Given parallel chords TEST 534

PERSPECTIVE ON HISTORY Circumference of the Earth By traveling around the earth at the equator, one would traverse the circumference of the earth. Early mathematicians attempted to discover the numerical circumference of the earth. But the best approximation of the circumference was due to the work of the Greek mathematician Eratosthenes 276–194 B . C .. In his day, Eratosthenes held the highly regarded post as the head of the museum at the university in Alexandria. What Eratosthenes did to calculate the earth’s circumference was based upon several assumptions. With the sun at a great distance from the earth, its rays would be parallel as they struck the earth. Because of parallel lines, the alternate interior angles shown in the diagram would have the same measure indicated by the Greek letter ␣. In Eratosthenes’ plan, an angle measurement in Alexandria would be determined when the sun was directly over the city of Syene. While the angle suggested at the center of the earth could not be measured, the angle in Alexandria formed by the vertical and the related shadow could be measured; in fact, the measure was . Eratosthenes’ solution to the problem was based upon this fact: The ratio comparing angle measures is equivalent to the ratio comparing land distances. The distance between Syene and Alexandria was approximately 5,000 stadia ␣ L 7.2° 1 stadium 516.73 ft. Where C is the circumference of the earth in stadia, this leads to the proportion Solving the proportion and converting to miles, Eratosthenes approximation of the earth’s circumference was about 24,662 mi, which is about 245 mi less than the actual circumference. Eratosthenes, a tireless student and teacher, lost his sight late in life. Unable to bear his loss of sight and lack of productivity, Eratosthenes committed suicide by refusing to eat. ␣ 360° = 5000 C or 7.2 360 = 5000 C L PERSPECTIVE ON APPLICATIONS Sum of Interior Angles of a Polygon Suppose that we had studied the circle before studying polygons. Our methods of proof and justifications would be greatly affected. In particular, suppose that you do not know the sum of interior angles of a triangle but that you do know these facts: 1. The sum of the arc measures of a circle is 360°. 2. The measure of an inscribed angle of a circle is the measure of its intercepted arc. Using these facts, we prove “The sum of the interior angles of a triangle is 180°.” Proof: In , , , and . Then 1 2 mBC ¬ + mAC ¬ + mAB ¬ = 1 2 360° = 180°. m ∠A + m∠B + m∠C = m ∠C = 1 2 m AB ¬ m ∠B = 1 2 mAC ¬ m ∠A = 1 2 mBC ¬ 䉭ABC 1 2 Alexandria Syene Figure 6.62 Using known facts 1 and 2, we can also show that “The sum of the interior angles of a quadrilateral is 360°.” However, we would complete our proof by utilizing a cyclic quadrilateral. The strategic ordering and association of terms leads to the desired result. Proof: For quadrilateral HJKL in Figure 6.64, or 1 2 360° + 1 2 360° 1 2 m LKJ ២ + m LHJ ២ + 1 2 mHLK ២ + m HJK ២ = 1 2 m LKJ ២ + 1 2 mHLK ២ + 1 2 m LHJ ២ + 1 2 m HJK ២ m ∠H + m∠J + m∠K + m∠L = B A C Figure 6.63