* CHAPTER XX.

ON THE AREA OF THE CIRCLE, THE CONSTRUCTION OF TRIGONOMETRICAL TABLES, &c.

283. Let r be the radius of the circle described about a regular polygon of n sides. Let O be the centre; HK one of the sides.

From

draw OD perpendicular to HK, bisecting both HK and the angle HOK. Then, since the polygon has n equal sides, n times the angle HOK make up four right angles

the perimeter of the polygon, which is n times HK,

= 2n, DH = 2n. OH sin DOH

the area of the polygon, which is n times the area of HOK,

284. Letr be the radius of the circle inscribed in a regular polygon of n sides. Let O be the centre, HK one of the sides.

Ha Ma K

From O draw OM perpendicular to HK, bisecting both HK and the angle HOK. Then since the polygon has n equal sides, n times the angle HOK make up four right angles,

the perimeter of the polygon, which is n times HK,

= 2n, MH = 2n. OM tan MOH

The area of the polygon, which is n times the area of HOK,

=n. OM. MH n. OM, OM tan MOH

the radius × half the perimeter.

285. Let a circle of radius r have a regular polygon of n sides circumscribed about it, and also a regular polygon of the same number (n) of sides inscribed in it.

The perimeter of the circumscribing polygon is 2nr tan

The perimeter of the inscribed polygon is 2nr sin

Π

The ratio of these perimeters is 1 : cos

The area of the circumscribed polygon is nr tan

The area of the inscribed polygon is nr2 sin

The ratio of these areas is 1 : cos2

T

n

π

n

286. We must assume the two following axioms:

π

n

(i) The circumference of the circle lies in magnitude between the perimeter of a circumscribed and that of an inscribed polygon.

(ii) The area of the circle lies in magnitude between the area of a circumscribed and that of an inscribed polygon.

π

Now, when n is increased, is diminished, and therefore

Hence, as the number of the sides of the two polygons in Art. 285 is increased, their perimeters approach to equality.

And since the circumference of the circle always lies in magnitude between them, each of these perimeters must approach the circumference of the circle.

Therefore, the circumference of a circle is that, to which the perimeter of a regular inscribed (or circumscribed) polygon approaches as the number of its sides is increased, and from which it can be made to differ by less than any assignable quantity however small. (Cf. Art. 32.)

287. In like manner it follows, from (ii) Art. 285, that the area of the circle is the 'limit' of the area of a regular inscribed (or circumscribed) polygon when the number of the sides is indefinitely increased.

Now, twice the area of a polygon circumscribed about a circle is equal to (the radius × the perimeter). [Art. 284]

This is true however great be the number of the sides. It is therefore true when the number of the sides is indefinitely increased.

Therefore it is true for the circle itself.

Hence,

twice the area of a circle = the radius × the circumference.

Or, the area of a circle = år (2πr)

=

[Art. 34]

EXAMPLES. LXXIV.

(1) Prove the surd expressions of Art. 37 for the ratios of the perimeters of certain regular polygons to the diameters of their circumscribing circles.

(2) Prove that the area of a regular polygon of twelve sides described about a circle whose radius is 1 foot is 3.215 sq. ft.

(3) Prove that the area of a square described about a circle is of the dodecagon inscribed in the same circle.

(4) Find the perimeter of a regular polygon of 100 sides (i) when described about a circle of 1 foot diameter, (ii) when inscribed in the same circle. Ans. (i) 3·14263, (ii) 3·14108 ft.

(5) An equilateral triangle and a regular hexagon have the same perimeter: show that the areas of their inscribed circles are as 4:9.

(6) Prove that the area of a regular polygon of n sides, each

(7) If the areas of a regular pentagon and decagon are equal, the ratio of their sides is 20: 1.

(8) If a be a side of a regular polygon and R and r the radii of the inscribed and described circles respectively, then 2R=

= a cosec and 2ra cot

n

π

n

(9) If R and r be the radii of the inscribed and circumscribed circles of a regular polygon of n sides, each=a,

(10) The triangle formed from one side each of a regular pentagon, hexagon and decagon inscribed in the same circle, is right-angled.

(11) Prove that the area of an irregular polygon described about a circle is equal to the product of the radius and half the perimeter of the polygon.

(12) The area of an irregular polygon of an even number of sides circumscribed about a circle is equal to the radius x the sum of every alternate side.

(13) The diameter of the dome of St Paul's is 108 feet; prove that it covers an area of 1018 sq. yds.

(14) The radius of the circle whose area is one acre is 39 yds.

(15) A length of 300 yards of paper, the thickness of which is the hundred and fiftieth part of an inch, is rolled up into a solid cylinder. Find approximately the diameter of the cylinder. Ans. 9.575 in.

(16) The diameter of a roll of carpet is 2 feet and the thickness of the carpet is the eighth of an inch. What is the length of the carpet? Ans. 3016 ft.