A complete and practical solution book for the common school teacher . 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.
A complete and practical solution book for the common school teacher . 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 globe; andAD = 4 ft., the distance the boyseye is above the ball. (2) 14 +DL = AL. (3) AF=VAL2 — LF2=4V6~ft. (4) Now, by the similar triangles ALF and ABC, AF : LF::AB : BC,or 4V6 : 10 :: 54 ^BC, fromwhich BC = 540+4 V6=135+V 6 =135V^6xi-=1-^^ ft. 6. (5) ?BC2 = A35V6 = area of the FIG. 31. surface hidden from the boy. PROBLEM 338. What is the area of the largest square that can be inscribed in asemicircle, the diameter of the circle being- 20 inches? MENSURA TION 159 Solution. (1) Let AB be the diameter of the circle. (2) Draw SB equal and perpen- dicular to AB, tangent tothe circle; E and SO to thecenter of the circle. (3) Let K be the point where SO cuts the circle. (4) From the point K, draw KC parallel to SB; TK parallelto AB, TX parallel to the square is com-pleted. (5) Then1^K^40C2-hOC2, or 50C2. (6) ButOK2=100; .\ 5OC2 = 100; OC2 or + in. (7) CX = the side of the square, or X 2=8 944+ in (8) CX2 =+ sq. in., area of the inscribed .square.
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