A complete and practical solution book for the common school teacher . he sides are 13 and 15:find the other side. Solution. (1) From Fig. 48, AB = 15 ft., CB = 13 ft. (2) CD=84X2-M5= ft. (3) CDB is a right triangle, with its hypothenuse 13, and altitude ft. (4) BD = y/(13)2 — ()2= ft. (5) AD-15— ft., the base of the right triangle ADC. (6) CA = V()2—()2=14 ft., the third side. PROBLEM a triangle whose sides are 13, 14, 15, required the length of eachline from the angles to a common point, which shall divide the tri-angle into three equal parts. Solution. (1
A complete and practical solution book for the common school teacher . he sides are 13 and 15:find the other side. Solution. (1) From Fig. 48, AB = 15 ft., CB = 13 ft. (2) CD=84X2-M5= ft. (3) CDB is a right triangle, with its hypothenuse 13, and altitude ft. (4) BD = y/(13)2 — ()2= ft. (5) AD-15— ft., the base of the right triangle ADC. (6) CA = V()2—()2=14 ft., the third side. PROBLEM a triangle whose sides are 13, 14, 15, required the length of eachline from the angles to a common point, which shall divide the tri-angle into three equal parts. Solution. (1) Let ABC be the triangle. (2) We know that the area of the triangle is 84; then each part is 28. (3) The common point is at the intersection of lines drawn parallel to the sides and one-third the distance fromthe side opposite angle. 174 FAIRCHILDS SOLUTION BOOK. (4) AC=13, CB = 15, and AB = 14. (5) EO = (28x2)+14=4. (6) FO=(28X2)+13=ff. (7) OD=(28x2)-M5=-ff (8) AH=Y- AX, or OH =Y; ???HE= l/(OH2—UEa)=fAE=y+j=Y; A0= T/(AE2_^.QE2)- £1/505=+. (9). (10) BC=V(BE» + UE») = 4^/673=+. (11) OG=£ofl5=5. FG=v/OG2-TyF2 (12) CF=«+V=W- (13) CO=T/[(W) + (»)a]=ii/592=+. FIG. 50. 8 8 PROBLEM 364. In a triangular garden whose sides are a=300 ft., 6=250 ft., and£=200 ft., is a spring equally distant from each corner: how far is itfrom each corner? Solution. (1) Let BAC be the garden. (2) CA=300 ft., BC=250 AB=200 ft., and O represents the position olthe spring. (3) BO, OA and OC are radii of the circumscribingcircle and the requireddistance. (4) Let g±|±£ =s; then, by 2formula. CD= -Vs{s- a)\s—b) (s—c) —
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