Take one of its sides equal 122.) Make a right angle. to M, the other equal to N. extremity, erect a perpendicular, and take in it a part equal to O. Draw Y. Upon Y, at its extremity, erect a perpendicular equal to P. Draw Z. Erect a perpendicular at the extremity of Z. Draw V. Now v2=Q2+Z3; Z2=P2+Y2; Y2=02+X2; X2= 2 2 2 2 2 2 2 2 2 2 2 M2+N2; consequently V2=M2+N2+02+P2+Q?. 224. To construct a square which shall be 4, 9, 16, 90, 100, 105, &c., times as great as a given square. Sol. Produce the sides of the given square, until the produced parts shall be 2, 3, 4, 10, &c., times as great as the sides themselves; upon these parts construct squares, and such squares will be quadruple, nine times &c., the given square. If it is required to construct a square which shall be 90 times as great as a given square, the operation will be somewhat more intricate. The shortest method is to take the next smallest square of a whole number, which is 81. Now 90 = 819. Construct squares 81 times and 9 times as great as the given square, and place the two squares together at right angles with each other. Draw a line connecting the sides of the two squares which include the right angle, and such line will be a side of the required square. For example, (fig. 123,) ACAB2+ BC-9 × 1+81 X 190 X 1; the sides of the given square being equal to 1 part, a square constructed upon AC will be the square required. Suppose it is required to construct a square 105 times as great as a given square. 105100+4+1. Make a right angle; take one of the sides 10 parts long; and the other side 2 parts long; draw the hypothenuse. At the 2 2 extremity of this erect a perpendicular 1 part long. Draw again an hypothenuse, and it will be a side of the required square. 225. To construct a square which shall be equal to the difference of two given squares. Sol. (Fig. 122.) Let N and X be sides of the given squares. Make a right angle. Take one side of it equal to N, a side of the lesser square. tremity of this side of the angle as a centre, with a radius equal to X, a side of the greater square, describe an arc, which will intersect the other side of the angle. Thus we get M, which is a side of the required square. For M2+N2X2; therefore M-X-N. 2 = 2 = 226. The angle formed by the intersection of two chords (produced) without a circle is equal to the difference between the two angles in the circumference standing upon the arcs intercepted by the two chords. Dem. Draw BC. (Fig. 124.) Angle ABC=BEC +BCE, (124,) therefore angle BEC― ABC — BCE. 227. An angle formed by the intersection of two chords, (AB and CD,) within a circle is equal to the sum of the angles in the circumference standing upon the arcs intercepted by the chords. Dem. (Fig. 125.) Draw the chords AC and BD. Angle AED = ACD + CAB, (124.) In like manner angle CEBABD+CDB. 228. To draw a tangent to a circle, whose centre is O, from a point D without the circle, (fig. 126.) Sol. Draw DO. Upon DO as a diameter describe a circumference, which will intersect the circumference of the given circle at two points, A and B. Draw DA and DB, both of which lines are tangents to the given circle. To demonstrate this let us draw the radii OA and OB. Now each of the angles OAD and OBD is a right angle, (174.) OA is perpendicular to AD and OB to BD; and consequently AD and BD are tangents, (170.) 229. Again, OD=OD; OA=OB; angles OAD and OBD are right angles; therefore triangle OAD= OBD, (135;) consequently AD=BD; that is, two tangents to a circle drawn from the same point, and not produced beyond the point of contact, are equal. 230. Since the triangles OAD and OBD are equal, angle ODA = ODB; that is, the angle formed by the intersection of two tangents to the same circle is bisected by a straight line drawn from the centre of that circle to the vertex of the angle. 231. To transform a rectangle into a square. Sol. Upon the greatest side AB of the rectangle ABCD, (fig. 127,) construct a square ABFG. Upon AF as a diameter describe a semi-circle. Produce CD to the circumference at E. Draw AE. A square constructed upon AE will be equivalent to the rectangle ABCD. 2 2 Dem. Draw EF. Angle AEF is a right angle, (174,) therefore AE2+ EF2= AF2= ABGF. ED is perpendicular to AF, therefore EA -2 = ABCD, (220.) 232. To draw a meridian line; that is, to draw a horizontal straight line in the direction North and South. At noon the shadows of objects fall in the direction of their meridian line. Such a line is used in the con struction of sun dials. Sol. Erect upon an exactly horizontal plane a perpendicular rod. From the foot of the rod, as a centre, with radii of any length, describe two or more concentric circles. The length of the shadow of the rod will become less as the sun becomes higher; therefore there will be a moment when the end point of the shadow will lie in the circumference of the greater circle. Mark this point in the circumference. In the afternoon the length of the shadow will again increase, and there will be a moment when the end of the shadow will again lie in the circumference of the greater circle. Mark this point. Connect by a straight line the two points thus found. Bisect this straight line, and at the middle point erect a perpendicular. Such perpendicular will mark the direction in which the shadows fall at noon; it is therefore the meridian line. By proceeding in the same manner with the second circle the accuracy of the first observation will be tested. 233. Upon a given straight line to describe a segment which shall contain a given angle; that is, a segment such that the angle having its vertex in the arc, and its sides standing upon the extremities of the segment, shall be equal to the given angle. Let AB be the given straight line and O the given angle, (fig. 129.) Sol. At the point A in line AB make an angle BAC equal to the given angle O. Upon AC at A erect a perpendicular. Bisect AB and at the middle point G erect a perpendicular. The two perpendiculars will intersect each other at D. From D as a centre with radius DA describe a circle: AEB is the required segment. Make any angle in the circumference of this segment, for instance, AEB; then is angle AEB = angle BAC= angle O. Dem. Angle DAC=1 R. A.; therefore angle DAG is <1 R. A. Angle DGA=1 R. A.; therefore angle DGA+ angle DAG <2 R. A. Consequently the perpendiculars intersect each other, (112.) GA=GB, GD=GD, angle DGA= angle DGB; therefore triangle DGA= triangle DGB, therefore DA = DB. Thus the circle which is described from D, with radius DA, passes also through B. Now AC is a tangent to the circle, (170;) therefore angle AEB= angle CAB, (178,) which was to be proved. 234. To construct a triangle of which the base, the altitude, and the angle opposite to the base are given. (Fig. 128.) Let M be the base, N the altitude, and O the angle opposite to the base. Sol. Draw a straight line; take ABM; describe upon AB a segment capable of containing the angle O. Upon AB at any point G erect a perpendicular, and in it take GCN. Through C draw a line parallel to AB. From the points D and E where this parallel cuts the arc of the segment draw the straight lines DA, DB, EA, EB; either of the triangles DAB and EAB is the triangle required. 235. In every circle equal chords are at an equal distance from the centre; and if two chords are at an equal distance from the centre they are equal. Dem. (Fig. 130.) 1. Let AB=CD. Draw the |