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EXAMPLES.

I. Find the area of a parabola whose base is 10 ft. 9 in., and axis 6 ft. 4 in. Ans. 45.4.8 sq. ft.

2. Find the area of a parabola whose base is 252 yards, and axis 242 yards. Ans. 8 a. I r. 24 p.

3. The base of a parabola is 16 feet, and the axis 24 feet; find the area of a portion cut off by a double ordinate whose abscissa is 6 feet. Ans. 32 sq. feet.

4. The base of a parabola is 20, and the axis 12; find the lengths of three equidistant ordinates to the axis. Ans. 5, 7071, 8·660.

EQUIDISTANT ORDINATES.
[WEDDLE'S METHOD.]

XX.

To find the area of a figure by means of seven equidistant ordinates.

To five times the sum of the even ordinates add the middle ordinate, and all the odd ordinates; multiply this sum by the common distance between the ordinates, and the product will be the area.

Ex. Find the content of an irregular area, by means of seven equidistant ordinates, whose lengths are 225, 231, 246, 273, 290, 306, 314 links; the common distance between them being 65 links.

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EXAMPLES.

I. In an irregular piece of ground the following ordinates were taken at intervals of 4 chains; namely, 2, 2, 4, 5, 7, 5, 3 chains; find the area. Ans. 10 a. I г. II p.

2. The following ordinates were taken at intervals of 5 yards, namely, 62, 6·8, 71, 74, 85, 72, 64 yards; find the area. Ans. 2139 sq. yds.

3. The diameter of a semicircle is 12; shew that the area calculated by means of seven equidistant ordinates differs from the true area by 1349.

4. The diameter of a semicircle is 36, which is divided into 12 equal parts; find the area contained between the second and the eighth ordinates. Ans. 304-6766.

5. The transverse and conjugate diameters of an ellipse are 40 and 32; the transverse diameter is divided into IO equal parts; find the area contained between the first and the seventh ordinates. Ans. 349'6468.

6. Find the content of an irregular area from the seven equidistant ordinates o, 40, 55, 78, 84, 33, o links; the distance between them being 25 links. Ans. 11P.

7. Find the content of an irregular area from the seven equidistant ordinates 21, 2, 3, 3, 41, 41, 5% chains; the distance between them being I chain. Ans. 2 a. or. 26 p.

8. The versed sine of a circular arc is 18, and the chord of the whole arc is 96; find the area.

Ordinates: 0, 2017123, 28.9827, 35'1141, 400998, 44°3283, 48. Area, 1175 3461.

9. The axis of a parabola is 30, and the base 24; shew that the area found by seven equidistant ordinates from the axis differs from the true area by 3'2857; but if calculated by seven equidistant ordinates from the base, we get the area exactly.

IO. The diameter of a semicircle is 30, and an ordinate divides it into two parts, 12 and 18. Calculate the areas of the two parts into which the semicircle is divided; and shew that their sum differs from the true area by 3·1063.

First ordinates: o, 7'4833, 10'1980, 12, 13.2665, 141421, 146969. Second ordinates: 0, 9, 12, 13'7477, 146969, 15, 14·6969.

First area, 130 9730; second area, 219°4920; sum, 350 4650; area of semicircle, 353'5714; difference, 3*1063.

REGULAR FIGURES.

XXI.

To find the area of a regular figure.

Multiply the square of the side by the corresponding number in the following table :

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Ex. Find the area of a regular decagon whose side is 7 ft. 3 in.

7:25

7:25

3625

1450

5075

52'5625

7.694

2102500

4730625

3153750

3679375

404*4158750

EXAMPLES.

I.

Find the area of a regular pentagon whose side is
Ans. 1230 76 sq. yds.

263 yards.

2.

Find the cost of boarding an heptagonal floor whose side is 18 feet, at is. 6d. per square yard.

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I. Find the value of 34. 7. 8 square feet of carpet at 35. 7 d. per foot. Ans. £6. 5s. 612d.

2. Find the side of a square the area of which is equal to a rectangle whose sides are 21 ft. 11 in. and 197 ft. 3 in. Ans. 65 ft. 9 in.

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