22. If ∆ABC is any triangle, show that the altitude CD = b sin A = a sin B. 23. Show from the results of Ex. 22 that 24. By drawing another altitude in A ABC, Ex. 22, show that b sin C = c sin B, and thus show that in any triangle This important relation is known as the Law of Sines. 25. In ABC, if_AB = 148, ∠A = 46°, and ∠B = 58°, find AC. SUGGESTION. - ∠C = 76°. Why? Hence by the law of sines in Ex. 24, C B 26. In surveying through great distances and over rough or mountainous country, a process called triangulation is employed. First a base line AB is carefully measured. Then a third point C is selected and ∠BAC and ∠ABC measured. By the law of sines CB is then computed. Another point D is then selected, and ∠BCD and ∠CBD measured. Then, by again applying the law of sines, CD is computed. Continuing, point E is selected, and ∠ECD and ∠CDE measured. From these measurements, ED is computed. In order to find ED it is only necessary to measure the one line AB. The process may be continued for any number A of steps. E C 650 80. 50 42° D 454701 B If AB = 2180 ft., and the angles measured are as indicated in the figure, find ED. NOTE. - Let the students measure by triangulation the distance between two points by measuring a distant base line such as AB in Exercise 26 For measuring the angles, use may be made of some instrumens such as that described in Exercise 16, page 40. 136. Theorem. The perimeters of two similar polygons have the same ratio as any two corresponding sides. Hypothesis. ABCD... and MNOP ... are similar polygons, with AB and MN as corresponding sides. Conclusion. AB+BC+ CD + etc. AB Proof. 1. ABCD and MNOP ... ... etc. MN are similar, with Нур. AB BC 2. .. = etc. Def. sim. poly. MNNO 137. Theorem. - Two similar polygons can be divided into the same number of triangles, similar each to each and similarly placed. D Hypothesis. ABCD... and MNOP ... are similar polygons. Conclusion. ABCD and MNOP into the same number of triangles, similar each to each and similarly placed. Suggestions. Draw all diagonals possible from two corresponding vertices A and M. Prove △ABC~△MNO by § 130. Then Z1 = 23. Why? Next, prove ∠ 2 = ∠4. Also = MONO Why? Hence then, is AACD ~ △ MOP? Write the proof in full. AC CD = Why? Why, Give the proof by drawing all possible diagonals from the corresponding vertices E and Q, instead of the diagonals from A and M. 138. Theorem. - If two polygons are composed of the same number of triangles, similar each to each and similarly placed, the polygons are similar. Suggestions. ∠B = ∠ N. and ∠2 = 24. ... ... are com Hypothesis. Polygons ABCD and MNOP posed of & ABC, ACD, etc., similar respectively to A MNO, MOP, etc., and similarly placed. Why? Similarly, ∠1=23 Hence ∠C=∠0. Why? Similarly, prove Give the proof by drawing diagonals from E and Q. 139. Construction. - Upon a given line-segment corresponding to a side of a given polygon, construct a polygon similar to the given polygon. Required to construct on MN as a side corresponding to AB, a polygon similar to ABCD .... Construction. 1. Draw all possible diagonals from A. 3. At N construct 24 = ∠w, giving point O. ... EXERCISES 1. The perimeters of two similar polygons are 144 yd. and 256 yd., respectively. A side of the first is 18 yd. Find the corresponding side of the second. 2. The sides of a polygon are 8 in., 12 in., 15 in., 6 in., and 20 in. The side of a similar polygon corresponding to the 8-in. side is 6 in. Find the perimeter of the similar polygon. 3. The perimeter of a polygon is p and one side is x. If the perimeter of a similar polygon is q, find the side corresponding to x. 4. A rectangular field is wyd. wide and lyd. long. Find the perimeter of a similar field 3 w yd. wide. 5. Prove that two corresponding diagonals of two similar polygons have the same ratio as a pair of corresponding sides. 6. Prove that if two polygons are each similar to a third polygon, they are similar to each other. 7. The vertices of a given polygon ABCDE are joined to a given point H. A line-segment MN is drawn parallel to AB so that Mis on AH and N on BH. Then NO || BC, OP || CD, PQ || DE, and QM are drawn as in the figure. Prove MNOPQ ~ ABCDE. proportions to show that SUGGESTION. - How may QM be proved parallel to EA? Derive = EH AH 8. Polygon ABCDE is formed by joining points of intersection of lines on squared paper. Polygon MNOPQ is formed by joining corresponding intersections of lines separated by only one half the interval. Prove that ABCDE ~ MNOPQ. 9. Explain how squared paper may be used to construct similar polygons the ratio of whose sides is 3; the ratio of whose sides is 5. 10. Draw any quadrilateral. Upon a given line-segment as a side corresponding to a given side of the quadrilateral, construct by use of compasses and straightedge a similar quadrilateral. E 140. Maps and plans. A map or plan is a figure similar to the figure formed by the object which it represents. Thus, a map of a state is a drawing similar to the figure formed by the state itself. The drawing in the margin shows an architect's floor plan of a house. A map or plan is always drawn to scale, i.e., in the map or plan the distances are made proportional to the actual distances which they represent. Laundry Pantry Kitchen Dining Porch Living Room |