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PROBLEM III.

From any given point in a right Line, to erect a

Perpendicular.

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1.-On each side of the point A, take equal distances, as 6 A, A c; from b and c, as centres, with any radius greater than 6 A or c A, describe arcs cutting each other in n; then will a line drawn from the point A through n be the perpendicular required. 2.-With

any

radius from the point o as a centre, describe the arc m B n, cutting the line in m and B; draw a line from the point m through the centre o to n; then a line drawn through the point n to B is the perpendicular required.

3.-On the point B as a centre, with any radius describe the arc l m n, cutting the line in l: with the same radius taking m and n as centres, describe arcs cutting each other in r; then a line drawn froin B through r will be the perpendicular required.

4.-From the point B, on the line A B, take three equal parts (as feet, inches, &c.) to m; and from m and B as centres, describe arcs cutting each other in n, making the distance from B to n four parts, and from m to n five parts, then will the line B n be the perpendicular required..

PROBLEM IV.

72

To divide a right Line into any number of equal parts,

On each side of the given line make any angle with the line, as A B m, A Bn; from each end of the line, and along the outside of each angle, with any distance, set off the proposed number of equal parts, as A, 1, 2, 3, &c.; B, 1, 2, 3, &c. ; join the parts, as A, 5; 1, 4; 2, 3, &c.; and the line A B will be divided as required.

PROBLEM V.

D

To find the Centre of a Circle already described. Upon any chord as A B, erect the perpendicular c D; continue c D to Ê, divide E D by problem first, then will the intersections of the two diameters at o be the centre of the circle required.

E

PROBLEM VI.

Through any three points out of a right Line, to

describe the circumference of a Circle. From the middle point as a centre, with any convenient distance, describe the circle or arcs of a circle, as A and B; and from the other points, with the same distance, describe arcs cutting the circle in C D and EF; draw lines through C D and EF, and where they intersect each other at o is the centre of the circle required.

PROBLEM VII.

B

Two right Lines being given, to find a Mean Pro

portional. Upon a right line as a diameter equal to both lines given, describe the semi-circle ABC, where the two lines meet, or, between their respective lengths, erect a perpendicular to A the semi-circle at B, and the perpedicular will be the mean proportional required.

PROBLEM VIII,

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To describe a Square upon any right Line. Let A B equal the side of the square; then from A and B as cen- 'D tres, with the distance A B, describe the arcs A C and B D, cutting each other in m; bisect A m, or B m, in n; and with the radius n m, A in m, as a centre describe arcs cut. ting A C and B D; then lines drawn from A to D, from D to C, and from C to B will be the

square required.

B

PROBLEM IX.

6P

D

C

To Circumscribe a Square about a given Circle. Draw two diameters at right angles as m n, and O P, from m n, O P as centres, with the radius of the circle, describe arcs cutting each other in A B C and D; join A B, BC, CD, D A, and ABCD will be the square required.

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А

B

And from A as a centre, with the distance A o cut the lines A B, A D, in 2 and 7; from B as a centre cut the lines B A,BC, in 1 and 4; from C as a centre cut the lines C B, C D, in 3 and 6; and from D as a centre cut the lines D C, D A, in 5 and 8; join 1, 8; 2,3; 4,5; and 6, 7; and 1, 2, 3, 4, 5, 6, 7, 8 will be a regular octagon.

PROBLEM X.

Upon a right Line, to describe an Octagon.

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On the extremities of one side A B, erect the perpendiculars A F and B E; continue the line A B, to A m, and B, n, forming the angles A, m, r, and B n, s; bisect the angles with the lines A Hand B C; make each of those lines equal to A B; make H G and C D the same length, and parallel to A F and B E; from G and D as centres with the radius A B, describe arcs cutting A F and B E; join G F, F E, and ED, then A B C D E F G H will be the octagon required.

m

A

B

PROBLEM XI.

In a given Circle, to inscribe any regular Polygon.

Divide the diameter A B into as many equal parts as the polygon is required to have sides ; from X and B as centres, with the distance AB, describe arcs cutting each other in C; draw a line through the second A division, meeting the circumference at D; join A D, and it will be the side of the polygon required.

B

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PROBLEM XII.

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To find the Side of a Square that shall be any

number of times the Area of a given Square. Let A B C D be the given square, then will the diagonal B D be the side of a square A E F G, double in G area to the given square A B C D; D and if the diagonal be drawn from B to G, it will be the side of a square A H K L, three times the area of the square A B C D, or the diagonal A BL will equal the side of a square four times the area of the square A B C D, &c.

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PROBLEM XIII.

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To find the Diameter of a Circle that shall be any num

ber of times the Area of a given Circle. Let ABCD be the given circle, draw the two diameters A B and CD at right angles to each other, and the chord A D will be the radius of the circle o P, twice the area of the given circle nearly; and half the chord will be the radius of a circle that will contain half the area, &c.

B

PROBLEM XIV.

To divide a given Circle into any Number of Co-centric

Parts equal to each other. Upon the radius A B describe the semi-circle A e d B; divide A B into the proposed number of equal parts, as 1, 2, &c.; erect the perpendiculars 1 e2, d, &c., meeting the semi-circlein e and d; then from the centre A, and radii A e A d, &c. describe circles; so shall the circle be divided into the proposed number of equal parts, as required.

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