Proposition 17. Problem.

461. To compute the ratio of the circumference of a circle to its diameter.

Given, the Oce C, and the radius R.

Required, to find the number π.

Cons. С 2πR.

(436)

Let R = 1, then π = = C.

That is, the number = semi Oce of radius 1.

Therefore, the semi-perimeter of each polygon inscribed in this Oce is an approximate value of : and as the number of sides of the polygons increases indefinitely, the lengths of the perimeters approach to that of the Oce as the limit.

(430)

Hence, by the process of (460), we may obtain a succession of nearer and nearer approximations to the length of the semi Oce.

It is convenient to begin the computation with the inscribed hexagon.

... making AB = 1, we have from (460), the following:

The last two results show that the first four decimals do

not change as the number of sides is increased.

Hence the approximate value of π is 3.1415, correct to

the fourth decimal place.

In practice we generally take 3.14159.

Q.E.F.

462. SCH. The above is called the method of perimeters. For the method of isoperimeters see Rouché et Comberousse, Edition of 1883, p. 194.

NOTE. The number is of such fundamental importance in geometry that mathematicians have sought for its value in a great variety of ways, all of which agree in the conclusion that it cannot be expressed exactly in decimals, but only approximately. Archimedes (born 287 B.C.) was the first to assign an approximate value of T. He proved that it is included between the numbers 24 and 349, or, in decimals, between 3.1428 and 3.1408; he therefore assigned its value correctly within a unit of the third decimal place. The number 34, or 4 is often used in rough computations.

Adrian Metius, a Dutch geometer of the 16th century, has given us the much more accurate value of , which is correct to within a half-millionth, and which is remarkable for the manner in which it is formed by the first three odd numbers 1, 3, 5, each written twice.

More recently, the value has been found to a great number of decimals by the aid of series. Dase and Clausen, German computers, carried the calculation to 200 decimal places, independently of each other, and their results agreed to the last figure.

The first 20 figures of their result are as follows:

3.14159 26535 89793 23846.

(437)

This result is far beyond all the wants of mathematics. Ten decimals are sufficient to give the circumference of the earth to the fraction of an inch, and thirty decimals would give the circumference of the whole visible universe to a quantity imperceptible with the most powerful microscope.*

EXERCISES.

1. The area of the regular inscribed hexagon is equal to twice the area of the inscribed equilateral triangle.

2. The square of a side of the inscribed equilateral triangle is three times the square of a side of the regular inscribed hexagon.

3. The area of a regular inscribed hexagon is half the area of the circumscribed equilateral triangle.

4. If the diameter of a circle be produced to C until the produced part is equal to the radius, the two tangents from C and their chord of contact form an equilateral triangle.

*Newcomb's Geom., p. 235.

5. Divide an angle of an equilateral triangle into five equal parts.

Describe a about the ▲, then use (453).

6. The square inscribed in a semicircle is equal to twofifths the square inscribed in the whole circle.

7. The area of a given circle is 314.16; if this circle be circumscribed by a square, find the area of the part between the circumference and the perimeter of the square.

8. The area of a circle is 40 feet; find the side of the inscribed square.

9. Find the angle subtended at the centre of a circle by an arc 6 feet long, if the radius is 8 feet long.

10. Find the length of the arc subtended by one side of a regular octagon inscribed in a circle whose radius is 20 feet.

11. Find the area of a circular sector whose arc contains 18°, the radius of the circle being 4 feet.

12. Find the area of a circular sector, the chord of half the arc being 20 inches, and the radius 45 inches.

13. The radius of a circle is 5 feet: find the area of a circle 7 times as large.

14. The radius of a circle is 7 feet: find the radius of a circle 16 times as large.

15. Find the height of an arc, the chord of half the arc being 10 feet, and the radius 16 feet.

16. Find the area of a segment whose height is 4 inches, and chord 30 inches.

17. Find the area of a segment whose height is 16 inches, the radius of the circle being 20 inches.

18. Find the area of a segment whose arc is 100°, the radius being 12 feet.

19. Find the area of a sector, the radius of the circle being 24 feet, and the angle at the centre 30°.

20. A circular park 500 feet in diameter has a carriageway around it 25 feet wide: find the area of the carriageway.

MAXIMA AND MINIMA.

463. DEF. A maximum quantity is the greatest quantity of the same kind; and a minimum quantity is the least quantity of the same kind.

Thus, the diameter of a circle is a maximum among all inscribed straight lines; and a perpendicular is a minimum among all the straight lines drawn from a given point to a given straight line.

464. Isoperimetric figures are those which have equal perimeters.

We give here a few simple but important propositions bearing on this part of Geometry.

Proposition 18. Theorem.

465. Of all triangles formed with two given sides, that in which these sides are perpendicular to each other is the maximum.

Hyp. Let ABC, A'BC be two As having the sides AB, BC, respectively equal to A'B, BC; and let ABC be

is the shortest distance from a pt. to a line, (58)

... A'B > A'D.

AB = A'B.

... AB > AʼD.

(Hyp.)

But AB and A'D are the altitudes respectively of the As ABC, A'BC.

... Δ ABC > Δ ABC.

(369)

Q.E.D.

Proposition 19. Theorem.

466. Of all triangles having equal perimeters and the same base, the isosceles triangle is the maximum.

Hyp. Let the As ABC, ABD

H

=

Then

ABH is a rt. ≤,

being inscribed in the

whose cent. is Cand radius is CB (240).

=

Produce HB, take DL DB, join AL, and draw CG, DM ||, and CE, DF L, to AB.

Then AD + DL = AD + DB = AC + CB = AH.

But BH = BG = CE, and BL = BMDF.

(61)

(369)

Q.E.D.

467. COR. Of all the triangles of the same perimeter, that which is equilateral is the maximum.

For, the maximum triangle having a given perimeter must be isosceles whichever side is taken as the base.