24. See Arithmetic, Art. 929, No. 8, for a rectangle, a rhombus, and a rhomboid, having equal bases and equal altitudes. No. 5 shows three rhomboids of equal bases and equal altitudes, but differing in shape. 25. See Art. 929, No. 8. 847. 1. (of 15 × 20) sq. in. The length of the third side does not enter into the computation. 6. Let the scholars find the area of the rectangle, 66 ft. by 63 ft., and the two triangles, 31 ft. each by 63 ft., and find the sum of the areas. Then lead them to see that bringing the righthand triangle to the left of the rhombus would make a rectangle 97 ft. by 63 ft., whose area is the sum above found. 7. Find the area in square meters, saying nothing more about the meter than that it is largely used on the continent of Europe, and is a little longer than a yard. 8-10. Give no rules yet for calculating the areas of trapezoids and trapeziums. Let the pupils ascertain the areas of the figures from the data supplied. XIII NOTES ON CHAPTER TEN The formal study of algebra belongs to the high-school; but some so-called arithmetical problems are so much simplified by the use of the equation that it is a mistake for a teacher not to avail herself of this means of lightening her pupils' burdens. In beginning this part of her mathematical instruction, the teacher should not bewilder her scholars with definitions. The necessary terms should be employed as occasion requires, and without any explanation beyond that which is absolutely necessary. 849. Very young pupils can give answers to most of these questions; so that there will be no need, for the present, at least, of introducing a number of axioms to enable the scholar to obtain a result that he can reach without them. 850. Pupils will learn how to work these problems by working a number of them. They may need to be told that x stands for 1x; and that, as a rule, only abstract numbers are used in the equations, the denomination-dollars, marbles, etc. being supplied afterwards. While the scholars should be required to furnish rather full solutions of the earlier problems, they should be permitted to shorten the work by degrees, writing only whatever may be necessary. 4. x+2x=54. 8. x+2x+6x=27000. 9. x+5x=72. 10. x+2x+3x=54. 11. x+6x=42. 12. 2x+10x=96. 13. Let x= the fourth; then 4 x = the third, 12x= the second, and 24 x the first. x+4x+12x + 24x=41. 14. x = the second, 2x = first, 9x= third. 15. 5x+4x=81. 16. 24 = = 456. 17. 4x=340. 19. 3x+4x=175. = 20. Let x each boy's share; 2x each girl's share. 2x+4x=240. 21. x = number of days son worked; 2x= number father worked. 3x=son's earnings; 8x=father's earnings. 3x+8x=165. = 22. x = number of dimes; 2x = number of nickels; 6x= number of cents. or (10 × x)+(5 × 2 x) +( 1 × 6 x) = 78, 23. 15x-12x=75. 10x+10x+6x=78. 24. x+4x+x+4x= = 250. 25. Let x= cost of speller; then 3x= cost of reader. 26. Let x = smaller; then 5x = larger. 27. Let x= Susan's number; 2x = Mary's; 3x=Jane's. 852. Pupils already know that means 3÷4, so that they 3 x can understand that 4 means 3x÷4, or 4 of 3x. When of something (3x) is 24, the whole thing (3x) must be 4 times 24, 3 x or 96; that is, when 4 From these examples can be formulated the rule for disposing of a fraction in one term of an equation, which is, to multiply 3 x = both terms by the denominator of the fraction. In changing the 24, to 3x, it has been multiplied by 4, so that the second term must also be multiplied by 4. first term of the equation, 4 853. In solving these examples by the algebraic method of clearing of fractions," attention may be called to its similarity to the arithmetical method. To find the value of y in 2, the pupil multiplies 8 by 5 and divides the product by 2; as an example in arithmetic, he would divide 8 by, that is, he would multiply 8 by ; the only difference being that by the latter method he would cancel. У While =8 may be changed to 2 = 4 by dividing both terms 248 5 5 by 2, beginners are usually advised to begin by "clearing of fractions," short methods being deferred to a later stage. 8. 27 x should be reduced to an improper fraction, making the 23x equation, =115. Make similar changes in 12, 14, 18, and 20. 8 = = denominator. 7x-5x=24; 2x 24; x=12. The numerator, 5x, will be 5 times 12, or 60; the denominator will be 84; and the fraction, . Ans. The employment of the latter plan does away with fractions Then 2x will be cost of plums, and 12x the cost of the peaches. 19. Let x price per yard of the 48-yard piece; 2x = price per yard of the 36-yard piece; 48x will be the total cost of one, and 72x, of the other. 48x+72x=240. 20. 160x120x = 840. 856. The pupils should be permitted to give these answers without assistance. In Art. 857 is explained what is meant by "transposing." |