A complete and practical solution book for the common school teacher . of the altitude. 5. Radius of circumscribed circle = two-thirds of the alti-tude. 6. All equilateral triangles are similar. 7. Each angle is J of a circle of 60°. 8. Side=the radius of the circumscribed circle multiplied by V3. Note—The student should commit the above rules; it will oftenbe the saving of much labor. 172 FAIRCHIL&S SOLUTION BOOK. III. SCALENE TRIANGLES. PROBLEM 360. What is the area of a triangle whose base is 20 ft., and altitude Solution. (1) 20 (2) 12 ft. = altitude. (3) (20X12)^=120
A complete and practical solution book for the common school teacher . of the altitude. 5. Radius of circumscribed circle = two-thirds of the alti-tude. 6. All equilateral triangles are similar. 7. Each angle is J of a circle of 60°. 8. Side=the radius of the circumscribed circle multiplied by V3. Note—The student should commit the above rules; it will oftenbe the saving of much labor. 172 FAIRCHIL&S SOLUTION BOOK. III. SCALENE TRIANGLES. PROBLEM 360. What is the area of a triangle whose base is 20 ft., and altitude Solution. (1) 20 (2) 12 ft. = altitude. (3) (20X12)^=120 .*. 120 sq. of the tri-angle ABC. FIG. 48 Bulo.—Multiply the base by the altitude and take half the product. PROBLEM 361. Find the area of a triangle whose sides are 13, 14 and 15 ft. respec-tiyely. Solution. (1) 13 ft.+14 ft., sum of the sides. (2) $ of 42=21 ft., half the sum of the sides. (3) 21 ft.—13 ft., 21 ft.—14 ft., 21 ft.—15 ft. (4) 21x8x7x6=7056, product of the half sum and the three (5) V 7056=84 sq. ft., area of ABC.
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