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thus a square may always be found equivalent to a given rectilineal figure, which operation is called squaring the rectilineal figure, or finding the quadrature of it.

The problem of the quadrature of the circle, consists in finding a square equivalent to a circle whose diameter is given.

PROBLEM IX.

To find the side of a square which shall be equivalent to the sum, or the difference of two given squares.

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right-angles to each other; take ED=A, and EG=B; join DG: this will be the side of the square required.

For the triangle DEG being right-angled, the square described upon DG is equivalent to the sum of the squares upon ED and EG. (11. 4.)

Second. If it is required to find a square equivalent to the difference of the given squares, form in the same manner the right-angle FEH; take GE equal to the shorter of the sides A and B; from the point G as a centre, with a radius GH, equal to the other side, describe an arc cutting EH in H: the square described upon EH will be equivalent to the difference of the squares described upon the lines A and B.

For the triangle GEH is right-angled, the hypothenuse GHA, and the side GE=B; hence the square described upon EH, is equal to the difference of the squares described upon B and A.

Scholium. A square may thus be found equivalent to the sum of any number of squares; for the construction which

reduces two of them to one, will reduce three of them to two, and these two to one, and so of others. It would be the same, if any of the squares were to be 'subtracted from the sum of the others.

PROBLEM X.

To construct a square which shall be to a given square ABCD as the line M is to the line N.

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point F erect the perpendicular FH. From the point H, draw the chords HG, HE, which produce indefinitely upon the first take HK equal to the side AB of the given square, and through the point K draw KI parallel to EG; HI will be the side of the square required.

For, by reason of the parallels KI, GE, we have, (15.4,) HI: HK:: HE: HG; hence HI2: HK2:: HE2: HG2: (20. 3) but in the right-angled triangle EHG, the square of HE is to the square of HG as the segment EF is to the segment FG, (11. 4. Cor. 3,) or as M is to N; hence HI2: HK2: :: M: N. But HK-AB; therefore the square described upon HI is to the square described upon AB as M is to N.

PROBLEM XI.

Upon the side FG, homologous to AB, to describe a polygon similar to the given polygon ABCDE.

In the given polygon, draw the diagonals AC, AD; at the

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other in H, and FGH will be a triangle similar to ABC. In the same manner upon FH, homologous to AC, describe the triangle FIH similar to ADC; and upon FI, homologous to AD, describe the triangle FIK similar to ADE. The polygon FGHIK will be similar to ABCDE, as required.

For, these two polygons are composed of the same number of triangles, which are similar, and similarly situated. (26. 4. Sch.)

PROBLEM XII.

To construct a rectangle equivalent to a given square C, and having its adjacent sides together equal to a given line AB.

Upon AB as a di

ameter, describe a semicircle; draw the

line DE parallel to the diameter, at a distance AD equal to the

D

E

FB

side of the given square C; from the point E, where the parallel cuts the circumference, draw EF perpendicular to the diameter; AF and FB will be the sides of the rectangle required.

For, their sum is equal to AB; and their rectangle AF.FB is equivalent to the square of EF, or to the square of AD; hence that rectangle is equivalent to the given square C.

Scholium. To render the problem possible, the distance

ÀD must not exceed the radius; that is, the side of the square C must not exceed the half of the line AB.

PROBLEM XIII.

To construct a rectangle that shall be equivalent to a given square C, and the difference of whose adajcent sides shall be equal to a given line AB.

Upon the given line AB as a diameter, describe a semicircle; at the extremity of the diameter draw the tangent AD, equal to the side of the square C; through the point D and the centre O draw the secant DF: then will DE and DF be the adjacent sides of the rectangle required.

A

D

0

B

F

For, first, the difference of their sides is equal to the diameter EF, or AB; second, the rectangle, DE.DF is equal to AD2; (30. 4;) hence that rectangle is equivalent to the given square C.

BOOK V.

REGULAR POLYGONS, AND THE MEASUREMENT

OF THE CIRCLE.

DEFINITION.

A polygon, which is at once equilateral and equiangular, is called a regular polygon.

Regular polygons may have any number of sides; the equilateral triangle is one of three sides, the square is one of four, &c.

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Regular polygons of the same number of sides are similar figures.

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in both figures, being in each equal to eight right-angles. (29. 1. Cor. 3.) The angle A is a sixth part of that sum; so is the angle a: hence the angles A and a are equal; (Ax. 1;) and for the same reason, the angles B and b, the angles C and c, &c., are equal.

Again, since the polygons are regular, the sides AB, BC, CD, &c., are all equal, and likewise the sides ab, bc, cd, &c; it is plain that AB: ab :: BC: bc:: CD : cd, &c., hence the two figures in question have their angles equal, and their homologous sides proportional; consequently they are similar. (Def. 3. 4.) Hence, All regular polygons, &c.

Cor. The perimeters of two regular polygons of the same number of sides, are to each other as the homologous sides, and their surfaces are as the squares of those sides. (27. 4.)

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