A draw CD perpendicular to OA, which will be the tangent required.

Problem 46.

When the point A is without the circle, Fig. 36.

From the centre O draw O A, and bisect it in n; from the point n, with the radius n A or n O, describe the semicircle AD O, cutting the given circle in D; through the points AD draw A B, the tangent required.

Problem 47.

To cut off from a circle a segment containing any proposed angle, as 60°, Fig. 37.

Let F be the point from whence it is required to draw a chord which shall contain 60 degrees; through F draw FR, a tangent to the circle; from F draw F A, making an angle of 60 degrees with the tangent FR, and FCA is the segment required.

Problem 48.

To find the diameter of a circle that shall be any number of times the area of a given circle, Fig. 38.

Let ABCD be the given circle, draw the two diameters A B and CD at right angles to each other; the chord AD 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.

Problem 49.

To divide a given circle into any number of concentric parts, equal in proportion to each other, Fig. 39, Plate 3.

Upon the radius A B describe the semicircle Aed B; divide A B into the proposed number of equal parts, as 1, 2, &c. ; erect perpendiculars from those points to the semicircle; then from the centre A and radius A e A d, &c., describe circles, which will be the proposed number of parts.

Problem 50.

Required the diameter of a pipe that shall be equal to three pipes of three inches diameter each, Fig. 40, Plate 3.

Rule. Multiply .7854 by the square of the given diameter; multiply that product by the number of times required to be enlarged; the square root of that product multiplied by 1.12837 will give the diameter of the pipe required.

Note. This problem shows the error committed by an engineer on the application of a water company for the above enlargement, on which a pipe was made three times the diameter, as shown by the diagram.

Ex. .7854 x 32 = 7.0686 x 3 =

And

21.2058, the area required 21.2058 4.60 × 1.12837 = 5.1905, the dia. CD

=

The area of the 3 inch pipe = 7.0686
The area of the 9 inch pipe = 63.617

=

Therefore the pipe 9 inches in diameter 63.617, was in excess by 58.426.

TO CONSTRUCT A PROTRACTOR, OR DIVIDE A CIRCLE INTO
360 DEGREES, MINUTES, ETC.

With a radius of 5 inches describe a circle; with the same radius mark off each length very minutely round the circle, which should be exactly 6 times, or 60° each.

Then set the compass to the natural sine of 15°, which to radius 5 will be equal to half the chord of 30°; this distance will bisect each 60°, and divide the circle into arcs of 30° each. This may be proved by setting the succeeding chords off each way from those points which they are intended to bisect; if any inaccuracy exists the bisector will not be perfect, and if the error is inconsiderable the middle point will be assumed as

correct.

Each 60° may next be trisected by setting off the natural sine 10°, equal the chord of 20° to our radius, which will divide the circle to every 10 degrees.

The natural sine of 7° 30', equal the chord of 15°, marked from the points already determined, will divide the circle to every 5th degree.

The natural sine of 3°, equal the chord of 6°, divided as before, divides 30° into 5 parts, and set off from the other divisions, divides the circle to single degrees.

Fifteen degrees bisected on the natural sine of 3° 45', equal the chord 7° 30', set off from the other divisions, divides the circle into half degrees.

The natural sine of 3° 20', equal the chord of 6° 40', divides 20° into three parts, and set off from the rest of the division, divides the whole circle to every 10 minutes, which is as minute as can possibly be obtained; smaller quantities must be subdivided by the eye.

The division should be numbered in like manner to the theodolite, that is, from 0° to 360°.

INSCRIBING AND CIRCUMSCRIBING FIGURES.

Problem 51.

To inscribe a circle in a given triangle A B C, Fig. 41, Plate 3. Bisect the angles A and B; from the point of intersection at O, let fall a perpendicular O N, it will be the radius of the required circle.

Problem 52.

To inscribe a pentagon, or hexagon, or a decagon in a given circle, Fig. 42.

Draw the diameters A B and CE at right angles to each other; bisect D B at G, on G with the radius G C describe the arc C F; join C and F, and the line C F will be one side of the pentagon.

The two sides DC, FD, of the triangle FD C, inscribe an hexagon or a decagon in the same circle; D C is the side of the hexagon, D F that of the decagon.

Problem 53.

To inscribe a square or an octagon in a given circle, Fig. 43. Draw the diameters A C, B D, at right angles to each other; draw the lines AD, BA, BC, C D, the square required.

For the octagon: Bisect the arc A B and A D of the square in the point F and H; draw lines from F and H through the centre, intersecting the circumference at E and G; draw lines from each point of intersection, which will be the required octagon.

Problem 54.

In a given circle to inscribe an equilateral triangle, an hexagon, or a dodecagon, Fig. 44.

For the equilateral triangle: From any point A as a centre, with a distance equal to the radius A O, describe the arc F O B; draw the line B F, and make B D equal to B F; join D F, and DBF will be the equilateral triangle required.

For the hexagon: Carry the radius A O six times round the circumference it will give the hexagon required.

For the dodecagon: Bisect the arc A B of the hexagon in the point n; and the line A n being carried twelve times round the circumference will form the dodecagon required.

Problem 55.

To inscribe a dodecagon in a circle, or to divide the circumference of a given circle into twelve parts, each being equal to 30 degrees, Fig. 45.

Draw the two diameters 12, 6, 3, 9, perpendicular to each other, with the radius of the circle; on the points 12, 6, 3, 9 as centres describe arcs through the centre, intersecting the circumferences in the points required, dividing it into twelve equal parts.

Problem 56.

In a given circle to inscribe any regular polygon, Fig. 46. Draw the diameter on A B; upon A B construct an equilateral triangle A B C; divide the diameter into as many equal parts as the polygon has sides; from C draw a line through the second division in the diameter intersecting the circumference at D; join the points A D, it will be one side the octagon.

Problem 57.

About a given triangle to circumscribe a circle, Fig. 47. Bisect any two sides A C, B C; draw the lines D O and F 0; the point of intersection at O is the centre; with the radius A O describe the circle required.

Problem 58.

About a given circle to circumscribe a pentagon, Fig. 48. Inscribe a pentagon within a circle; through the middle of each side draw the lines O A, O C, O D, O E, and OB; through the point n draw the tangent A B, meeting OA and O B in A and B; through the points A and m, draw the line Am C, meeting O C in C; in like manner draw the lines CD, DE, EB; and A B C D E will be the pentagon required.

Problem 59.

On a given line A B to form a regular octagon, Fig. 49. On the extremities of a given line A B erect perpendiculars AF, BE; produce A B both ways to s and w; bisect the angles n A s and o Bw by the lines A H, BC; make A H and B C each equal to A B, and draw the line H C; make o v equal to on, and through v draw G D parallel to H C; draw H G and CD parallel to AF and BE, and make v E equal to v D; through E, draw E F parallel to A B, and join the points G F and D E, and ABCDEFGH will be the octagon required.

Problem 60.

To circumscribe a circle about a given square A B C D, and construct an octagon, Fig. 50.

For the circle: Describe the square A B C D; draw the two diagonals A C and D B, intersecting each other at O; with the distance O A or O B, describe the circle required.

For the octagon: Bisect the sides A B C D; draw lines from a to c and b to d; join the lines a d, d c, c b, and b a, forming an