8.0

40

Here =.2, the quotient or tabular height; and the corresponding Area Seg. is .111823; hence, .111823 × 402.111823 x 1600 178.9168, the area of the segment required.

2. What is the area of a segment whose height is 9, and the diameter of the circle 25?

Ans. 159.09375. 3. Find the area of a segment whose height is 25, and the diameter of the circle 55.

Ans. 1050.60065. 4. What is the area of a segment, greater than a semicircle, whose height is 66 feet, and the chord of the whole are 60 feet 10 inches ?

Ans. 4435 ft. 8 in. 11 pa. 5. The base of a stone column is the greater segment of a circle whose chord measures 2 feet, and versed sine 1 foot 4 inches; what is its area?

PROBLEM XVIII.

Ans. 2.304 feet.

To find the area of a circular zone, or the space included between any two parallel chords and their intercepted arcs.

RULE.

Find the area of that part of the zone forming the trapezoid ABCD, to which add twice the area of the segment AED; and the sum will be the area of the zone required. (See the next figure.)

Note 1. When great accuracy is required, the area of the segment should be found by Rule 3, Problem XVII.

2. The chord AD and versed sine Em may be found from the parallel sides AB, DC, and the perpendicular distance DF, by the help of Theorem 12, Part I.; much calculation may, however, be saved, by measuring the chord and versed sine in taking the dimensions of the zone. (See the Key to the Mensuration, page 22.)

9. When the parallel sides and their perpendicular distance are given, the zone may be constructed in the following manner: draw the side AB, make AF equal to half the difference between AB and DC, and erect the perpendicular FD. From the point D, draw DC parallel to AB, and join AD. Bisect AB and AD with the perpendiculars mo, no; and o will be the centre of the circle of which the zone is a part: thus you will determine whether the centre of the circle fails within or without the zone.

EXAMPLES.

1. The greater side AB measures 40 feet, the less DC 30 feet, the perpendicular FD 35 feet, the chord AD 35 feet 3 inches, and the versed sine Em 7 feet 3 inches; what is the area of the zone ABCD?

By Problem VIII. we have 40 + 30 × 35 = 70

× 35 = 2450; and

2450
2

=1225, the area of the

trapezoid ABCD.

Also, by Problem XVII. Rule 2, we have 35.25 × 7.25

x+

= 255.5625 ×+

381.078125

70.5

7.253 35.25 × 2 ≈ 170.375 + 5.40536 = 175.78036, the area of the segment AED; hence, 1225 + 175.78036 x 2 = 1225351.56072 = 1576.56072 feet 1576 ft. 6in. 8 pa. the area of the zone ABCD, as required.

2. The greater side is 120 feet, the less 75 feet, the chord 39 feet 6 inches, and the versed sine 3 feet 3 inches; what is the area of the zone?

Ans. 3337 ft. 4 in. 10 pa. 3. What is the area of a zone whose greater side measures 72 feet, the less 45 feet, and its breadth 19 feet 6 inches? Ans. 1201 ft. 11 in. 1 pa. 4. The greater side is 80, the less 60, and their distance 70; what is the area of the zone?

Ans. 6326.96.

PROBLEM XIX.

To find the area of a circular ring, or the space included between the circumferences of two concentric circles.

RULE.

Multiply the sum of the diameters by their difference, and this product by .7854; and it will give the area required. Or, the difference of the areas of the two circles will be the area of the ring.

Note 1. The area of a circular ring may also be found by multiplying half the sum of the circumferences by half the difference of the diameters.

2. The area of part of a ring, or the segment of a sector, may be found by multiplying half the sum of the bounding arcs by the nearest distance between them.

EXAMPLES.

1. The diameters AB and CD are 30 and 20; what is the area of the circular ring?

Here 30+ 20 × 30 20 X 7854 = 50 × 10 × .7854 = 500 × .7854 = 392.7, the area of the ring required.

2. The diameters of two concentric circles are 35 and 23; what is the area of the ring formed by the circumferences of those circles? Ans. 546.6384. 3. The inner diameter of a circular building is 73 feet 3 inches, and the thickness of the wall 1 foot 9 inches; how many square feet of ground does the wall occupy? Ans. 412.335 feet.

4. What was the expense of making a moat round a circular island, at 2s. 6d. per square yard; the diame ter of the island being 525 feet, and the breadth of the moat 21 feet 6 inches? Ans. £512 13s. 7 d.

5. What is the area of the front of a circular arch, built with stones, each 3 feet 6 inches long; the length

of the upper bounding arc being 35 feet 3 inches, and the length of the lower 24 feet 9 inches? Ans. 105 feet.

PROBLEM XX.

To find the area of a lune, or the space included between the intersecting arcs of two eccentric circles.

RULE.

Find the areas of the two segments forming the lune, and their difference will be the area required.

Note 1. If ABCD be a square, and circles be described from the points B and C, with the radii BD D and CD; the area of the lune DGEFD will be equal to the area of the square ABCD,

2. If ABC be a right-angled triangle, and semicircles be described on the three sides, as diameters; then will the area of the said triangle be equal to the sum of the areas of the two lunes D and E.

G

F

E

D

Several other curious properties of lunes may be seen in Dr. Hutton's Recreations, and Mathematical Dictionary,

EXAMPLES.

1. The length of the chord AB is 30, the height DC 12, and DE 5; what is the area of the lune ACBEA?

By Rule 2, Problem XVII. we have 30 × 12 ×

123

30 X 2

= 360 × +

1728
60

=240+28.8 =

+
268.8, the area of the segment ACB.

1002.08333, = 102.08333, the area of the

ment AEB; hence, 268.8 102.08333 the area of the lune required.

seg

166.71667,

2. The chord is 24, and the heights of the segments 9 and 4: what is the area of the lune? Ans. 93.8542.

3. The length of the chord is 48, and the versed sines of the segments 18 and 8; what is the area of the lune? Ans. 374.775.

Note. The answer to this question was obtained by the assistance of Rule 2, Problem XIV. and Rule 3, Problem XVII.

PROBLEM XXI.

To find the area of an ellipse.

RULE.

Multiply continually together the two diameters and the number .7854, and the product will be the area of the ellipse.

Note 1. The area of an elliptical ring, or the space included between the circumferences of two concentric similar or dissimilar ellipses may be found thus from the product of the two diameters of the greater ellipse subtract the product of the two diameters of the less; multiply the remainder by .7854, and the product will be the area of the ring. Or, subtract the area of the less ellipse from that of the greater, and the remainder will be the area of the ring.

2. If the sum of the two diameters of an ellipse he multiplied by 1.5708, (half of 3.1416,) the product will be the circumference, exact enough for most practical purposes.

3. To half the sum of the two diameters add the square root of half the sum of their squares; multiply the last sum by 1.570s, and the product will be the circumference, extremely near."

4. The ellipse is equal to a circle whose diameter is a mean proportional between the two axes; hence we obtain the above Rule.

EXAMPLES.

1. What is the area of the ellipse ABCD, whose transverse diameter AB is 34, and conjugate CD 25?

C