153.9379 square inches, the square root of which is 12.4 inches. This applied as a chord to the circle, will subtend an angle at the centre of nearly 125.°

ANOTHER METHOD.

Having found the side of the square, (12.4 inches,) say as 12.4 is to 14, or as 6.2 is to 7, so is the side of the square to the radius of the circle, nearly.

PROBLEM LXXIII.

To make a rectangle (one of its longest sides being given) equal to a square.

1. Construct a square A B C D, say each side one inch. 2. Produce the base A B to E the length required, say A E 2 inches.

3. Produce DC, and draw E F parallel to BC cutting DC produced in F.

4. Draw the diagonal AF cutting BC in G.

5. Through G draw HI parallel to A E, cutting AD in H, and E F in I; when A HIE shall be the rectangle required.

Reason because the complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another. Vide Prop. XLIII, Bk. I. Euclid; (i. e. A F being the diameter,) the complement BEIG is equal to the complement HGCD; therefore the sum of BE IG and ABGH, is equal to the sum of ABG H and HGCD.

If this problem be understood, all others arising out of it will be equally clear.

EXAMPLE.

Make a rectangle whose length is to be 3 inches, equal to a square of 2 inches each side.

PROBLEM LXXIV.

To make a triangle equal to any sector of a circle.

1. Let AFBD A be the given sector.

2. Complete the circle AFB A, and make a right-angled triangle equal to it by Prob. LXXÍ.

3. Measure the angle A D B. The measure of this angle will also be the measure of the arc AB, (suppose it to be 90 degrees,) then say as the measure of the arc (90°) is to the circumference, (360°,) so is the sector ADBA to the circle AF BA. Therefore, (in this case,) dividing 360 by 90 gives one-fourth, as the ratio of the sector ADBA to the circle.

4. By Prob. XVI. cut off from either end of the base of the triangle the above ratio, (in this case one-fourth.) Draw a line to the vertex including this fourth, and the remaining three-fourths will be the triangle required.

PROBLEM LXXV.

To make a square equal to the sector of any circle.

1. By Prob. LXXIV. make a triangle equal to the sector. 2. By Prob. LXIX. make a square equal to the triangle, and the work is done.

Excellent exercises in transformations may be obtained, by drawing figures similar to the outlines of Egyptian letters, as LK A (at least 3 inches high,) and making triangles equal in area to them.

ADDITION.

Addition and Multiplication are in principle nearly identical both in arithmetic and geometry; the former being employed to combine quantities of various values; and the latter providing an expeditious mode of combining quantities of equal value. For example, if in arithmetic the sum of 3, 4, and 5 were required, the numbers would be added, making together 12; but if the sum of 3, 3, 3, 3, and 3 were required, the same effect as addition would be obtained by multiplying 3 by 5, producing 15. So in geometry unequal planes can be added together, but only equal planes can be multiplied.

PROBLEM LXXVI.

To make a triangle that shall be equal to any number of triangles.

CASE I.

When the triangles have equal altitudes.

1. Draw a line as AB in Prob. LXIV., equal to the sum of the bases of the given triangles. Make BC perpendicular to A B, and equal to the altitude of one of the triangles. 2. Draw CA, when A CB will be the triangle required.

CASE II.

When the triangles have unequal altitudes.

Make the altitudes of all the triangles equal by Prob. LXIII. after which proceed as in Case I.

Reason: "Triangles (or parallelograms) of the same altitude are to one another as their bases."

PROBLEM LXXVII.

To make a square equal to the sum of two or more

squares.

1. Let AB, E O, Prob. LXII., be the sides of the squares given.

2. From B make BC equal to EO, and perpendicular to AB. Draw a line from A to C, when A C shall be the side of a square equal to the two given squares.

Reason: "In any right-angled triangle, the square which is described upon the side subtending the right-angle, is equal to the squares described upon the sides which contain the right angle."

If a third square had been given, a side of it would have been applied from B on BA (produced if necessary,) and AC applied from B on B C produced; and a hypothenuse being drawn to meet the extremities of these measures, would have been the side of the square, equal to the sum of the squares of A C, and the third line; and therefore equal to the three given squares.

Precisely the same process must be employed (as in this problem) to make a circle equal to the sum of two or more circles, considering AB and BC, as the diameters of two circles given, and AC as the diameter of a circle equal to their sum. For circles being to each other as the squares of their diameters, (vide Prop. II. Bk. XII. Euclid,) any circle

must be equal to two other circles, the sum of the squares of whose diameters is equal to the square of the diameter of the first circle.

Again, to make an equilateral triangle equal to two or more equilateral triangles, or to make a regular polygon equal in area to two or more similar regular polygons, or to make an irregular polygon equal to two or more similar irregular polygons-in all these cases, taking AB as above, equal to the side of one polygon, and B C equal to the homologous side of another, A C will be the homologous side of a similar polygon, equal to the sum of the given polygons. For," universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides."

Duplicate ratio is the ratio of the square of one quantity to the square of another.

EXAMPLES.

1. Make a square equal to the sum of three squares, whose sides are 1, 2, and 3 inches.

2. A circle is required, equal in area to four circles, whose diameters are 1, 2, 3, and 4 inches.

3. Describe an equilateral triangle equal to any three equilateral triangles.

4. Make a pentagon equal to the sum of four unequal pentagons.

5. The two similar figures in Prob. LVI. being given, another similar figure is demanded, equal to the sum of their

areas.

Other exercises in Addition may be selected from the Problems in Transformation.

SUBTRACTION

Teaches the art of taking one quantity from another quantity of greater value, so as to ascertain the remainder, or difference of the quantity subtracted and the quantity it is subtracted from. In principle it is nearly the same as division, since all the operations of division could be effected by subtraction; but division is employed as a much more expeditious method of gaining the result desired, when any quantity is to be taken as many times as possible from another quantity.

PROBLEM LXXVIII.

To subtract one or more triangles from another

triangle.

CASE I.

When the given triangles are of the same altitude.

1. Let three triangles be given, whose altitudes are 2 inches each, and their bases, 2, and 1 inches; it is required to make a triangle equal to the difference of their sum and another triangle ACB, whose altitude is 2 inches, and its base 2 inches.

2. On the base A B of the greater triangle, set off from A to D the sum of the bases of the other triangles. Draw a line from D to C; when the triangle DCB will be the difference required.

For reason, vide Prob. LXXVI.

CASE II.

When the given triangles are of unequal altitudes.

Reduce all the triangles to the same altitude, and proceed as in Case I.

PROBLEM LXXIX.

To make a square equal to the difference of two

squares.

1. Let OG, be the side of the greater A

of the given squares.

2. From O as a centre, with OG as a radius describe a semicircle. Take OF equal to the side of the second square. From F draw FH perpendicular to O G, when FH will be the side of the square required.

For reason, vide Prob. LXXVII.

PROBLEM LXXX.

B

D

H

G

Two triangles being given, to make a square equal to their difference.

Transform the given triangles to squares; and make a square equal to their difference, by Prob. LXXIX.

F