EQUIVALENT SQUARES. Problem 34. 45 157. To find the side of a square equal to half of a given square. Draw the diagonals of the given square: they bisect each other at right angles. The square of the half diagonal will be equal to half of the given square. Or, bisect the side of the given square, and at the point of bisection raise a perpendicular equal to the half side. The line joining the end of the perpendicular to either end of the side, will be the side of a square equal to half of the given square. Problem 35. 158. To make a square equal to a rectangle of which the sides are given. FIG. 40 E Let AB, BC, the sides of the given rectangle, be placed in one straight line AC. Bisect AC in the point D, and from the centre D, with the radius AD, on AC describe a semicircle. At B draw BE at right angles to AC, meeting the arc in E. The square of BE is equal to 1 D B the rectangle contained by the lines AB, BC. Note. If the lengths of the sides are given in numbers, multiply the sides, and extract the square root of the product. This will give the length of the side of a square equal to a rectangle under the given sides. Thus, let the sides of a rectangle be respectively 4 and 25. The product of 4 and 25 is 100; the square root of 100 is 10. Then a square of which the side is 10 is exactly equal to a rectangle, of which the sides are 4 and 25; or equal to any rectangle, the product of whose sides is 100; as, sides 8 and 121; 5 and 20 ; 16 and 61; and so on. 4. PROBLEMS RELATING TO THE CIRCLE.. Problem 36. 159. To find the centre of a given circle. Draw any chord; bisect it. Through the point of bisection, draw a chord at right angles to it. The second chord will be a diameter, and its middle point will be the centre of the circle. 160. Note.-A straight line drawn from the centre of a circle to the middle point of any chord, is at right angles to it. If a straight line be drawn from the centre at right angles to a chord, it will also bisect the chord. Problem 37. 161. To describe a circle through any three given points, not in the same straight line. FIG. 41 B Let A, B, D be the three given points. Join any one to the other two, as B to A and D, forming the straight lines BA, BD. Bisect these lines in E and F. Draw perpendiculars to them at the points of bisection, E and F, meeting in the point C. C will be equidistant from the three, A, B, and D; and a circle drawn from C as centre, with CA, CB, or CD as radius, will pass through all three points. 162. Note 1.-This problem shows how to find a point equidistant from three given points not in the same straight line. 163. Note 2.-It shows also how to describe a circle about a given triangle; that is, passing through its angular points. Bisect any two sides, and so on as above. Problem 38. 164. To find the centre of a circle, only an arc of it being given. Take any three points in the arc, and proceed with these as in Problem 37. The point in which the perpendiculars meet will be the centre of the circle. Problem 39. 165. To draw a tangent to a circle at a given point in the circumference. Let B be the given point. Find the centre of the circle, C. Join CB. A perpendicular to CB at B will be a tangent to the circle at B; that is, will touch the circle at the point B, every other point in the line being without the circle. 166. Note.—Any straight line from the centre of a circle to the point of contact of a tangent is at right angles to the tangent; and any straight line at right angles to a tangent at its point of contact passes through the centre. Problem 40. 167. To draw a tangent to a circle from a given point without it. Let B be the point without the circle (fig. 42). Join BC, and bisect that line The straight line BE is a tangent to the circle, and it is drawn from the point B. Note 1.-From the same point, another tangent may evidently be drawn, to the point where a semicircle on the other side of BC would cut the circle. tangents would be equal. These two 168. Note 2.-If EC were drawn, the angle BEC would be a right angle, being the angle in the semicircle BEC (72);-a line from the centre of a circle is always perpendicular to a tangent at the point of contact. Problem 41. 169. On a given straight line to describe a segment of a circle containing an angle equal to a given angle. Let AB be the given straight line, and H the given angle. At B make the angle ABD equal to H. Bisect AB in E. At E and B draw perpendiculars to AB and BD meeting in C. From C, with the radius CB, describe the circle BFAG. The segment AFB will contain an angle equal to H; that is, lines from any point in the arc BFA to A and B, as FA, FB, will contain an angle equal to H. SIMILAR FIGURES. 49 170. Note.-This illustrates two remarkable properties of the circle. FIG. 43 H B 1. The angle (ABD) made by a chord (AB) and a tangent (BD) is equal to any angle (F) in the alternate segment of the circle (BFA). 2. Angles in the same segment of a circle are equal to another; that is, the angles formed by lines to A and B, from any points in the arc of the segment BFA, are equal. 5. PROBLEMS RELATING TO THE PROPORTIONS OF LINES AND FIGURES. DEFINITIONS-THIRD SERIES. (First, read Article 274, &c., on Proportion, in the Algebra.) 171. SIMILAR FIGURES are those which have the same shape or form, though they may differ in magnitude. D |