24. 20 acres The diagonal

= 3200

=3

sq. rd. Each side measures √3200 rd. the square root of the sum of the squares of two

equal sides. The square of each=3200.

The diagonal√3200+ 3200 rd.

√6400 rd. = 80 rd.

NOTE. By constructing a square on the diagonal of a square, the pupils will see that the former will be twice as large as the latter; that is, that a square on the diagonal of the above will contain 6400 sq. rd., making each side 80 rd.

In the above example, the square root of 3200 should not be extracted.

1107. 6. Find the proceeds of $1572.50 for 81 da. (Omitting days of grace, the proceeds for 78 da. = $1552.06.)

9. One costs 851% of $1500; the other costs 1021% of $1300.

10. ($2562.50÷1.025) $62.50 number of acres.

=

1. It is frequently difficult, for various reasons, to measure the altitude of a triangular field. On this account, a method of determining the area when the lengths of the sides are given, is useful, even though the underlying principles be not understood. A pupil can satisfy himself as to its accuracy, by calculating the area of the right-angled triangle in 3, and of the isosceles triangle in 5.

2. The half sum of (35 +84 +91) = 105. The remainders are: (105-35) 70, (10584) 21, and (105-91) 14. Area √105 × 70 × 21 × 14 ft. sq. = 1470 sq. ft.

=

3. This is a right-angled triangle, since 212+28 352; its area, therefore, is of (21 × 28) sq. rd., or 294 sq. rd. By the other method, the area = √42 × 21 × 14 × 7 sq. rd., or 294 sq. rd.

4. The sides of one triangle measure 39, 52, and 65 rd. respectively. Its area 1014 sq. √78 x 39 x 26 x 13 sq. rd. rd. The sides of the other are 25, 60, and 65 rd., respectively; and its area = √75 × 50 × 15 × 10 750 sq. rd. The area of the quadrilateral = 1014 sq. rd. + 750 sq. rd. 1764 sq. rd.

rd. sq.

=

=

Each of these triangles is right-angled, AC being their common hypotenuse. Their areas are of (39 × 52) sq. rd., and 1⁄2 of (25 × 60) sq. rd., respectively.

5. Since the altitude AC, Fig. 2, Arithmetic, Art. 1263, of an isosceles triangle divides the base into two equal parts, BC 15 yd. AC2 AB-BC2625-225=400; AC= 20 yd.

=

The area =

=

=

of 600 sq. rd. = 300 sq. rd.

The area by the other method = √40 x 10 x 15 x 15 sq. yd. 6. In the right-angled triangle ACB (Art. 1263, Fig. 2), BC 48 ft; AB 64 ft.; The area = The area = Area √9 x 3 x 3 x 3 sq. ft.

of 96 ft. √4096-2304

1792 ft.

7. sum

= 9 ft.

8. The base

sq. ft.

9.

=

=

AC= √642

√702422 ft. 56 ft.

48' ft. =

×

of (96 x 42.332) sq. ft.

Area = 1 (42 × 56)

50-482-14, the number of feet in one-half the base.

See Fig. 2, Arithmetic, Art. 1263.

10. The common base of the two triangles will be one diagonal, 2 in. The altitude of a triangle will be one-half the other diagonal = √2'-1' in. = √3 in. = 1.732 in. The diago12 nal= 3.464 in. The area of each triangle = (2 × 1.732) sq. in. 1.732 sq. in. The area of the rhombus = 3.464 sq. in.

11. The scholars should be led to ascertain for themselves the approximate relation between the diameter of a circle and its circumference. Place a point marked on the circumference of a spool, or something similar, on a given point on the surface of a sheet of paper. Roll the spool until the point on the circumference again touches the surface of the paper. The distance between the two points on the paper will be equal to the circumference of the circle. Measure this distance very carefully, also the diameter of the circle, and ascertain the ratio.

12. See Art. 1274, 6-14, for the reason for the rule for determining the area of a circle.

13. The circumference = (2x × 3.1416) = 3.1416x; the diameter = x; the area = 3.1416 x x x = = 3.1416x2.

1108. The balance, $851.72, is found by adding the credits, and subtracting from $2535.35 in one operation. See Art. 384.

1110. 1. Amount of $500, July 25 to

April 1, 250 da.,

Amount of $100, Sept. 18 to April 1, 195 da.,

$520.83

$103.25

201.83 305.08

$215.75

$635.40

$307.60

201.90 509.50

$125.90

Due at settlement,

5. In the debit column, place the interest for 329 da., $46.06. The total of this column is written on the same line as the total of the credit column, and is $886.06. The amount is also written as the total of credits. The total interest ($28.38) on $500 for 297 da., $24.75, and on $200 for 109 da., $3.63, is written among the credits; and the cash payment is ascertained by writing in its place the sum necessary ($157.68) to make the total, $ 886.06.

See Art. 384.

6. From the amount of $725 for 234 da. + the amount of $603 for 174 da., take the amount of $600 for 183 da. + the amount of $300 for 37 da.

1113. 3. The field contains 160 sq. rd. x 74. Its length is (160 × 730) rd. 40 rd. The diagonal √402 + 302 rd. = 50 rd.

=

4. The rate per cent that will produce $36 interest in 13 yr. is 7. The interest on $212.50 at 71% for the given time $52.02. Ans.

Or, the problem may be solved as follows (Arithmetic, Art.

$36 x 212.50 x 40 $36 x 212.50 x 204

300 x 20

=

300 x 20 x 5

5. The date not being given, the number of days is taken as 120+ 3. The proceeds for 120 da. = $490.

8. The interest on $635 for 205 da. at 5% = $18.08.

XVII

NOTES ON CHAPTER FOURTEEN

1115. 1. The total interest on the given sums of money is equal to the interest of $3000 for 1 mo. As the total sum is $1000, the interest of $3000 for 1 mo. equals the interest on $1000 for 3 mo.

1116. 3. Since the time is required, the products may be expressed in years (or months or days), and their total divided by the total of the multipliers.

5. [(15 da. ×200) + (30 da. × 300) + (45 da. × 400)] ÷ (200 +300+400).

6. [(1 mo. x 210) + (2 mo. x 210) + (3 mo. x 210)+(4 mo. × 210)] ÷ 840.

Since the sum due at each period is the same, the 210 may be omitted; (1 mo. + 2 mo. + 3 mo. + 4 mo.) ÷ 4.

7. [(2 mo. × 320) + (4 mo. × 160) + (5 mo. × 240)+(6 mo. × 240)]+960.

=

Or, (2 mo. × 1) + (4 mo. × 1) + (5 mo. × 1) + (6 mo. × 1) 4 mo. Ans.

8. (2 mo. x)+(3 mo. × })+(4 mo. × 1) + (12 mo. ×1) = 718 mo. = 7 mo. 26 da. Ans.