A complete and practical solution book for the common school teacher (1899)
Fairchild, J[ohn] T[heodore] [from old catalog] · Crawfis College, Ohio, The author · leaf 169
ing, 200^=12000. We find *=60. (5) AL = AE, LO=OD = LC, and EB = BD. EB = AB—AE, _or 100—AE, or 60=40. (6) EB2+OE2, or 402+r2=20V5 = OD2. (7) r=20. AL+LC=80. BD-fDC = 80. /. ABC is a 60, 80 and 100 triangle.Note.—d, or 20V5= VaO2 — OB3. PROBLEM 330. The perimeter of a right-angled triangle is 240… … MENSURA TION. 155 (2) LC+CD=40, and 240—40=200, the sum of the lines AL, AE, EB and BD, which are tangents to the same circle. (3) Hence, AL=AE,EB = BD, and AL+DB = AE+EB=100, the hypothenuse. (4) AAOB=AAOL+ABOD, because they have the same bases and altitudes. (5) P being the middle point of AB, CP…
Arithmetic · Geometry
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