Find the perimeter of a regular octagon if the length of

PERSPECTIVE ON HISTORY The Value of ␲ In geometry, any two figures that have the same shape are described as similar. Because all circles have the same shape, we say that all circles are similar to each other. Just as a proportionality exists among the corresponding sides of similar triangles, we can demonstrate a proportionality among the circumferences distances around and diameters distances across of circles. By representing the circumferences of the circles in Figure 7.33 by C 1 , C 2 , and C 3 and their corresponding lengths of diameters by d 1 , d 2 , and d 3 , we claim that for some constant of proportionality k. C 1 d 1 = C 2 d 2 = C 3 d 3 = k We denote the constant k described above by the Greek letter ␲. Thus, in any circle. It follows that or C = ␲d ␲ = C d C 1 d 1 C 2 d 2 C 3 d 3 Figure 7.33 because d 2r in any circle. In applying these formulas for the circumference of a circle, we often leave ␲ in the answer so that the result is exact. When an approximation for the circumference and later for the area of a circle is needed, several common substitutions are used for ␲. Among these are and . A calculator may display the value . Because ␲ is needed in many applications involving the circumference or area of a circle, its approximation is often necessary; but finding an accurate approximation of ␲ was not quickly or easily done. The formula for circumference can be expressed as , but the formula for the area of the circle is . This and other area formulas will be given more attention in Chapter 8. Several references to the value of ␲ are made in literature. One of the earliest comes from the Bible; the passage from I Kings, Chapter 7, verse 23, describes the distance around a vat as three times the distance across the vat which suggests that ␲ equals 3, a very rough approximation. Perhaps no greater accuracy was needed in some applications of that time. A = ␲r 2 C = 2␲r ␲ L 3.1415926535 ␲ L 3.14 ␲ L 22 7 C = 2␲r In the content of the Rhind papyrus a document over 3000 years old, the Egyptian scribe Ahmes gives the formula for the area of a circle as . To determine the Egyptian approximation of ␲, we need to expand this expression as follows: In the formula for the area of the circle, the value of ␲ is the multiplier coefficient of . Because this coefficient is which has the decimal equivalent of 3.1604, the Egyptians had a better approximation of ␲ than was given in the book of I Kings. Archimedes, the brilliant Greek geometer, knew that the formula for the area of a circle was with C the circumference and r the length of radius. His formula was equivalent to the one we use today and is developed as follows: The second proposition of Archimedes’ work Measure of the Circle develops a relationship between the area of a circle and the area of the square in which it is inscribed. See Figure 7.34. Specifically, Archimedes claimed that the ratio of the area of the circle to that of the square was 11:14. This leads to the following set of equations and to an approximation of the value of ␲. Archimedes later improved his approximation of ␲ by showing that Today’s calculators provide excellent approximations for the irrational number ␲. We should recall, however, that ␲ is an irrational number that can be expressed as an exact value only by the symbol ␲. 3 10 71 6 ␲ 6 3 1 7 ␲ L 4 11 14 L 22 7 ␲ 4 L 11 14 ␲ r 2 4r 2 L 11 14 ␲ r 2 2r 2 L 11 14 A = 1 2 Cr = 1 2 2␲rr = ␲r 2 A = 1 2 Cr 256 81 r 2 ad - 1 9 d b 2 = a 8 9 d b 2 = a 8 9 2r b 2 = a 16 9 r b 2 = 256 81 r 2 Ad - 1 9 d B 2 r Figure 7.34 Figure 7.37 PERSPECTIVE ON APPLICATION The Nine-Point Circle In the study of geometry, there is a curiosity known as the Nine-Point Circle—a curiosity because its practical value consists of the reasoning needed to verify its plausibility. In , in Figure 7.35 we locate these points: M, N, and P, the midpoints of the sides of , D, E, and F, points on determined by its altitudes, and X, Y, and Z, the midpoints of the line segments determined by orthocenter O and the vertices of . Through these nine points, it is possible to draw or construct the circle shown in Figure 7.35. 䉭ABC 䉭ABC 䉭ABC 䉭ABC Although we do not provide the details, it can also be shown that quadrilateral XZPN of Figure 7.36 is a rectangle as well. Further, is also a diagonal of rectangle XZPN. Then we can choose the radius of the circumscribed circle for rectangle XZPN to have the length Because it has the same center G and the same length of radius r as the circle that was circumscribed about rectangle NMZY, we see that the same circle must contain points N, X, M, Z, P, and Y Finally, we need to show that the circle in Figure 7.37 with center G and radius will contain the points D, E , and F. This can be done by an indirect argument. If we suppose that these points do not lie on the circle, then we contradict the fact that an angle inscribed in a semicircle must be a right angle. Of course, , and were altitudes of , so inscribed angles at D, E, and F must measure 90°; in turn, these angles must lie inside semicircles. In Figure 7.35, intercepts an arc a semicircle determined by diameter . So D, E, and F are on the same circle that has center G and radius r. Thus, the circle described in the preceding paragraphs is the anticipated nine- point circle NZ ∠NFZ 䉭ABC CE BF , AD r = 1 2 NZ r = 1 2 NZ = NG. NZ Then must be parallel to With , it follows that must be perpendicular to as well. In turn, , and NMZY is a rectangle in Figure 7.35. It is possible to circumscribe a circle about any rectangle; in fact, the length of the radius of the circumscribed circle is one-half the length of a diagonal of the rectangle, so we choose This circle certainly contains the points N , M, Z, and Y. r = 1 2 NZ = NG. MZ ⬜ NM AD NM CB ⬜ AD MZ . NY To understand why the nine-point circle can be drawn, we show that the quadrilateral NMZY is both a parallelogram and a rectangle. Because joins the midpoints of and , we know that and Likewise, Y and Z are midpoints of the sides of , so and By Theorem 4.2.1, NMZY is a parallelogram. YZ = 1 2 CB. YZ 7 CB 䉭OBC NM = 1 2 CB. NM 7 CB AB AC NM A E F N C B P D M Z Y G O X A E F N C B P D M Z Y G O X A E F N C B P D M Z Y G O X Figure 7.35 Figure 7.36