A complete and practical solution book for the common school teacher . f the circumscribing circle is ABxBCX AC = diameterx2 times the triangle; or diameter=(ABBCX AC)-i-twice the area of the triangle, or 5. (6) Hence, the sum of the diameters=7. (7) In the problem it is 1^L. .. The required is \l, and the sides are 30f, 41 and 51 \. NOTE.—We should remember that the continued product of thethree sides of a triangle divided by twice the area of the triangle isequal to the diameter of the circumscribing circle. PROBLEM men, A and B, started from the same point at the same time;A travele
A complete and practical solution book for the common school teacher . f the circumscribing circle is ABxBCX AC = diameterx2 times the triangle; or diameter=(ABBCX AC)-i-twice the area of the triangle, or 5. (6) Hence, the sum of the diameters=7. (7) In the problem it is 1^L. .. The required is \l, and the sides are 30f, 41 and 51 \. NOTE.—We should remember that the continued product of thethree sides of a triangle divided by twice the area of the triangle isequal to the diameter of the circumscribing circle. PROBLEM men, A and B, started from the same point at the same time;A traveled southeast for 10 hr., and at the rate of 10 mi. per hr., andB traveled due south for the same time, going 6 mi. per hr : theyturned and traveled directly towards each other at the same ratesrespectively, till they met: how far did each man travel? Solution. (1) Let C be the starting point, CA=the distanceA travels southeast, andCB = the distance Btraveled south. (2) Then CA = 100 mi., and CB = 60 mi. (3) Now draw AD perpendic- ular to CB produced to D. KT(;. 158 FAIRCHILDS SOLUTION BOOK. (4) As the Z D=a right angle, and L C=45°, then CD=AD% (5) 2AD2=AC2 = 1002, whence AD=50V*2, and BD=50V2 —60. (6) In the right triangle ADB, AB = V (50y2)2 + (50y2—60)- = V13600 — 6000V 2=+ mi. (7) A and B together travel 16 miles per hour, and the time required, until they meet in traveling AB, is TV of AB =?+ hr. (8) .. A traveled + mi., and B, + mi. of the distance AB. .*. The total distance traveled by A is + mi.,and by B, + mi. Note.—This solution was prepared for the American MathematicalMonthly by G. B. M. Zerr, A. M., Ph. D., President of Russell College,Lebanon, Va. PROBLEM 337. On a globe 20 ft. in diameter, 30 ft. above a plane, there is a bo\r 4 to his eyes: how much surface on the plane is hidden from him? Solution. (1) 30 ft. = EB, the distance the globeis above the plane; 20 ft.—DE,the diameter of the g
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