To construct a Polygon of a given Area and similar to a given Polygon.

found equivalent to any given polygon, by finding, by § 289, the triangle which is equivalent to the polygon, and, by § 291, the square which is equivalent to this triangle.

295. Problem.

To construct a polygon equivalent to a given circle or polygon, P, and similar to a given. polygon, Q.

Solution. Find, by §§ 293 or 294, M the side of a square equivalent to P, and N the side of a square equivalent to Q. Let A be one of the sides of Q. Find, by 165, a fourth proportional X, to N, M, A. The polygon constructed by § 194, similar to Q upon X, homologous to A, is the required polygon.

Proof. Let Y be the polygon constructed upon X, we

have only to prove that it is equivalent to P.

and leaving out the common ratio A2 : X2,

Solution. Find, by § 294, M the side of a square equiv

alent to the given polygon.

of a square which is to the

Find, by § 265, R the side

given square in the ratio of

the diameter of a circle to its circumference. R is the radius of the required circle.

To construct a Parallelogram equivalent to a given Square.

Proof. Using as in § 237, we have, by construction,

That is, by § 280, the circle of which R is the radius is equal to the given polygon..

297. Problem. To construct a parallelogram, equivalent to a given square, and having the sum of its base and altitude equal to a given line.

Solution. Upon the given line AB (fig. 138) as a diameter describe a semicircle. At A, erect the perpendicular AC equal to the side of the given square. Draw CD parallel to AB, to meet the circumference at D. Draw DE perpendicular to AB; AE and EB are the required base and altitude.

Proof. For AE+EBAB, and by § 290,

AE XEB =DE2: = AC2.

298. Problem. To construct a parallelogram, equivalent to a given square, and having the difference of its base and altitude equal to a given line.

Solution. Upon the given line AB (fig. 139) as a diameter describe a circle. At A draw the tangent AC equal to the side of the given square. Through the centre O of the circle, draw the secant COE. CD and CE are the required base and altitude.

Ratio of a Circumference to its Diameter.

299. Lemma. If in a circle, whose radius is R, C is the chord of an arc and C' the chord of half the arc; C, C and R will always satisfy the equation

C2 = 2 R2-R √√ (4 R2 — C2).

Proof. Let AB (fig. 140) be the chord C and let AA' be C'; OMA' is, by § 117, perpendicular to AB, and the triangle OMA gives

R2

QM2=OA2-AM2 R2 (} C)2= R2— C2

Hence, by § 187,
A'M

AO-OM-R-√(R2 — ↓ C2)
C/2AA12 A'M × A'D'

=

301. Problem. To find the ratio of the circumference of a circle to its diameter.

Solution. This ratio has been denoted, in § 237, by л; it does not admit of being expressed in numbers, and can only be obtained approximately. The principle of approximation consists in supposing the circumference to be equal to the perimeter of some one of its inscribed polygons: : and the error of this hypothesis is the less, the greater the number of sides of the polygon.

First Approximation. Let the radius AO (fig. 140) of the circle be unity, and its circumference is, by § 238, 2 π. If, now, the hexagon ABCD &c. is inscribed in the circle, we have, by § 214, for its side,

Ratio of a Circumference to its Diameter.

and for its perimeter

6 X AB=

= 6;

so that, by supposing this perimeter to be equal to the circumference, we have for a first approximation

Second Approximation. Bisect the arcs AB, BC &c by the radii OA', OB' &c. Join AA', A'B &c., and we have an inscribed polygon of 12 sides, and, by § 300,

And, if this is assumed for the circumference, we have,

for the second approximation,

π= 3.102 nearly.

Third Approximation. If now we consider AB as the side of the inscribed polygon of 12 sides, AA′ is the side of the polygon of 24 sides, and we have for AB,

AA2—2—√4— AB2 = 2−√2+ √3=0·068.

=

and, by assuming this perimeter for the circumference, we have

π= = 3.13 nearly.

Further approximations might be obtained by supposing AB successively to be the side of an inscribed polygon of 24, 48, &c. sides, and by carrying the calculation to a

Value of π.

greater number of decimals. But it is useless to extend this process any further, as much more expeditious methods of calculating the value of are obtained from the higher branches of mathematics, by means of which it has been calculated to 140 places of decimals.

For almost all practical purposes, the value of

is sufficiently accurate.

π= =3.1416,

CHAPTER XIV.

ISOPERIMETRICAL FIGURES.

302. Definitions. Those figures which have equal perimeters are called isoperimetrical figures.

Among quantities of the same kind, that which is greatest is called a maximum; and that which is smallest a minimum.

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.

303. Theorem. The maximum of isoperimetrical triangles of the same base is that triangle in which the two undetermined sides are equal.

Proof. Let the two triangles ACB and ADB (fig. 141) have the same base AB, and the same perimeter, that is, ABAC+BC=AB+AD +- BD,