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2. The greater chord is 96, the lesser 60, and the breadth 26; what is the area of the zone ?
Ans. 2136.7500. 3. One end of a circular zone is 48, the other end is 30, and the breadth is 13; what is the area of the zone?
PROBLEM XV. To find the area of a circular ring, or the space include between the circumference of two concentric circles.
RULE.* The difference between the areas of the two circles will be the area of the ring.
Or, multiply the sum of diameters by their difference, and this product again by .7854, and it will give the area required.
1. The diameters AB and CD are 20 and 15; required the area of the circular ring, or the space included between the circumferences of those circles.
* Demon. The area of the circle AEBA=AB? X.7854, and the area of the small circle cd is=CD? X.7854; there. fore the area of the ring=ABP X.7854—CD2 X.7854= AB+CDX AR-CD X.7854. Q. E. D.
Coroll. If cE be a perpendicular at the point c, the area of the ring will be equal to that of a circle whose radius is CE.
Rule 2. Multiply half the sum of the circumferences by half th difference of the diameters, and the product will be the area.
This rule will also serve for any part of the ring, using half tn": sum of the intercepted arcs for half the sum of the circumferences.
Here AB+CDX AB-CD=35*5=175; and 175 x 7854=137.4450=area of the ring required.
2. The diameters of two concentric circles are 16 and 10; what is the area of the ring formed by those circles?
Ans. 122.5224. 3. The two diameters are 21.75 and 9.5, required the area of the circular ring.
Ans. 300.6609. 4. Required the area of the ring, the diameters of whose bounding circles are 6 and 4.
To find the areas of lunes, or the spaces between the inter
secting arcs of two eccentric circles.
Find the areas of the two segments from which the lune 18 formed, and their difference will be the area required.
* Whoever wishes to be acquainted with the properties of lunes, and the various theorems arising from them, may consult Mr. Wiston's Commentary on Tacquet's Euclid, where they will find this subject very ingeniously managed. The following property is one of the most curious :
If ABC be a right angled triangle; and semicircles be described on the three sides as diameters, then will the said triangle be equal to the two lunes D and F taken together.
B For the semicircles described on AC and BC=the one described on AB (31.66,) from each take the segments cut off by AC and BC, then will the lunes AFCE and BDCG=the triangle ACB. Q. E D.
The length of the chord AB is 40, the height DC 10 and DE 4; required the area of the lune ACBEA.
Ꭰ . By note, page 74, the diameter of the circle of which
202 +102 ACB is a part=
10 And the diameter of the circle of which AEB is a parl 202 +42
=104. 4 Now having the diameter and versed sines, we find by Prob. XIII. Rule III.
The area of seg. ACB=.111823 x 502=279.5575
Their difference is the area of the
=171.8842 lune AEBCA, required, 2. The chord is 20, and the heights of the segment 10 and 2 ; required the area of the lune. Ans. 130.287.
3. The length of the chord is 48, and the heights of the segments 18 and 7; required the area of the lune.
To find the area of an irregular polygon, or a figure of
any number of sides.
1. Divide the figure into triangles and trapeziums, and find the area of each separately.
2. Add these areas together, and the sum will be equal to the area of the whole polygon.
1. Required the area of the irregular figure ABCDEFGA, the following lines being given.
GB=30.5 An=11.2, CO=6
* When any part of the figure is bounded by a curve, the area may be found as follows:
Rule 1. Erect any number of perpendiculars upon the base, at equal distances, and find their lengths.
2 Add the lengths of the perpendiculars, thus found, together, and Che sum divided by their number will give the mean breadth.
3. Multiply the mean breadth by the length of the base, and it will give the area of that part of the figure required.
x 30.5 = 8.6 X 30.5 = 2
x 29= 8.8 x 29=255 2 2
2 =area of the trapezium GCDF.
FDx Ep_24.8 x 4_99.2 Also
=49.6=aren of the tri 2 2
2 angle FDE.
Whence 262.3+255.2+49.63567.1=area of the whole figure required.
2. *In a pentangular field, beginning with the south side and measuring round towards the east, the first, or south side, is 2735 links, the second 3115, the third 2370, the fourth 2925, and the fifth 2220; also the diagonal from the first angle to the third is 3800 links, and that from the third to the fifth 4010; required the area of the field.
Ans. 117 ac. 2 ro. 39 po.
Promiscuous Questions concerning Lines and Areas.
1. Given AC=32, AD=3, EC=8, the perpendicular DG=4, and the perpendicular EF=6, to find the area of the triangle ABC and the sides AB and BC.
To find the area of mixed or compound figures, or such as are com. posed of rectilineal and curvilineal figures together; the rule is to find the area of the several figures of which the whole figure is composed, then add all the areas together, and the sum will be the area of the whole compound figure. And in the same manner may any irregular field or piece of land be measured, by dividing it into irapeziums and triangles, and finding the area of each separately.
* Note. As this figure consists of three triangles, all of whose sides are given, by calculating their areas according to Problem III. the su n will be the area of the whole figure accurately, without drawing pe. pendiculars from the angles to the diagonals.
The same thing may also be done in most other cases of this kind,