which shews that the ratio between the circumference and diameter is the same for all circles (is constant); this ratio, which is an abstract number, is denoted by the letter л, so that P πο 2πR.

It therefore depends upon, once for all, to calculate this number; when it is known, we can always of the three quantities P, D, and R calculate the two, when we know the third. As the ratio is the same for all circles, we take radius equal to 1, therefore the diameter equal to 2, and calculate the perimeter of an inscribed polygon with several sides; if the number of sides be n, we have

=

=

We have ke -I and found from this (104) k12 V2 −√3 0.517638090...; from this we find by degrees k24, k48... for example, to k788; by multiplying the value found thus by 768, we get the perimeter of the 768 sided figure (nk,) equal to 6.283160...; after division by 2, we therefore have

3.141580 <π.

From 768 we thereupon calculate p768 (103) and from this again 1768; if this be multiplied by 768, we get the perimeter of the circumscribed 768 sided figure (nt,); for this we find the numbet 6.283212..., and get by dividing by 2

π <3.141606.

As π lies between the values found thus, we have with 4 correct decimals

By easier methods,

which cannot be shewn here, π has been found with several hundred decimals; the first are

I

108. As an arc of 1° is of the circumference, its length

360

and consequently the length b of an arc of go

By the help of this equation, when two of the three quantities b, r, and g are known, we can calculate the third. If g be expressed in minutes or seconds, the denominator must be multiplied by 60 or 602 respectively.

EXAMPLES.

184. How great is the circumference of a circle, when the radius is 2'4", and how great is the radius, when the circumference is 3' 6"?

==

AC.

185. A circle touches another circle internally in A and passes through its centre. Through this centre a line is drawn cutting the circles in B and C; shew that AB 186. Two circles with radii R and r touch a circle with radius R+r internally respectively in A and B, whilst C is one of their points of intersection. Prove that AB ~ AC+~CB.

=

187. How long is an arc of 12° 13′ 20′′, when the radius is

I' 8"?

188. How many degrees are contained in an arc, the length of which is 3", when the radius is 2“?

189. A row of semicircles is so placed that the diameters are prolongations of each other. Prove that the sum of the lengths of the circumferences of all the semicircles is equal to the length of a semicircle, the diameter of which is the sum of the given diameters.

190. Find the number of degrees contained in an arc, the length of which equals the radius.

191. How great is the radius of the earth, when one degree of latitude is 60 geographical miles?

192. O is the centre of a circle and A a point, the distance of which from 0 is n times the radius. On OA as di

1

ameter a circle is described. Two lines from A intercept arcs on the first circle, having the lengths b1 and b2; how long is the arc intercepted by the two lines on the second circle?

193. An undulating line is formed by dividing a line a into n equal parts, and on each of the parts as chord describing an arc of 60°, alternately upwards and downwards. How long is the undulating line? What does this length become, and what becomes of the undulating line, when n grows infinitely, whilst a remains unchanged?

IV.

AREA.

109. The area of a figure can be measured by the area of a square, the side of which is the unit of length.

According as this is foot, inch, &c., the unit of surface is called square foot, square inch, &c.; the designations are the same as for lineal measure, but with the addition of the sign D.

110. Rectangles with equal bases are to each other as their altitudes.

If the altitudes are incommensurable, the proposition also holds good and is proved as ß in 83.

111. Rectangles with equal altitudes are to each other as their bases.

This proposition is the same as the preceding one, for any of the sides we please can be taken as base.

112. The ratio between any two rectangles is the product of the ratio between the altitudes and the ratio between the bases.

Let the rectangles be R, with altitude H and base G, and r, with altitude h and base g; let us then imagine a third rectangle P, with altitude H and base g.

If r denote the unit of surface, therefore h and g the units of length, this equation shews that:

The number of units of surface in a rectangle is found by multiplying the number of units of length in the altitude by the number of units of length in the base; this is written R H.G,

where R, H, and G denote abstract numbers, and we ourselves must remember the denominations; to prevent this, we have adopted the phrase, that foot multiplied by foot gives square foot &c., by which we are enabled to calculate with concrete numbers; we therefore say, that

The area of a rectangle equals the product of the altitude and the base.

The area of a square is therefore the square of the side; we therefore have '

decimal measure, 1"☐

144"

100"

and "O = 144" Duo

Decimal measure.

113. A parallelogram is equal to a rectangle with the same altitude and the same base.

For we have

A

E

FB

C

ΔΕΑΒ Ω Δ FDC.

By subtracting these triangles one at a time from the whole figure, the par

allelogram AC and the rectangle AF will be left respectively; they are therefore equal.

From this it follows that the area of a parallelogram is the product of the altitude and the base.

S

114. The area of a triangle is half the product of the altitude and the base.

C

For the triangle ABC is half of the parallelogram AB, which has the same base and the same altitude. The area of a rightangled triangle is half the product of the sides containing the right angle.

In an equilateral triangle with the side s the altitude is

115. The area of a figure in which a circle can be inscribed is half the product of the radius of the circle and the perimeter of the figure.

d

a

γ b

From the centre draw lines to the angular points; thereby the figure is divided into triangles, which all have the radius of the circle for their altitude, and the bases of which are the sides of the figure a, b, c..., we therefore have the area of the figure

A

1ra + 1 rb + rc....

=

when Ρ is the perimeter.

The radius of a circle inscribed in a triangle is therefore equal to the area divided by half the perimeter.