3rd. Find the area of each triangle by the rule already given; their sum is the area of the whole figure.

EXAMPLE IX.-In an irregular four-sided plate of trapezium form, the distance, l

6 inches.

5 feet, h1

=

2 feet, and h2

=

I foot

Find the superficial area of the plate.

Area =

1 (h,+ h) = 5(2 + 1'5) = 8'75 square feet.

1

2

2

7. To Find the Area of any Rectilineal Figure or Polygon.

DEFINITION.—A rectilineal figure is a plane figure bounded by straight lines. The sides may be of any number, according to the shape of the figure.

RULE.--Divide the figure into convenient parts. Find the area of each part, and the sum of the parts will be the area of the whole figure. Note. In general, the parts into which the rectilineal figure can be most conveniently divided will be triangles; but in certain cases a square, a parallelogram, or a trapezoid may form one or more of the parts.

EXAMPLE X.-Referring to the first figure above, let it be divided as shown by the dotted lines, and let

6 feet, h1 the figure.

Area

I foot, h2

=

3 feet, and h

2 feet.

=2'5 + 6 × 2'5

=175 square feet.

EXAMPLE XI.-Referring to the last figure, take the dimensions and working_out given in Todhunter's Mensuration for Beginners, where ABCDEF is a six-sided figure: BK, CL, EM, and FN are perpendiculars on AD. The following lengths are in feet:

BK = 3, CL = 4, EM = 4'7, FN = 5'1.

Also AK = 3'4, KL=3°2, LD=4'1, AN=3'3, NM = 5'3.

It follows from these lengths that AD = 10°7, and that AM = 8'6; hence MD = 10'7 — 8'6 = 2.1.

LECTURE I.-QUESTIONS.

1. If 7 bolts are pitched at 5 inches apart in a straight line, what is the distance between the centres of the outside bolts? Ans. 2 feet 6 inches.

2. There are 9 bolts in a straight line; the distance between the centres of the outside bolts is 3 feet 6 inches. What must be the pitch? Ans. 5'25 inches.

3. The distance between the centres of the outside bolts is 5 feet 10 inches, the pitch is 6 inches. How many bolts must there be in a straight line? Ans. 12 bolts.

4. A sheet of lead is 10' 4" square. Find its weight at 7 lbs. per square foot. Ans. 747:44 lbs.

5. What is the number of square inches in a rectangle which measures 24 inches long by 18 inches broad? Ans. 432.

6. Find the area of a rectangle which is 3 feet wide and 15 feet 4 inches long? Ans. 46 square feet.

7. A fire-grate is 5' 6" long and 3' 1" wide. How many square feet are there in it? Ans. 16.958.

8. A boiler plate is 13′ 6′′ long and 2' 11" wide. 35 lbs. per square foot? Ans. 1378 125 lbs.

What will it weigh at

9. A rectangular tank is 10' long, 7' wide, and 7' high. What would be the cost of cementing the two sides and the top at 78. 6d. per square yard? Ans. £8 158.

10. The base of a triangular plate is 2' 6" and the height is also 2′ 6′′. What is the area in square feet and in square inches? Ans. 3'125 square feet, 450 square inches.

II. Find the area of a trapezoid, the parallel sides being 5 and 4 feet respectively, and the perpendicular distance between them being 2 feet. Ana. 9 square feet.

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LECTURE II.

CONTENTS.-To Find the Area of a Four-sided Figure, one of whose Sides is a Curve-Simpson's Rule -To Find the Circumference of a CircleThe Superficial Area of a Cylinder--The Area of a Circle-The Circumference of an Ellipse-The Superficial Area of an Elliptical Cylinder The Area of an Ellipse; with Examples.

1. To Find the Area of a Four-sided Figure, one of whose Sides is a Curve.

hn+

'n + 1

FIRST RULE.-Divide the length, l, of the figure into, n, equal parts by lines h1, h2 hn+1 drawn at right angles to the base line; add together the first and the last (h1+hn+1); call this sum, A; add together all the intermediate ones (h2+h+h, &c.), and call this sum, B.

Then:

L

n

Where is the interval, or pitch, or length of each of the parts into which the figure is divided; for example, let the length L= 20 inches, and the number of equal parts, n, 10.

=

This rule is simply an approximate method, based upon the fact that the whole figure is divided into a number of little trapezoids. The curved part of each trapezoid, being short, is assumed to be a straight line; then the sum of the areas of the several trapezoids will be equal to the area of the whole figure.

Of course, the greater the number of equal parts into which the length of the figure is divided, the nearer will the answer be to the true area of the figure.

EXAMPLE I.—-See the four-sided figure (p. 119) of Watt's "diagram of work," which is divided into ten equal parts.

L

n

Let the whole length of the figure, L, be 100 inches; then = 10 inches as the interval between each division, and let the numbers representing the perpendiculars, or ordinates, be also reckoned in inches.

Let A = the sum of the 1st and last perpendiculars or ordinates, = 100+25=125.

Let B

= the sum of the intermediate ordinates,

= 100+100+83.3+625+50+41·6+357 +3125+277533'23. .. 2B=533°23 × 2 = 1066'46.

Note.--The mean length of all the ordinates is evidently—

which is very near the arithmetical mean worked out at p. 119.