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929. 1. The pupils should not be shown how to calculate these areas.
2. If any difficulty is experienced in finding the areas of these triangles, the pupils should be referred to 1; after which they should be led to deduce the rule. Thus the area of the second triangle may be calculated from the figure in 1 by adding
of (60 X 50) to £ of (60 X 50); that of the third by adding 1 of (60 x 60) to tof (60 x 40); and that of the fourth by adding f of (60 x 70) to £ of (60 x 30). Each of these will be found equal to ļof (60 x 100).
3. The second rectangle is divided into three triangles, two of them right-angled. By deducting from the area of the rectangle the sum of the areas of the two right-angled triangles, they will obtain the area of the remaining triangle.
4. The area of each of these triangles can be ascertained by referring to the corresponding triangle of 3. Let the scholars do this for themselves.
. : 5. The areas of the oblique-angled triangles constituting the first and second quadrilaterals, can be calculated by the pupils that have benefited by the work in 4. If they see that the area of each triangle of a parallelogram is equal to 1 (base x altitude), the area of the latter is equal to base x altitude.
For definitions of quadrilaterals see Art. 1265.
6. The area of the first is equal to the area of the rectangle, (50 x 60), plus the area of the triangle, f of (50 x 60); or 4500 sq. ft.
The second is made up of a rectangle and of two triangles ; its area is also 4500 sq. ft.
The pupils should be led to see that if, in the fourth, the upper left triangle were cut off and placed below, and if the lower right triangle were cut off and placed above, as indicated by the dotted lines, the resultant figure would be a rectangle 60 x 75.
Cutting off both triangles in the third, and placing them above, will make a 60 x 75 rectangle.
The area of each trapezoid is equal to [1 of (50 + 100)] x 60.
7. The area of each of these quadrilaterals equals of (30 x 100) +1 of (40 x 100), or [1 of (30+40)] x 100.
The first three quadrilaterals are trapeziums. The last is a trapezoid. Which are the parallel sides ?
8. A strip of paper of any uniform width may be used. Carefully cut a rectangle by making square corners with a card. Using the base of the rectangle as a measure, place two dots on the lower edge of the strip to mark the extremities of the base of a parallelogram equal in length to the base of the rectangle, and above these, at any convenient distance to the right or to the left, two others to mark the extremities of the opposite side of the parallelogram. Draw lines forming the right and left sides, and cut along these lines. That the parallelogram is equal in area to the rectangle, may be shown by carefully drawing a perpendicular at one corner; cutting off the triangle thus made, and placing it, in a reversed position, on the opposite side of the parallelogram.
9. See 6, third and fourth trapezoid.
930. 2. Four faces will measure 6 ft. by 41 ft. each, and two will measure 41 ft. by 41 ft. each.
4. Dimensions of floor, 57 ft. by 18 ft., or 19 yd. by 6 yd.
5. Volume in cubic feet, 4 x 4 x 12. Multiply by 1000 to get the weight in ounces of an equal volume of water. Multiply by 2.8 to get weight of marble in ounces. Divide by 16 x 2000 to reduce to tons.
4 X 4 x 12 x 1000 x 2.8
16 X 2000
9. Outer dimensions, 14 x 14 x 14, or 2744 cu. in. See if the same number of cubic inches of wood is obtained by calculating the volume of the wood in 6-2 pieces, 12 x 12, 1 in. thick ; 2 pieces, 12 x 14, 1 in. thick; 2 pieces, 14 x 14, 1 in. thick.
10. A cube of water 2 ft. long contains (2 x 2 x 2) cu. ft., or 8 cu. ft. At 1000 oz. to a cubic foot, it weighs 1980 lb. X8, or 500 lb. The cube of iron weighs 8 times as much as an equal volume of water.
A cube of iron 1 ft. long also weighs 8 times as much as a corresponding cube of water, or 8 times 146° lb. = 500 lb., or 1 ton.
A 3 ft. cube of iron contains 27 cu. ft., weighing 8 times as much as a corresponding cube of water, or 216 times 1000 oz. = 6 tons.
NOTES ON CHAPTER TWELVE
931. While problems requiring the pupil to find the principal, the rate, or the time have very little “practical" value, they can be so readily taught by the algebraic method that the time spent upon them need not be very great. A pupil that is able to calculate one of a series of related items is benefited by being required to calculate the others, if he is not compelled to resort to a series of ill-understood rules in order to obtain the results.
Although there is no real difference between the algebraic method and the arithmetical one, a great number of scholars fail to obtain a thorough understanding of the latter. They can work a number of examples, following a model solution at the head of the lesson ; but they fail to grasp the underlying principles. By the algebraic method, x is used to represent the number of years or dollars, or the rate, instead of the 1 year, $1, or 1%, of the other; but this method seems to require the formulation of a number of rules, as against practically none in the case of the other.
After pupils have learned to work examples by the algebraic method, they can be encouraged to discontinue the use of the x; but they should not be taught both methods at one time.
933. Represent the required item by x. Simplify the first member before proceeding to solve the equation.
1. 2000 x 160 *3=300.
= x= Ans.
100 x 360 15. Principal = $ 97.57 – $ 7.57 = $ 90.
90 X 100 X x = 7.57. 21. Let x = principal.
w X X 4 X 846 x+
100 360 The interest is then found by subtracting from $ 196.92, the value obtained for x.
22. First find the principal (2). 25. See 15.
934. The recommendation so frequently made, that all written work be preceded by oral exercises of the same character, should not be followed without some modifications. Oral work is necessarily accompanied by a number of devices that tend to simplify the task of handling numbers that are not seen ; written work should follow general rules in order to be learned by a majority of the pupils, although later they may adopt some of the short-cuts of their oral exercises. Even the oral addition of two numbers of two figures each, is done in a manner different from the ordinary slate method, the operation in the former case being commenced generally with the tens' figures, and in the