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ILLUSTRATIONS. – BD and AC are diagonals of figure 3; Z Y and XP are diagonals of figure 4; F K, FJ, FI, and F E are diagonals of figure 6.

(n.) The Area of a square or of a rectangle equals its length multiplied by its breadth.

(o.) The area of a triangle equals half the product of its basa by its altitude.

ILLUSTRATION. — In figure 1, if the base, MN, is 8 ft., and the altitude, RS, is 5 ft., the area will be 1 of 8 times 5 sq. ft. ft.

2 sq.

(p.) The area of a triangle is also equal to the square root of the product obtained by multiplying half the sum of its three sides by the three remainders found by subtracting each sido separately from the half sum of the sides.

10

ILLUSTRATION. – To find the area of figure 1, when MN = 8 ft., Ripor 5 ft., and MR 7 ft., we proceed thus: 1 of the sum of the three sides 1 of 8 + 5 + 7 = 1 of 20 = 10. This half sum minus the first side 8 2. This half sum minus the second side = 10 5 = 5. The half sum minus the third side 10 — 7 = 3. The product of the half sum and the three remainders 10 X 2 X 5 X 3 = 300, and the square root of 300 17.3205 +. Hence, the area of the triangle is 17.3205 sq. ft.

(q.) The area of a parallelogram equals the product of its base by its altitude.

ILLUSTRATION. — If, in figure 4, the base, X Y, equals 12 ft., and the altitude, ZQ, equals 8 ft., the area will equal 12 times 8 sq. ft. = 96 sq. ft.

(r.) The area of a trapezoid equals half the product of its altitude by the sum of its parallel bases.

ILLUSTRATION.-If the lower base, MN, of figure 5 equals 12 ft., and the upper base, R 0, equals 9 ft., and the altitude, U V, equals 8 ft., the area will equal 1 of the product of 8 multiplied by the sum of 12 and 9 - 1 of 8 times 21 84 sq. ft.

(s.) The area of an irregular polygon can be found by dividing it into triangles.

ILLUSTRATION. - The area of figure 6 equals the sum of the triangles FHG, FHI, FJI, FJ K, and FKE, into which it is divided.

(t.) The areas of different triangles, squares, and parallelograms are to each other as the product of their bases by their altitudes.

ILLUSTRATION. — The area of the triangle MRN, figure 1, is to the area of A CB, figure 2, as M N times R S is to A B times A C.

(u.) The areas of similar polygons are to each other as the squares of their like dimensions.

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ILLUSTRATION.-If the two triangles A BC and E F G are similar, the area of A B C will be to the area of E F G as A B’ is to E F, or as B C is to FG?, or as A C2 is to E G’, or as A D® is to E H”. If the side A B equals 5 ft., and the corresponding side, EF, equals 4 ft., the area of the first triangle will be to that of the second as 25 is to 16,

the area of A B C will be is of the area of E F G.

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(v.) The circumference of a circle equals very nearly 3.1416 * times its diameter.

ILLUSTRATION.— The circumference of a circle 9 ft. in diameter is 3.1416 times 9 ft. The diameter of a circle 20 ft. in circumference is

1

3.1416 of 20 ft.

(w.) The circumferences of circles are to each other as their diameters or radii.

ILLUSTRATION.— The circumference of a circle 4 ft. in diameter is of the circumference of a circle 5 ft. in diameter.

(x.) The area of a circle equals half the product of its circumference by its radius, or quarter the product of its circumference by its diameter.

ILLUSTRATION.-If the circumference of a circle is 25.1328 ft., its diameter will be 8 ft. and its radius 4 ft. Its area then will equal of 8 times 25.1328 sq. ft., or } of 4 times 25.1328 sq. ft.

50.2656 sq. ft.

* More accurately, 3.141592653589; but the above is sufficiently exact for most purposes. Indeed, 34 is sometimes used where no great degree of accuracy is required.

(y.) The area of a circle also equals the square of its radius multiplied by 3.1416.

ILLUSTRATION. — If the radius of a circle is 4 ft., its area will equal 4' X 3.1416 sq. ft. = 16 X 3.1416 sq. ft. 50.2656

sq.

(z.) The areas of circles are to each other as the squares of their diameters or of their radii.

ILLUSTRATION. — The area of a circle 6 ft. in diameter is 4 times the area of a circle 3 ft. in diameter, because the square of 6 is 4 times the square of 3. The area of a circle of 5 ft. radius is of the area of a circle of 2 ft. radius, because the square of 5 is of the square of 2.

