Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Boot, that shall be the side of the square which may be inscribed in that circle. So if the diameter of a circle be 1, then the half is,5 which squared is,25 and this doubled is 5, whose square root is ,7071 the side of the square inscribed. Let it be required to find the area of a circle, whose diameter is an unit. By the proportion of Van Culen, if the diameter be 1, the circumference will be 3,1415926, &c. Then multiply half the circumference by half the diameter, and the product is the area.

EXAMPLES.

Problem 1. Having the diameter and circumference to find the area.

RULE. Every circle is equal to a parallelogram, whose length is equal to half the circumference, and the breadth equal to half the diameter; therefore multiply half the circumference by half the diameter, and the product is the area of the circle.

1. Let the diameter of a circle be 22,6 and the circumference 71, what is the area of said circle? Ans. 401,15.

Problem 2. Having the diameter of a circle to find the circumference.

2. Let the diameter of a circle be 22,6 feet, required the circumference thereof? Ans. 71+ feet. Problem 3. Having the circumference of a circle to find the diameter.

3. Let the circumference of a circle be 71 feet, required the diameter ? Ans. 22,6 feet. Problem 4. Having the diameter of a circle to find the All circles are in proportion, one to another, as are the squares of their diameters, (by Euclid, 12, 2.)

area.

In the Key I will give Van Culen's proportion, and also the proportions of Metius and Archimedes.

4. Let the diameter of a circle be 22,6 feet, required the area ? According to Van Culen 401,15 feet.

Metius

401,15 feet. Archimedes 401,31 feet.

Ans.

Problem 5. Having the circumference of a circle to find the area.

Demonstration by Van Culen.-Because the diameters of circles are proportional to their circumferences, that is, as the diameter of one circle is to its circumference, so is the diameter of another circle to its oircumference; therefore the areas of circles are to one another as, the squares of their cirtumferences.

5. Let the circumference of a circle be 71 feet, required the area? Ans. 401,15 feet.

Problem 6. By having the diameter to find the side of a square equal thereunto, that is, equal in area to that circle.

6. Let the diameter of a circle be 22,6 feet, required the side of a square equal in area to the circle, of which the diameter is here given? Ans. 20,02872 side.

Problem 7. By having the circumference to find the side of the square equal.

7. Let the circumference of a circle be 71 feet, required the side of a square which will be equal in area?

Ans. 20,0291 feet.

Problem 8. Having the diameter to find the side of a square which may be inscribed in that circle.

8. Let the diameter of a circle be 22,6 feet, required the side of a square equal, when inscribed in the circle, of which the diameter is here given? Ans. 15,98046 feet.

Problem 9. Having the circumference to find the side of a square which may be inscribed.

9. Let the circumference of a circle be 71 feet, required the side of an inscribed square equal?

Ans. 15,9821 feet.

Problem 10. Having the area to find the diameter. 10. Let the area of a circle be 401,15 feet, required the diameter of that circle? Ans. 22,599 diameter.

Problem 11. Having the area to find the circumference. 11. Let the area of a circle be 401,15 feet, required the circumference of that circle? Ans. 70,999+ feet.

Problem 12. Having the area to find the side of a square inscribed.

12. Let the area of a circle be 401,15 feet, required the side of a square inscribed, which will be equal in area? Ans. 15,98 feet.

Problem 13. Having the side of a square to find the diameter of the circumscribing circle.

13. Let the side of a square be 15,98 feet, required the diameter of a circle which shall circumscribe that given square? Ans. 22,598916 feet.

Problem 14. Having the side of a square to find the diameter of a circle equal.

14. Let the side of a square be 20,0291 feet, required the diameter of a circle equal in area to that square? Ans. 22,5928248 feet.

Problem 15. Having the side of a square to find the circumference of a circumscribing figure.

15. Let the side of a square be 15,98 feet, required the circumference of a circle that will encompass that square. Ans. 70,999+ cir.

Problem 16. Having the side of a square to find the circumference of a circle that will be equal thereunto.

16. Let the side of a square be 20,0291 feet, required the circumference that will be equal in area to that given square? Ans. 71,0051595 feet cir.

