Problem 4. On a given line-segment as a side corresponding to a given side of a given polygon, to construct a polygon similar to that given polygon. Given the polygon ABCD and the line-segment A'B'. to construct on A'B' as a side corresponding to AB, a polygon A'B'C'D' ABCD. S Construction. 1. In Fig. 1, place A'B' II AB. I, pr. 6 2. Draw AA', BB', meeting at O; draw OC, OD. Post. 2 Proof. 1. OA: OA' OB: OB'- OC: OC' OD: OD'. Th. 1 2. ... D'A' II DA. 3. A'B': ABOB': OB = B'C': BC= Th. 3 Th. 1, cor. 2 4. And C'B'A' CBA, and so for the other. 5. = ... A'B'C'D' I, th. 17, cor. 2; ax. 3 ABCD. Th. 12 6. If A'B' AB, as in Fig. 2, draw from C, D IIs to AA'; otherwise. the construction is as above. It is left to the student to prove D'A' II DA, and A'B'C'D' ABCD. .. A'B'C'D' ABCD by the corollary 1 under the definition of similar figures. NOTE. It should be noticed that as the point O recedes, AA' and BB' approach the condition of being parallel, Fig. 1 approaching Fig. 2 in form. BOOK V. MENSURATION OF PLANE FIGURES. REGULAR POLYGONS AND THE CIRCLE. Section 1. Mensuration of Plane Figures. Theorem 1. Two rectangles having equal altitudes are proportional to their bases. Given two rectangles R and R', with altitude a, and with Proof. 1. Suppose b and b' divided into equal segments, I, and suppose b = nl, and b'n'l. (In the figures, n = 6, n' = 4.) 2. Then if I are erected from the points of division, Rn congruent rectangles al, and R' = n' 66 66 NOTE. The above proof assumes that b and b' are commensurable, and hence that they can be divided into equal segments I. The proposition is, however, entirely general. The proof on p. 171 is valid if b and b' are incommensurable. 2. Then if I are erected from the points of division, R = n congruent rectangles al, and R'n' 66 ax, such that ax < al. 66 al + a remainder 3. Then b' lies between n'l and (n'+1) I, Why? and R' lies between n' al and (n'+1) · al. Why? 1 6. And ... can be made smaller than any assumed 7. n difference, by increasing n, .. to assume any difference leads to an absurdity. COROLLARIES. 1. Rectangles having equal bases are proportional to their altitudes. For they can be turned through 90° so as to interchange base and altitude. 2. Triangles having equal altitudes are proportional to their bases; having equal bases, to their altitudes. (Why?) EXERCISE. 439. Prove that any quadrilateral is divided by its interior diagonals into four triangles which form a proportion. Theorem 2. Two rectangles have the same ratio as the products of (the numerical measures of) their bases and altitudes. Given two rectangles R, R', with bases b, b', and altitudes Proof. 1. Let X be a rectangle of altitude a and base b'. NOTE. Thus again appears the Law of Homology (p. 86), that to the product of two numbers corresponds the rectangle of two lines. DEFINITION. As already mentioned (pp. 100, 143), to measure a surface is to find its ratio to some unit. The unit of measure, multiplied by this ratio, is called the area. Thus in a surface 4 ft. long by 2 ft. broad, the ratio of the surface to 1 sq. ft. is 8, and 8 sq. ft. is the area. COROLLARIES. 1. Parallelograms (or triangles) having equal bases have the same ratio as their altitudes; having equal altitudes, as their bases. For by II, th. 1, cor. 1, and II, th. 2, cor. 1, they have the same ratio as rectangles of those bases and altitudes. 2. Parallelograms (or triangles) have the same ratio as the products of their bases and altitudes. (Why?) 3. The area of a rectangle equals the product of its base and altitude. That is, the number which represents its square units of area is the product of the two numbers which represent its base and altitude. = For in th. 2, if R' 1, the square unit of area, then a' and b' must each equal 1, the unit of length. Hence, R/1 ab/1, or R = ab. = 4. The area of a parallelogram equals the product of its base and altitude; of a triangle, half that product. 5. The area of a square equals the second power of its side. This is the reason that the second power of a number is called its square. 6. The area of a trapezoid equals the product of its altitude and half the sum of its bases. (Why?) Theorem 3. Triangles, or parallelograms, which have an angle in one equal to an angle in the other, have the same ratio as the products of the including sides. Given two triangles ABC, AB'C', having an angle, A, of one equal to an angle, A, of the other. 4. their bases being AC, AC'. . B ..A ABC: A AB'C' AB AC: AB'· AC'. 5. And the Why? Why? Ax. 6 in the figure are double the ▲, the theorem is true for parallelograms. COROLLARY. Similar triangles have the same ratio as the squares of their corresponding sides. For if the are similar, BC | B'C', and the ratio AB AB' equals, and may be substituted for, the ratio AC AC', thus making the second member of step 4, AB2: AB'2. |