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r : r ; thus, s : 32.5 s 2 Q R sr 1 2 Q R 5
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A s 2 Q R . Then sr 10,260
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10.5 Circumference and Area of a Circle
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p (pi) is the ratio of the circumference C of any circle to its diameter d; that is, p C pd or
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C/d. Hence,
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Approximate values for p are 3.1416 or 3.14 or 7 . Unless you are told otherwise, we shall use 3.14 for p in solving problems. A circle may be regarded as a regular polygon having an infinite number of sides. If a square is inscribed in a circle, and the number of sides is continually doubled (to form an octagon, a 16-gon, and so on), the perimeters of the resulting polygons will get closer and closer to the circumference of the circle (Fig. 10-7).
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Fig. 10-7
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Thus to find the area of a circle, the formula A we get A
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can be used with C substituted for p; doing so, pr2
1 2 (2pr) (r)
All circles are similar figures, since they have the same shape. Because they are similar figures, (1) corresponding segments of circles are in proportion and (2) the areas of two circles are to each other as the squares of their radii or circumferences.
SOLVED PROBLEMS
10.8 Finding the circumference and area of a circle In a circle, (a) find the circumference and area if the radius is 6; (b) find the radius and area if the circumference is 18p; (c) find the radius and circumference if the area is 144p. (Answer both in terms of p and to the nearest integer.)
CHAPTER 10 Regular Polygons and the Circle
Solutions
(a) r (b) C (c) A 6. Then C 2pr 12p 38 and A pr2 36p 36(3.14) 9. Then A 12. Then C 113. pr2 2pr 81p 24p 254. 75.
18p. Since C 144p. Since A
2pr, we have 18p pr2, we have 144p
2pr and r pr2 and r
10.9 Circumference and area of circumscribed and inscribed circles Find the circumference and area of the circumscribed circle and inscribed circle (a) of a regular hexagon whose side has length 8; (b) of an equilateral triangle whose altitude has length 9 23. (See Fig. 10-8.)
Solutions
(a) Here R Also r (b) Here R Also r s 8. Then C 2pR
1 4 23. Then 2 R 23 2 h 6 23. Then C 3 1 3 23. Then C 3h
16p and A pR2 64p. C 2pr 8p 23 and A pr2
48p.
2pR 12p 23 and A pR2 108p. 2pr 6p 23 and A pr2 27p.
Fig. 10-8
10.10 Ratios of segments and areas in circles (a) If the circumferences of two circles are in the ratio 2:3, find the ratio of the diameters and the ratio of the areas. (b) If the areas of two circles are in the ratio 1:25, find the ratio of the diameters and the ratio of the circumferences.
Solutions
(a) d dr C Cr A Ar 2 A and 3 Ar d 2 1 Q R , 25 dr Q C 2 R Cr 2 2 Q R 3 4 . 9 1 C . Also, 5 Cr d dr 1 . 5
(b) Since
d 2 d Q R and dr dr
10.6 Length of an Arc; Area of a Sector and a Segment
A sector of a circle is a part of a circle bounded by two radii and their intercepted arc. Thus in Fig. 10-9, the shaded section of circle O is sector OAB.
CHAPTER 10 Regular Polygons and the Circle
A segment of a circle is a part of a circle bounded by a chord and its arc. A minor segment of a circle is the smaller of the two segments thus formed. Thus in Fig. 10-10, the shaded section of circle Q is minor segment ACB.
Fig. 10-9
PRINCIPLE 1:
Fig. 10-10
PRINCIPLE 2:
n In a circle of radius r, the length l of an arc of measure n equals of the circumference 360 pnr n of the circle, or l . 2pr 360 180 n In a circle of radius r, the area K of a sector of measure n equals of the area of the 360 n 2. circle, or K pr 360 Area of a sector of n Area of the circle length of an arc of measure n circumference of the circle n 360
PRINCIPLE 3: PRINCIPLE 4:
The area of a minor segment of a circle equals the area of its sector less the area of the triangle formed by its radii and chord. If a regular polygon is inscribed in a circle, each segment cut off by the polygon has area equal to the difference between the area of the circle and the area of the polygon divided by the number of sides.
PRINCIPLE 5:
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