. 1),

ACDE AEFK :: AC: AK;

(1)

if we take AE and AM as bases, the rectangles AF and AL will have the same altitude AK, and we shall have