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its diameter into 0.7854, which is the area of a circle whose diameter is 1.

Ex. 1. What is the area of a circle whose diameter is 623 feet? Ans. 304836 square feet. 2. How many acres are there in a circular island whose diameter is 124 rods? Ans. 75 acres, and 76 rods.

3. If the diameter of a circle be 113, and the circumference 355, what is the area? Ans. 10029.

4. How many square yards are there in a circle whose diameter is 7 feet?

31. If the circumference of a circle be given, the area may be obtained, by first finding the diameter; or, without finding the diameter, by multiplying the square of the circumference by .07958.

14

=

For, if the circumference of a circle be 1, the diameter 1+3.14159-0.31831; and the product of this into the circumference is .07958 the area. But the areas of different circles, being as the squares of their diameters, are also as the squares of their circumferences. (Sup. Euc. 8, 1,)

Ex. 1. If the circumference of a circle be 136 feet, what is the area? Ans. 1472 feet.

2. What is the surface of a circular fish-pond, which is 10 rods in circumference?

32. If the area of a circle be given, the diameter may be found, by dividing the area by .7854, and extracting the square root of the quotient.

This is reversing the rule in art. 30.

Ex. 1. What is the diameter of a circle whose area is 380,1336 feet?

Ans. 380,1336.7854-484. And v 484-22.

2. What is the diameter of a circle whose area is 19.635?

33. The area of a circle, is to the area of the circumscribed square; as .7854 to 1; and to that of the inscribed square as .7854 to 2.

Let ABDF (Fig. 10.) be the inscribed square and LMNO, the circumscribed square, of the circle ABDF. The area of the circle is equal to AD2x.7854. (Art. 30.) But the area of the circumscribed square (Art. 4.) is equal to ON=AD?,

And the smaller square is half of the larger one. For the latter contains 8 equal triangles, of which the former contains only 4.

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Ex. What is the area of a square inscribed in a circle whose area is 159? Ans. .7854 :: 159: 101.22.

PROBLEM V.

To find the area of a SECTOR of a circle.

34. MULTIPLY THE RADIUS INTO HALF THE LENGTH OF THE ARC.

Or,

As 360, TO THE NUMBER OF DEGREES IN THE ARC;

SO IS THE AREA OF THE CIRCLE, TO THE AREA OF THE

SECTOR.

It is evident, that the area of the sector has the same ratio to the area of the circle, which the length of the arc has to the length of the whole circumference; or which the number of degrees in the arc has to the number of degrees in the circumference.

Ex. 1. If the arc AOB (Fig. 9.) be 120°, and the diameter of the circle 226; what is the area of the sector AOBC?

The area of the whole circle is 40115. (Art. 30.) And 360° : 120°:: 40115 : 13371%, the area of the sector. 2. What is the area of a quadrant whose radius is 621? 3. What is the area of a semi-circle whose diameter is 328? 4. What is the area of a sector which is less than a semicircle, if the radius be 15, and the chord of its arc 12?

Half the chord is the sine of 23° 34' nearly. (Art. 28.) The whole arc, then, is

The area of the circle is

And 360° 47° 9:: 706.86

470911

706.86

92.6 the area of the sector.

5. If the arc ADB (Fig. 9.) be 240 degrees, and the radius of the circle 113, what is the area of the sector ADBC?

PROBLEM VI.

To find the area of a SEGMENT of a circle.

35. FIND THE AREA OF THE SECTOR WHICH HAS THE SAME ARC, AND ALSO THE AREA OF THE TRIANGLE FORMED BY THE CHORD OF THE SEGMENT AND THE RADII OF THE SECTOR.

THEN, IF THE SEGMENT BE LESS THAN A SEMI-CIRCLE,

SUBTRACT THE AREA OF THE TRIANGLE FROM THE AREA

OF THE SECTOR. BUT, IF THE SEGMENT BE GREATER THAN A SEMI-CIRCLE, ADD THE AREA OF THE TRIANGLE TO THE AREA OF THE SECTOR.

If the triangle ABC, (Fig. 9.) be taken from the sector AOBC, it is evident the difference will be the segment AOBP, less than a semi-circle. And if the same triangle be added to the sector ADBC, the sum will be the segment ADBP, greater than a semi-circle.

