Then, as 360:

18

Or, as 20: 1 :: 70686: 35343 answer.

3. What is the area of the sector, whose radius is 10, and arc 20? 1

Ans. 100.

4. What is the area of the sector, whose radius is 9, and the chord of its arc 6?

Ans. 27.52678.

5. Required the area of a sector, whose radius is 25, its arc containing 147 degrees 29 minutes.

Ans. 804°4017.

6. To find the area of a quadrant and a semicircle, to the radius 13.

Ans. 1327326 and 265°4652.

PROBLEM XIII.

To find the area of a segment of a circle.

RULE I.

1. Find the area of the sector, having the same arc with the segment, by the last Problem.

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

3. Then the sum of these two will be the answer, when the segment is greater than a semicircle; but the difference will be the answer, when it is less than a semicircle.

EXAMPLE.

EXAMPLE.

Required the area of the segment ACBA, its chord AB being 12, and the radius EA or CE 10.

RULE 2.

1. To the chord of the whole arc add

of half the arc, or add the latter chord and

of the chord

of it more.

2. Multiply the sum by the versed sine or height of the of the product will be the area of the

segment, and

ย

its fluent, or the value of the half segment,

5d

ข

28d2

28d2

&c. Consequently its double 2v/dvx : }

&c. is the value of the whole segment..........

Now suppose the segment 2CD x m x BD+n× BC =

2v X m√ dvv2 + n√ dv = 2v √ dv × m√ 1.

7 +

Then let the coefficients of the corresponding terms be equated,

and we shall have m+n, and

172

=; whence m}, and n=}

Then, by substituting these values of m and n in the assumed. quantity, we shall have 2CD x mx BD+n × BC = 2CD

xBD+BC=

CD x 2BD+4BC, which is the rule.

EXAMPLE.

Take the same example, in which the radius is 10, and the chord AB 12.

Then, as before, åre found CD2, and the chord of the half arc AC.

1. Divide the height of the segment by the diameter, and find the quotient in the column of heights or versed sines, in the table of the areas of the segments of a circle.

2. Take out the corresponding area in the next column on the right hand, and multiply it by the square of the diameter, for the answer.

EXAMPLE.

*The table, to which this rule refers, is formed of the areas of the segments of a circle, whose diameter is ; and which is supposed to be divided by perpendicular chords into 1000 equal parts, and is at the end of MENSURATION.

The reason of the rule itself depends upon this property→→→ That the versed sines of similar segments are as the diameters of the circles, to which they belong, and the areas of those segments as the squares of the diameters; which may be thus proved :

Let

EXAMPLE.

The example being the same as before, we have CD equal to 2, and the diameter 20.

Then 20)2(1

And to I answers '040875

Square of diam.

400

Ans. 16.3500.

OTHER

Let ADBA and adba be any two similar segments, cut off from the similar sectors ADBCA and adbCa by the chords AB and ab; and let fali the perpendicular CD.

Then, by similar triangles, DB: db :: BC ; ¿C, and DB : db :: Dm : dn; whence, by equality, BC: C :: Dm : da, or 2BC: 2bC :: Dm : dn.

Again, since similar segments are as the squares of their chords, it will be AB : ab2 :: ADBA: adba; but AB : ab :: CB: C¿1, and therefore, by equality, ADBA: adba :: CB1 : C', or ADBA adba :: 4CB 4C6. Q. E. D.

Now, if d be put equal to any diameter, and v the versed sine, it will be d: 0 :: 1 (diameter in the table) := versed sine

ข

of a similar segment in the table, whose area let be called 4.

Then 1: d::a: ad2 = area of the segment, whose height is v, and diameter, as in the rule.