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Examples. 1. What is the area of a circle whose circumference is 44 rods?

44x44x.079585154.06688, Ans.

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The area of a circle given to find the diameter. Rule:--Divide the area by 7854, and extract the square root of the quotient; the root is the diame ter sought. (See application of square root, case 2.)

PROBLEM VII.

The area given to find the circumference. RULE.Divide the area by 07958 and extract the square root of the quotient, and the root will be the circumference sought.

CASE XI.

To measure a sector of a circle. DEFINITION.—The sector of a circle is a part of a circle bounded by an arc and two radii drawn to the extremities, see the figure A C B D. A C B the arc ; À D, or D B che radii.

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RULE.- Find the length of the arc A CB; to do which say as 180°, is to the number of degrees in the arc, so is the radius multiplied by 3•1416, to the length of the arc : having found the length of the arc, multiply the radius by half the arc, the product is area required.

Examples. 1. What is the superficial content of the sector A CB D, the radius A D, or D B being 10 feet, and the arc 4 C B containing 135°?

ft. As 180: 135 :: 10x3.1416:23.562 length of the arc.

ft. then 10x234562-117.81 Ans. RULE--Find the superficial content of a circle having the same radiús; then say, as 360° (the degrees in a circle) are to the area of the whole circle; so are the number of degrees in the arc of the sector, to the superficial content of the same.

Examples 1. át is area of a sector of a circle whose arc contains 45 degrees, and the radius of the same is 20 feet?

Ans. 157 sq. ft. Note. This method is exactly right, for it is evident that if a circle of 40 feet diameter contain 1256 sq. ft. that a sector of 45° with the same radius will be of the circle.

CASE XII. To measure the segment of a circle. A segment of a circle is a part of a circle bounded by part of the circle's periphery, and a chord connecting the two extremities of the periphery, ABC is a segment

See the figure.

B

NOTE.-The quantity of the angle ADC is 820, instead of 28°

RULE.--Measure the sector ABCD by case 10, and also measure the triangle A DC, and subtract the area of the triangle from the area of the sector; and the remainder will be the area of the segment A B C.

Examples 1. What is the area of the segment A B C, whose arch contains 82° : and its chord AC 17°5 feet : and the perpendicular of its triangle D E 10•4 ft : and semidiameter A D, or D C 137 feet?

First, find the area of a circle whose diameter is 27-4 feet.

As 113 : 355 :: 27.4 : 86f circumference. 27•4X862356:4+1=589.1 area of circle.

To find the area of the sector, say

area. As 360° : 589:1 : : 82° : to 134.1 the area of the 'sector A B C D.

To find the area of the triangle. The cord A Č=17-5, perpendicular D E 10-4X 17.5 182+1=91 feet area of the triangle AD C.

area of the sector A B C D=134-1
Area of the triangle A DCS 910

Area of the segment A B C= 431 ft. Ans. ) 기

CASE XIII.

To find the area of a polygon. DEFINITION.-A polygon is a figure having equal sides and equal angles, see gures 1. 2. 3.

RULE.-Multiply half the length of all the sides by the nearest distance from the centre to

any

of the sides ; or multiply, the area of one of the triangles by the number of triangles contained in the figure, the product is the answer.

Figures.

B

In measuring figure first, it is evident that mul. tiplying the distance from the centre to any one side, by half the length of all the sides is measur. ing 5 equal triangles; the sides represent the bases, and the distance from the centre, to a side the perpendicular.

Examples

1. What is the area of figure first A C equal to 8 ft. ; B D equal to 5.8 ft.?

8 X5.854604;} 23:2 area of B A C. there being 5 triangles contained in this figure the area of one viz. 23.2 multiplied by 5 will give the area of the polygon. 23:2X5116.0 Ans.

-2. What is the area of figure second, the distance from the centre to a side is 6 ft. ; the sides are 6, and their length 7 ft. ? Ans. 126 ft.

3. What is the area of figure third, the distance from the centre to a side 6.2 ft. ; number of sides 7 ; their length 5.9 ft. ?

Ans. 128.03 ft.

CASE XIV.

To describe and find the area of an ellipsis, or oval.

First, to describe an oval or elipsis. Rule-Draw a line and set one foot of the dividers on said line as a centre, and describe a çira cle, and move the dividers to some other point on the given line (less than the semidiameter) and describe another circle of the same radius, and in the two points where the circle's peripheries intersect, set the dividers and complete the sides of the oval, and through these two points, draw the con jugate diameter, crossing the transverse diameter in the centre of the oval,

NOTE. -The longest diameter of an oval is called the transverse, and tlie shortest, the conjugate diameter.

Figures. 1st.

2d.

To find the area of an ellipsis. Rule.-Multiply the transverse by the congugate diameter, and this again by .7854, and the last product is the area; or multiply the two diameters together and 11 of the product is the answer.

Eramples. 1. What is the area of figure first, the longest diameter is 17.5 and shortest 13. ?

17:5X13X.7854-178.67850 area Ans. 2. What is the area of figure 2d, the longest diameter being 21 and the shortest 17 ?

By rule first. 21 X 17X78543280•3878'area. By rule second, 21x17=357-11-2801 Ans.

CASE XV. To find the area of a globe or sphere. DÉFINITION.A globe or sphere is bounded by a circumference every way equally distant from a point within called the centre, thus a cannon ball may be called a globe or sphere.

RULE.Multiply the diameter and circumference together, the product is the area.

Examples. 1. What is the superficial area of a globe whose circumference is 44, diameter 14?

44X14=616 area Ans.

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