The area of the triangle (Art. 8.) is equal to the product of half the chord AB into CP, which is the difference between the radius and PO the height of the segment. Or CP is the cosine of half the arc BOA. If this cosine, and the chord of the segment are not given, they may be found from the arc

and the radius.

Ex. 1. If the arc AOB (Fig. 9.) be 120°, and the radius of the circle be 113 feet, what is the area of the segment AOBP?

In the right angled triangle BCP,

R: BC:: sin BCO : BP=97.86, half the chord. (Art. 28.)

The cosine PC-CO (Trig. 96, Cor.)
The area of the sector AOBC (Art. 34.)
The area of the triangle ABC=BP-PC
The area of the segment, therefore,

= 56.5
= 13371.67
5528.97

=

7842.7

2. If the base of a segment, less than a semi-circle, be 10 feet, and the radius of the circle 12 feet, what is the area of the segment?

The arc of the segment contains 491 degrees. (Art. 28.)

The area of the sector

The area of the triangle

And the area of the segment

=61.89

(Art. 34.)

=54.54

7.35 square feet.

3. What is the area of a circular segment, whose height is

19.2 and base 70?

Ans. 947.86.

4. What is the area of the segment ADBP, (Fig. 9.) if the base AB be 195.7, and the height PD 169.5?

Ans. 32272.*

36. The area of any figure which is bounded partly by arcs of circles, and partly by right lines, may be calculated, by finding the areas of the segments under the arcs, and then the area of the rectilinear space between the chords of the arcs and the other right lines.

Thus, the Gothic arch ACB, (Fig. 11.) contains the two segments ACH, BCD, and the plane triangle ABC.

Ex. If AB (Fig. 11.) be 110, each of the lines AC and BC 100, and the height of each of the segments ACH, BCD 10.435; what is the area of the whole figure?

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37. FROM THE AREA OF THE WHOLE CIRCLE, SUBTRACT THE TWO SEGMENTS ON THE SIDES OF THE ZONE.

If from the whole circle (Fig. 12.) there be taken the two segments ABC and DFH, there will remain the zone ACDH.

Or, the area of the zone may be found, by subtracting the segment ABC from the segment HBD: Or, by adding the two small segments GAH and VDC, to the trapezoid ACDH. See art. 36.

The latter method is rather the most expeditious in practice, as the two segments at the end of the zone are equal.

Ex. 1. What is the area of the zone ACDH, (Fig. 12.) if AC is 7.75, DH 6.93, and the diamater of the circle 8?

* For the method of finding the areas of segments by a table, see note D.

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2. What is the area of a zone, one side of which is 23.25, and the other side 20.8, in a circle whose diameter is 24? Ans. 208.

38. If the diameter of the circle is not given, it may be found from the sides and the breadth of the zone.

Let the center of the circle be at O. (Fig. 12.) Draw ON perpendicular to AH, NM perpendicular to LR, and HP perpendicular to AL. Then,

AN AH, (Euc. 3. 3.)
LM-LR, (Euc. 2. 6.)

MN={(LA+RH)
PA-LA-RH.

The triangles APH and OMN are similar, because the sides of one are perpendicular to those of the other, each to each. Therefore,

PH PA:: MN: MO

MO being found, we have ML-MO-OL.

And the radius CO= √OL2+CL2. (Euc. 47. 1.)

Ex. If the breadth of the zone ACDH (Fig. 12.) be 6.4, and the sides 6.8 and 6; what is the radius of the circle?

PA-3.4-3=0.4. And, MN (3.4+3)=3.2.

=

Then, 6.4 0.4 3.2 0.2-MO. And, 3.2-0.2—3—OL And the radius CO=v32+(3.4)2=4.534.

PROBLEM VIII.

To find the area of a LUNE or crescent.

39. FIND THE DIFFERENCE OF THE TWO SEGMENTS WHICH ARE BETWEEN THE ARCS OF THE CRESCENT and ITS CHORD.

If the segment ABC, (Fig. 14.) be taken from the segment ABD; there will remain the lune or crescent ACBD.

Ex. If the chord AB be 88, the height CH 20, and the height DH 40; what is the area of the crescent ACBD?

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