In several of the foregoing problems, where the diameter and circumference are required, the answers are not exactly the same as the diameter and circumference of the given circle, but are sometimes too much, and sometimes too little, as in the two last problems, where the answers in each should be 71, the one being too much, and the other too little. The reason of this is, the small defect that happens to be in the decimal fraction, they being sometimes too great, and sometimes too little; yet the defect is so small, that it is needless to calculate them to more exactness.

10th. OF A SEMICIRCLE.

To find the area of a semicircle, this is the

RULE. Multiply the fourth part of the circumference of the whole circle, (that is, half the arch line,) by the semidiameter, and the product is the area.

NOTE. The semidiameter, or radius, is a straight line, drawn off the centre of any circle to the circumference of the same.

Let the diameter of a semicircle be 22,6 feet, and the circumference 35,5 feet; required the area of the semicircle? Ans. 200,575 feet.

If only the diameter of the semicircle be given, you may say, by the Rule of Three, as 1 is to,3927 so is the square of the diameter to the area.

11th. OF A QUADRANT.

To find the area of a quadrant, or fourth part of a circle, this is the

RULE. Multiply half the arch line of the quadrant, (that is, the eighth part of the circumference of the whole circle,) by the semidiameter, and the product is the area of the quadrant.

Required the area of a quadrant, whose semidiameter is 11,3 and the arch line 17,75 feet? Ans. 100,2875 area.

These are the rules and ways commonly given for finding the area of a semicircle and quadrant, but I think it is as good a way to find the area of the whole circle, and then take half that area for the semicircle, and a fourth part for the quadrant. Before I proceed to shew how to find the area of the sector, and segment of a circle, I shall shew how to find the length of the arch line arithmetically.

To find the length of the arch line arithmetically.

RULE. Multiply the chord of half the segment by 8, and from the product subtract the chord of the whole segment, and divide the remainder by 3, and the quotient is the arch line sought.

[blocks in formation]

How to find the diameter of a circle by having the chord and versed sine of the segment arithmetically.

Let a segment be given, whose chord is 36, and the versed sine 6; required the diameter of that circle?

[blocks in formation]

12th. OF THE SECTOR OF A CIRCLE.

A sector of a circle is comprehended under two radii, or semidiameters, which are supposed not to make one right

line, and a part of the circumference, whence a sector may be either less or greater than a semicircle. To find the area, or superficial content thereof, this is the

RULE. Multiply half the arch line by the semidiameter, and the product is the area.

Let the sector of a circle be given, whose semidiameter is 24,5 feet, and the arch line 45,6 feet; required the area of the sector? Ans. 558,6 feet,

15th. OF THE SEGMENT OF A CIRCLE.

A segment of a circle is a part terminating by a right line, less than the diameter, called a chord, and by a part of the circumference. To find the area of the segment of a circle, this is the

RULE. Find the semidiameter of the circle by what has been shewn, and the arch line; then multiply half the arch line by the semidiameter, so you have the area of the sector. Then subtract the versed sine from the semidiameter, the remainder is the perpendicular of the triangle, and multiply the half chord by the perpendicular, and the product is the area of the triangle. Then subtract the area of the triangle from the area of the sector, and the remainder is the area of the segment.

Let it be required to find the area of the segment of a circle, whose chord line is 35, and the versed sine 9,6 feet. Ans. 236,060 the area.

14th. OF COMPOUND FIGURES.

Mixed, or compound figures, are such as are composed of rectilineal and curvilineal figures together.

To find the area of such figures, you must find the area of the several figures, of which the whole compound figure is composed, and add all the areas together, and the sum will be the area of the whole compound figure.

NOTE. As those compound figures are best understood by having the figure of such before the learner, and as there are none in this treatise, I will not say any thing more concerning them, but this, that the area of such figures is found in the same manner as a trapezium.

15th. OF AN ELLIPSIS.

An ellipsis, or oval, is a figure bounded by a regular curve line returning into itself, but of its two diameters, cutting each other in the centre; one is longer than the other, in which it differs from the circle. To find the area thereof,

« ΠροηγούμενηΣυνέχεια »