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3. What is the area of a circle whose diameter is 1, and circumference 3.1416 ?

Ans. .7854. 4. What is the area of a circle whose diameter is 7, and circumference 22?

Ans. 383. RULE II.* Multiply the square of the diameter by .7854, and the product will be the area ; or,

Multiply the square of the circumference by .07958, and the product will be the area.

* Demon. All circles are to each other as the squares of their diameters. (Euc. XII. 2.) But the area of a circle whose diameter is 1, is .7854. &c. (by

.7854, &c.x d2 Rule 1.) Therefore 1:d2 ::.7854, &c. :

1 785, &c.x d2=area of a circle whose diameter is d. Q. E. D. The following proportions are those of Metius and Archimedes.

As 452 : 355 :: square of the diameter : area.

As 14:11 :: square of the diameter : area. If the circumference be given, instead of the diameter, the area may be found as follows:

The square of the circumference x .07958=area.
As 88 : 7 :: square of the circumference : area.

As 1420 : 113 :: square of the circumference : area. And if d be the diameter, c the circumference, a the area, and p 3.14159, &c. then:

4a
1. d=-=-=2v-
P

р

4a
2. c=pd=-=2 pa

d
pd2c2
3. a=

4 4p 4 The following table will also show most of the useful problema relating to the circle and its equal or inscribed square.

1. diameter x .8862=side of an equal square.
2. circumf.X.2821=side of an equal square.
3. diameter x .7071=side of the inscribed square.
4. circumf.X.2251=side of the inscribed square.

a

с

dc

EXAMPLES.

1. What is the area of a circle whose diameter is 5 ?

7854

25=square of the diameter.

39270 15708

19.6350=the answer. 2. What is the area of a circle whose diameter is 7?

Ans. 38.4846. 3. What is the area of a circle whose diameter is 4.5 ?

Ans. 15.9043. 4. How many square yards are there in a circle whose diameter is 31 feet?

Ans. 1.0690. 5. How many square feet are there in a circle whose circumference is 10.9956 yards?

Ans. 86.5933. 6. How many square perches are there in a circle whose circumference is 7 miles?

Ans. 399300.6080.

PROBLEM XII.

To find the area of a sector, or that part of a circle which is bounded by any two radii and their included arc.

RULE.* Find the length of the arc by Problem X. then multiply the radius, or half the diameter, by the length of the arc of the sector, and half the product will be the area.

5. area X.6366=side of the inscribed square. 6. side of a square x 1.4142=diam. of its circums. circle. 7. side of a square x 4.443=circumf. of its circums.circle. 8. side of a square x 1.128=diameter of an equal circle. 9. side of a square x 3.545=circumf. of an equal circle.

* The rule for finding the area of the sector, is evidently the same as that for finding the area of the whole circle.

G

Note. If the diameter or radius is not given, add the square of half the chord of the arc, to the square of the versed sine of half the arc; this sum being divided by the versed sine, will give the diameter.

EXAMPLES.

1. The radius AB is 40, and the chord BC of the whole arc 50, required the area of the sector.

[blocks in formation]

80—V802-502

=8.7750=the versed sine of half the arc 2 80 x 60–8.7750 X 27=4563.0750=the reserved number. 2x V 8.7750 x 80=52.9906=twice the chord of haif the arc. 52.9906 X 8.7750 x 10

=1.0190 which added to twice the 4563.0750 chord of half the arc gives 54.0096 the length of the arc.

54.0096 x 40 And

=1080.1920=area of the sector re

2 quared.

2. Find the area of the sector, the chord of whose arc is 40, and the versed sine of half the arc 15.

Ans. 558.1250 3 Required the area of the sector, the chord of half the arc being 30, and the diameter of the circle 100.

Ans. 1523.4500

4. Given the diameter of the circle 50, and the versed sine 18, to find the area of the sector. Ans. 803.69875.

RULE II.*

As 360 is to the degrees in the arc of a sector, so is the area of the whole circle, whose radius is equal to that of the sector, to the area of the sector required.

Note. For a semicircle, a quadrant, &c. take one half, one quarter, &c, of the whole area.

EXAMPLES.

1. The radius of a sector of a circle is 20, and the de. grees in its arc 22; what is the area of the sector ?

Here the diameter is 40.

Hence, by Rule II. Prob. XII. the area of the circle= 402 X.7854=1600X.7854=1256.64.

Now, 360° : 22° :: 1256.64 : 76.7947=area of the sector.

* Demon. Let r=radius, d=number of degrees in the arc of the sector, and A=its area.

Then will 4r? X.7854=p2x 3.1416=area of the whole circle, and 2rx 3.1416=its circumference.

2dr x 3.1416 Also 360 : 2rx 3.1416::d:

length of 360

2dr x 3.1416 dr? x 3.1416 the arc of the sector. But

xixr 360

360 =A, by the last rule. And consequently 360 :d :: ? x 3.1416 : A.

Q. E. D.

2. Required the area of a sector whose radius is 25, and the length of its arc 147 degrees 29 minutes.

Ans. 804.3987. 3. Required the area of a semicircle whose radius is 13.

Ans. 265.4652 4. Required the area of a quadrant whose radius is 21.

Ans. 346.3614

PROBLEM XIII.

To find the area of a segment of a circle.

RULE I.*

I find the area of the sector, having the same arc with the sent, by the last problem.

2. Find the area of the triangle formed by the chord of the segnient, and the radii of the sector.

3. Then the sum, or difference, of these areas, according as the segment is greater or less than a semicircle, will be the area required.

Note.—The difference between the versed sine and radius, multiplied by half the chord of the arc, will give the area of the triangle.

EXAMPLES.

1. The radius OB is 10, and the chord AC 10; what is the area of the segment ABC?

* This rule is too evident to need a demonstration.

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