Note.—The student should draw figures to correspond with the problems. When only approximate values can be obtained, the work may be carried out to 4 decimal places. (aa.) Find the area of

1. A triangle whose base is 8 ft. and altitude is 6 ft. 2. A triangle whose base is 17 ft. and altitude is 11 ft. 3. A triangle of which the sides are 12 ft. 8 ft. and 7 ft. 4. A triangle of which the sides are 8 ft. 3 ft. and 10 ft. 5. An equilateral triangle 8 ft. on a side ? 6. An equilateral triangle 13 ft. on a side. 7. A parallelogram whose base is 18 ft. and altitude is 9 ft. 8. A parallelogram whose base is 163 ft. and altitude is 5f ft.

9. A trapezoid whose bases are 8 ft. and 7 ft. and altitude is 6 ft.

10. A trapezoid whose bases are 27 ft. and 26 ft., and altitude is 13 ft.

11. A rectangle 64 ft. long and 39 ft. wide.
12. A square 67 ft. on a side.
13. A circle 12 ft. in diameter.
14. A circle 100 ft. in diameter.
15. A circle of 8 ft. radius.
16. A circle of 100 ft. radius.
17. What is the side of a square containing 8649 sq. ft. ?
18. What is the side of a square containing 379 sq. ft. ?

19. What is the base of a triangle 100 ft. in area and 20 ft. in altitude ?

20. A rectangle containing 15138 sq. ft. is twice as long as it is wide. What is its length ?

SUGGESTION. — Divide it into squares.

21. A rectangle containing 41067 ft. is three times as long as it is wide. What is its length ?

28. A rectangle containing 384 sq. ft. is as wide as it is long. How long is it?

23. Find the side of a square equal in area to a triangle having a base of 50 ft. and an altitude of 72 ft. ?

24. What is the circumference of a circle 50 ft. in diameter ? 25. What is the diameter of a circle 34.5576 ft. in circumference ? 26. What is the diameter of a circle 50 ft. in circumference ? 27. What is the area of a circle 31.416 ft. in circumference ?

SUGGESTION. - First find the diameter or the radius.

28. What is the area of a circle 4084.08 ft. in circumference ? 29. What is the area of a circle 50 ft. in circumference ? 30. What is the area of a circle 72 ft. in circumference ? 81. What is the diameter of a circle containing 78.54 sq. ft. ? 32. What is the diameter of a circle containing 109956 sq. ft. ?

33. A parallelogram 20 ft. in altitude contains as much surface as a circle 20 ft. in diameter. What is its base ?

34. The base of a triangle is 60 ft., and it contains as much surface as a circle 75 ft. in circumference. What is its altitude ?

35. What is the side of a square containing 4 times as many sq. ft. as a circle 36 ft. in diameter ?

36. What is the radius of a circle containing 9 times as much surface as a circle 10 ft. in diameter ?

37. How will the area of a circle 6 inches in diameter compare with the area of a circle 2 inches in diameter ?

38. If an iron rod 1 inch in diameter and 10 ft. long weighs 25.5664 Ib., how much will an iron rod 2 inches in diameter and 5 ft. long weigh?

39. If a circular garden 100 ft. in diameter costs $100, what will a circular garden 300 ft. in diameter cost at the same rate ?

40. If a pipe 1 inch in diameter will empty a cask in 1 minute, how long will it take a pipe 1 inch in diameter to empty it?

41. I bought 5 lots of land, each containing 1 acre. The first was in form a circle ; the second was a square; the third was a rectangle twice as long as wide; the fourth was a triangle whose base was equal to its altitude; and the fifth was a trapezoid whose altitude was equal to į the sum of its parallel bases. It is required to find in feet the diameter of the circle, the side of the square, the sides of the rectangle, the base and altitude of the triangle, and the altitude of the trapezoid.

142. Properties of the Right-angled Triangle, with

Problems.

B

(a.) The side opposite the right angle of a right-angled triangle

is called the hypothenuse. The square of the hypothenuse equals the sum of the squares of the other two sides.

Note. — This is illustrated by the annexed figure, and can be rigidly demonstrated by geometry.

A B C represents a right-angled triangle, right angled at B. Then will A B + BC A C. If A B equals 4 ft. and B C equals 3 ft., A OS will equal 4' + 39 = 16 + 9 = 25, and A Cʻ will equal 16 + 9

✓ 25 5 ft. (b.) What is the missing side of a right-angled triangle whose1. Base is 8 ft. and altitude 6 ft. ?

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