A complete and practical solution book for the common school teacher . FIG. 38. rd. PROBLEM 348. A goat is tethered by a cord 30 ft. long, to the corner of a shed intthe form of an equilateral triangle, side 20 ft.: find area over which itmay graze outdide of the shed. Solution. (1) Let ABC be the shed. (2) The goat can graze with the 30 ft. cord over a cir-cle, except a sector of60°, or angle A. (3) EMD=300°, or JU=f oi a circle, which has 302?rX|= sq. ft. (4) Angle DCS=120°, and angle EBS = 120°. (5) Angle EBS+angle DCS= 240°, or | of circle with aradius of 10 ft. (6) .. The area of BE
A complete and practical solution book for the common school teacher . FIG. 38. rd. PROBLEM 348. A goat is tethered by a cord 30 ft. long, to the corner of a shed intthe form of an equilateral triangle, side 20 ft.: find area over which itmay graze outdide of the shed. Solution. (1) Let ABC be the shed. (2) The goat can graze with the 30 ft. cord over a cir-cle, except a sector of60°, or angle A. (3) EMD=300°, or JU=f oi a circle, which has 302?rX|= sq. ft. (4) Angle DCS=120°, and angle EBS = 120°. (5) Angle EBS+angle DCS= 240°, or | of circle with aradius of 10 ft. (6) .. The area of BES+SCD=10*irX|-= sq. FIG. 39. MENSURATION. 165 (7) Whole area grazed over is sq. ft.+ sq. sq. ft. PROBLEM 80 cents a rod, what would be the cost of fencing- a field in theform of an equilateral triangle, its altitude being 6 Solution. (1) Let ABC be the triangle. (2) AO bisects the angle A, then angle OAD=30°. Then AODis a right-angled triangle. (3) Now, OD+DE = CO, or the ra- dius of the circle. (4) CO or AO=| of 6 rd. = 4 rd. (5) AB = AOV3, or 4V3= rd. (6) The perimeter is rd. (7) X.$80 = $, cost of fencing the triangle.
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Keywords: ., bookcentury1800, bookdecade1890, booksubject, booksubjectgeometry
