A complete and practical solution book for the common school teacher . MENSURATION. 179 triangles CDB and BDE are similar. (3) We have CD : DB::DB : DE, or 2 : 6::6 : DE, or DE— 18 in. (4) DE+DC = EC, or 20 in. .*. The diameter was 20 in. PROBLEM 371. Three equal circles touch each other externally, and thus encloseone acre of ground: what is the diameter in rods of each of these circles? Solution. (1) Draw three equal circles to touch each other externallyand join the three centersABC, thus forming an equi-lateral triangle. (2) Let R represent the radius of the three equal circles;then it is
A complete and practical solution book for the common school teacher . MENSURATION. 179 triangles CDB and BDE are similar. (3) We have CD : DB::DB : DE, or 2 : 6::6 : DE, or DE— 18 in. (4) DE+DC = EC, or 20 in. .*. The diameter was 20 in. PROBLEM 371. Three equal circles touch each other externally, and thus encloseone acre of ground: what is the diameter in rods of each of these circles? Solution. (1) Draw three equal circles to touch each other externallyand join the three centersABC, thus forming an equi-lateral triangle. (2) Let R represent the radius of the three equal circles;then it is obvious that eachside of the triangle is equalto 2R. (3) This triangle encloses the given area, and three equalsectors. As each sector isa third of two right angles,the three sectors are equal to a semicircle. (4) The area of a semicircle whose radius is R =. FIG. 58. TrR (5) The area of the whole triangle is ?R2 2 160. (6) But the area of the equilateral triangle is R2j/8. (Fig. 38.)?R* (7) .\ R»V3 = (8) R*= 2320 f 160, or R*(2v/3— tt)= 2V3— (9) .. R = +rd. PROBLEM 372. Find the length of the longest and shortest chords that can bedrawn through a point 6 in. from the center with a radius of 12 inches- 180 FAIRCHILDS SOLUTION BOOK. Solution. (1) The longest possible chord through any point is the diameter drawn throughit, and the shortest is one perpendic-ular to the diameter. (2) Therefore, AB is the longest, and KP the shortest chord. (3) FB=FO+OB = 18in. AF=6 in. (4) The products of the segments AF and FB = the product of KF and FP. (5) 6Xl8=i/i08=+ in., or \ of KP. (6) = in., the shortest chord.
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