ART. 14. To find the Area of a Circle.

RULE. Multiply half the diameter by half the circumference and the product is the area.

If the diameter be given, find the circumference by Art. 12. If the circumference be given, find the diameter by Art. 13. EXAMP. A circle whose diameter is 12, and circumference is 37.7, given, to find the area?

113.10 area of the given circle.

Note. A circular ring is the figure contained between the peripheries of two concentric circles. Hence, the area of a circular ring must be the difference of the areas of the two circles.

ART. 15. The Diameter being given to find the Area of a Circle without finding the Circumference.

RULE. Multiply the square of the diameter by 7854,* and the product will be the area of the circle, whose diameter was given. EXAMP. The diameter of a circle being 12, to find the area?

⚫7854

12x12 144

31416

31416

7854

113 0976 area.

When the diameter is 1, the area is found to be 7854, and as the areas of

circles are as the squares of their diameters, the rule is evident.

By the Sliding Rule.

Set 1 on A to the diameter on B, then find 7854 (which expresses the area of a circle whose diameter is 1) on A, against which on B is a 4th number, then find this 4th number on A, against which on B is the area.

By Gunter.

The extent from 1 to the length of the diameter reaches from 7854 to a 4th number, and from that 4th number to the area.

ART, 16. The Circumference of a Circle being given, to find the Area without finding the Diameter.

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

EXAMP. The circumference of a circle being 37-7, to find the

ART. 17. The Dimensions of any of the parts of a Circle being given, to find the side of a Square equal to the Circle.

RULE. If the area of the circle be given, extract the square root of the area, which will be the side of a square equal to the circle : If the diameter or circumference be given, find the area by Art. 15 or 16, and then extract the square root, as before. And this is a general rule to find the side of a square equal to any superficial figure, regular or irregular: for the square root of the area of any figure whatever, is the side of a square equal to the given figure. But with regard to circles, if the diameter be given; multiply it by 886, and the product will be the side of an equal square or, as 13.545 is to 12, or 1354 to 1200: so is the diameter of a circle to the side of a square equal to the given circle. And, if the circumference be given, multiply it by 282 for the side of an equal square. Or, divide it by 3545, and the quotient will be the side of an equal square.

Let the diameter of a circle be The circumference being 37·7 12, to find the side of a square to find the side of an equal equal to the circle? square?

986×12=10.632=side of the

spuare.

Or, as 13.545: 12 :: 12: 10 631 =the side.

37.7X282=10-631-side

of

ART. 18. The Area of a Circle being given, to find the Diameter.

RULE. Multiply the given area by 1.2732, and the product will be the square of the diameter; then, extracting the square root of the product, you will have the diameter.*

EXAMP. The area of a circle being 113 09, to find the diameter.

ART. 19. The Area of a Circle being given, to find the circumference. RULE. Multiply the given area by 12.566, and extract the square root of the product, which root will be the circumference required. EXAMP. The area of a circle being 113-03 to find the circumfe

ART. 20. The Side of a Square being given, to find the Diameter of a circle equal to the Square, whose Side is given.

RULE. Multiply the given side by 1-128, and the product will be the diameter of a circle, whose area is equal to the area of the

* As the area of a circle, whose diameter is 1, is 7854, the area divided by 7854 must give the square of the diameter; but as 1-2732 is the reciprocal of 7854, the rule is evident.

given square. Or, if the side of the square be divided by 886, the quotient will be the diameter. Or, as 12 to 13.54, so is the side of any square to the diameter of an equal circle.

EXAMP. The side of a square being 10-635, to find the diameter of a circle equal to that square?

10.635X1·128=12 nearly. Or, 10·635-886=12=diameter. Or, as 12: 13-54 :: 10·635 : 12 nearly.

ART. 21. The Side af a Square being given, to find the circumference of a Circle equal to the given Square.

RULE. Multiply the given side by 3.545 and the product will be the circumference required. Or, divide it by 282, and the quotient will be the circumference.

EXAMP. The side of a square being 10.631, to find the circumference of a circle equal to that square.

10-631x3 545-37-686-circum. Or, 282) 10.631(37-698 circum.

ART. 22. To find the Area of a Semicircle, the Diameter being given. RULE. Find the area of the circle by Art. 15, and take the half of it.

In the same manner may the area of a quadrant, or a quarter of a circle, be found, by taking a fourth part of the area of the whole circle.

But with regard to measuring a sector, or a segment of a circle, it will be necessary first to show how to find the length of the arch line of a sector, and the diameter of the circle to a given segment.

ART. 23. A Segment of a Circle being given, to find the length of the Arch Line.

RULE. Divide the segment into two equal parts; then measure the chord of the half arch, from the double of which subtract the chord of the whole segment; and one third of that difference, being added to the double of the chord of the half arch, will give the length of the arch line.

EXAMP. In the segment ABCD, the whole chord ADC is 216, and the chord AB or BC 126, to find the arch line ABC.

ART. 24. The Chord and versed Sine of a Segment being given, to find the Diameter of a Circle.

RULE. Multiply half the chord by itself, and divide the product by the versed sine; then add the quotient to the versed sine, and the sum will be the diameter of the circle.

Definition. A sector is a part of a circle, contained between an arch line, and two radii or semidiameters of the circle.

RULE. Find the length of half the arch by Art. 23: Then multiply this by the radius or semidiameter, and the product will be

the area.

D