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1. Divide 360° by the number of sides, and make an angle AOB, at the centre, whose measure shall be equal to the degrees in the quotient.
2. Join the points AB, and apply the chord AB to the circumference the given number of times, and it will form the polygon required.
PROBLEM XXXVII. On a given line AB to form a regular polygon of any pro
posed number of sides.
1. Divide 360° by the number of sides, and subtract the quotient from 180 degrees.
2. Make the angles ABO and BAO each equal to half the difference last found.
3. From the point of intersection 0, with the distance OA or OB, describe a circle.
4. Apply the chord AB to the circumference the proposed number of times, and it will form the polygon re. quired.*
* By this method the circumference of a circle may aiso be divi. ded into any number of equal parts; for if 360 be divided by the
Upon a right line AB to describe a regular pentagon.
1. Produce AB towards n, and at the point B make the perpendicular Bm equal to AB.
2. Bisect AB in r, and from r as a centre, with the radius rm, describe the arc mn, cutting AB in n.
3. From the points A and B, with the radius An, describe arcs cutting each other in D.
4. And from the points A, D, and B, D, with the radius AB, describe arcs cutting each other in C and E.
5. Join BC, DC, DE, and EA, and ABCDE will be the pentagon required.
This method differs but little from that of Problem XXVIII, and is equally easy and convenient in practice.
PROBLEM XXXIX. To raise a perpendicular from any point A in a given line
number of parts, and the angle AOB be made equal to the degrees in he quotient, the arc AB will be one of the equal parts required.
* This and the following problem were not given in the first edition of this work, but are now added on account of their elegance and utility The second is derived from the 47th Prop. B. I. Euclid's Elements and the first is proposed for a demonstration in the Ladies' Diary for the year 1786.
1. From any scale of equal parts take a distance equal to 3 divisions, and set it from B to m.
2. And from the points B and m, with the distances 4 and 5, taken from the same scale, describe arcs cutting each other in n.
3. Through the points n, B, draw the line BC, and it will be the perpendicular required.
Explanation of the characters made use of in the following
part of this work.
1. An angle in a semicircle is a right angle. 2. All angles in the same segment of a circle are equal.
3. Triangles that have all the three angles of the one respectively equal to the three angles of the other, are called equiangular triangles, or similar triangles.
4. In similar triangles the like sides, or sides opposite to the equal angles, are proportional.
5. T'he areas or spaces of similar triangles are to each other as the squares of their like sides.
6. The areas of circles are to each other as the squares of their diameters, radii, or circumferences.
7. Similar figures are such as have the same number of sides, and the angles contained by those sides respectively qual.
8. The areas of similar figures are to each other as the squares of their like sides.
MENSURATION OF SUPERFICIES
The area of any figure is the measure of its surface, or
contained within the bounds of that surface, with. out any regard to thickness.
A square whose side is one inch, one foot, or one yard, &c. is called the measuring unit, and the area or content of any figure is computed by the number of those squares contained in that figure.
N. B. In all questions involving decimals, they are carried out to the fourth place inclusive, and then taken to the nearest figure; that is, if it be found that, by extending the operation, the next figure would be 5, or upwards, the fourth decimal figure is increased by 1. The student by observ. ing this rule will generally find his results to agree with those given in the book.
PROBLEM I. To find the area of a parallelogram ; whether it be a square, a rectangle, a rhombus, or a rhomboides.
RULE.* Multiply the length by the perpendicular height, and the product will be the area.
D * Take any rectangle ABCD, and di. vide each of its sides respectively, into as many equal parts as are expressed by the number of times they contain the linear measuring unit, and let all the opposite mints of division be connected by right lines. Then it is evident, that these lines divide the rectangle into a number of A