A complete and practical solution book for the common school teacher . 12 43560 (8) Clearing of fractions, 12R=348480, and R=29040. (9) -jy- = 193607r, by substituting the value of 193607r = no. of acres. PROBLEM 369. A gardener is desirous of laying- out a flower garden containing10 sq. rd. in the form of a crescent, bounded by a quadrant and semi-circle: what is the diameter of the semi-circle? Solution. (1) The center of the large circle, O, will be on the circum- ference of the smaller circle, ADC. (2) ACDF is a quadrant. The right triangle ADC=10 sq. rd., or AD=the side of a square c
A complete and practical solution book for the common school teacher . 12 43560 (8) Clearing of fractions, 12R=348480, and R=29040. (9) -jy- = 193607r, by substituting the value of 193607r = no. of acres. PROBLEM 369. A gardener is desirous of laying- out a flower garden containing10 sq. rd. in the form of a crescent, bounded by a quadrant and semi-circle: what is the diameter of the semi-circle? Solution. (1) The center of the large circle, O, will be on the circum- ference of the smaller circle, ADC. (2) ACDF is a quadrant. The right triangle ADC=10 sq. rd., or AD=the side of a square containing 20 sq. rd. 178 FAIRCHILDS SOLUTION BOOK. (3) TheluneorcrescentCFAM will contain 10 sq. rd.,for the quadrant ACDFis equal to the semicir-cle CAMC. (4) Now, taking away the com- mon segment X, we haveleft the triangle ACD =the lune CAMC. (5) As the triangle is equal to 10 sq. rd., then the cres-cent contains 10 sq. FIG. 55.
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