§ 163. The arcs AO, OB, etc., and the circle with center O', are constructed as in Ex. 28. The small circle with center at Q is constructed as in the preceding example. 32. ABC is an equilateral arch and AEB is a semicircle. (a) If AB (the span of the arch) is 10 feet, find the radius of the small circle tangent to the semicircle and the arcs of the arch. (b) Find the radius OE of the circle if AB (c) If OE = 2, find AB. = S. A C E 33. The accompanying design consists of three semicircles and three circles related as shown in the figure. (a) If AB = 12, find OQ by using the triangle OSD. method is similar to that used in Ex. 30. (b) Using the right triangle DRO', find O'x. Notice that in order to have the circles with the centers O and O' tangent to each other the sum of their radii must be equal to RD. (c) If ABs show that OQ The E F -- R D B = and 6 34. Upon a given segment AB construct the design shown in the figure. Notice that it consists of the two preceding figures put together. Compare the radii of the circles in this design. Ospedale Maggiore, Milan. 35. Between two parallel lines construct semicircles and circles as shown in the figure. SOLUTION. Suppose the construction made. OQr. Using the triangle EHO, show that struct the figure. 36. Between two parallel lines construct circles and equilateral arches as shown in the figure. Now divide AC into two segments proportional to 1 and √5, and so construct the figure. 37. ABC is an equilateral arch. AD = DB, AF = EB = } AB. Semicircles are constructed with E and F as centers, and radius FB. (a) If AB =s, find OH and construct the figure. SOLUTION. Let OH = r. Then BOS A r, FB B=3, DF = Ꭶ = 2 (2) 2 From ▲ ODF, OD2 = OF2 – DF2 = ( §3 + r)2 – (¿) 2. 2 2 From (1) and (2) (8 − r)2 − ( :-)2 = (¦ +r)2 - (6) 2. (b) If in the preceding OH = 4 feet, find AB and AF. 38. Between two parallel lines construct circles and semicircles having equal radii as shown in the figure. U PH SL Prove that the ratio = RLK being HQ RS' A WQ T an equilateral triangle and KS = SL. Divide CA in the ratio of SL : RS. K SAL 39. Through a fixed point on a circle chords are drawn and each extended to twice its length. Find the locus of the end-points of these segments. Compare Ex. 6, § 214. 40. If a quadrilateral is circumscribed about a circle, show that the sums of its pairs of opposite sides are equal. 41. On a diameter produced of a given circle, find a point from which the tangents to the circle are of a given length. Solve this problem by construction, and also algebraically. 42. Compare the perimeters of equilateral triangles circumscribed about and inscribed in the same circle. 43. In a given square construct semicircles each tangent to two sides of the square and terminating on the diameters of the square. CONSTRUCTION. Connect E and F two ex- D tremities of diameters and on EF as a diame- E B H PROOF. Draw O'L || FE. We need to prove that O"K = 0"L. CHAPTER IV. AREAS OF POLYGONS. AREAS OF RECTANGLES. 300. Heretofore certain properties of plane figures have been studied, such as congruence and similarity, but no attempt has been made to measure the extent of surface inclosed by such figures. For this purpose we first consider the rectangle. 301. The surface inclosed by a rectangle is said to be exactly measured when we find how many times some unit square is contained in it. E.g. if the base of a rectangle is five units long and its altitude three units, its surface con. tains a square one unit on a side fifteen times. 5 units 302. The number of times which a unit square is contained in the surface of a rectangle is called the numerical measure of the surface, or its area. We distinguish three cases. 303. Case 1. If the sides of the given rectangle are integral multiples of the sides of the unit square, then the area of the rectangle is determined by finding into how many unit squares it can be divided. Thus if the sides of a rectangle are m and n units respectively, then it can be divided into m rows of unit squares, each row containing n squares. Hence the area of such a rectangle is m x n unit squares, that is, in this case, area = base x altitude. (1) 304. Case 2. If the sides of the rectangle are not inte gral multiples of the side of the chosen unit square, but if the side of this square can be divided into equal parts such that the sides of the rectangle are integral multiples of one of these parts, then the area of the rectangle may be expressed integrally in terms of this smaller unit square, and fractionally in terms of the original unit. For example, if the base is 3.4 decimeters and the altitude 2.6 decimeters, then the rectangle cannot be exactly divided into square decimeters, but it can be exactly divided into square centimeters. Each row contains 34 centimeters and there are 26 such rows. Hence, the area is 34 x 26 = 884 small squares or 8.84 square decimeters. But 3.4 x 2.6 8.84. Hence, in this case also, area base x altitude. 84 c.m. (2) 305. Case 3. If a rectangle is such that there exists no common measure whatever of its base and altitude, then there is no surface unit in terms of which its area can be exactly expressed. But by choosing a unit sufficiently small we may determine the area of a rectangle which differs as little as we please from the given rectangle. E.g. if the base is 5 inches and the altitude is √5 inches, then the rectangle cannot be exactly divided into equal squares, however small. But since √5 = 2.2361..., if we take as a unit of area a square whose side is one one-thousandth of an inch, then the rectangle whose base is 5 inches and whose altitude is 2.236 inches can be exactly measured as in cases 1 and 2, and its area is 5 x 2.236-11.18 square inches. The small strip by which this rectangle differs from the given rectangle is less than .0002 of an inch in width, and its area is less than 5 x .0002.001 of a square inch. By expressing √5 to further places of decimals and thus using smaller units of area, successive rectangles may be found which differ less and less from the given rectangle. |