A complete and practical solution book for the common school teacher . find the first point, we have 202-f;ra=60a-f->2. (3) *a+^=100; then * = 66,j = 34. (4) To find the second point, we have 60a+#2 =80a+#2. (5) a-H?=100; then, ?< = 64, and ,g=36. (6) V (80H--2+72) = V (60a+*a+jpa) = V (20a+«a+*a) = 78852 = + ft., the length of the ladder. (7) From the first tower, the foot of the ladder is x/8852—202 = + ft. (8) From the second it is V8852+602 = 7i,.470rt21 + ft. (9) From the third it is V8852—802 = + ft. (10) The height of a tower on the fourth corner, that t
A complete and practical solution book for the common school teacher . find the first point, we have 202-f;ra=60a-f->2. (3) *a+^=100; then * = 66,j = 34. (4) To find the second point, we have 60a+#2 =80a+#2. (5) a-H?=100; then, ?< = 64, and ,g=36. (6) V (80H--2+72) = V (60a+*a+jpa) = V (20a+«a+*a) = 78852 = + ft., the length of the ladder. (7) From the first tower, the foot of the ladder is x/8852—202 = + ft. (8) From the second it is V8852+602 = 7i,.470rt21 + ft. (9) From the third it is V8852—802 = + ft. (10) The height of a tower on the fourth corner, that theladder will reach is V8852— (j =40^% or + ft. PROBLEM 419. A rectangular field is 64 rd. long and 50 rd. wide: if from the mid-dle of either side we set out a right line to its opposite, where must itintersect in order to cut the field inthe ratio of 1 to 2? Solution. (1) Let ABCD be the field. and P the point of theside AH. (2) Draw the line PE perpen- dicular from P to the op-posite side. (3) Draw PF which shall di- vide the field as required. fic. 212 FAIRCHILDS SOLUTION BOOK. (4) The area of the field is 64X50, or 3200 sq. id. (5) APED=£ of the field, and PBCF=| of the field. (6) The triangle EPF=|—J=* of the field, or \of 3200 sq. rd. (7) Since DA=50 rd., EF=(533£X2)-^50=21£rd. (8) EC=32 rd., half the side. (9) FC=CE=FE = 32-21i=10f rd., the distance from the corner of the field.
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