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10.2 Relationships of Segments in Regular Polygons of 3, 4, and 6 Sides
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In the regular hexagon, square, and equilateral triangle, special right triangles are formed when the apothem r and a radius R terminating in the same side are drawn. In the case of the square we obtain a 45 -45 -90 triangle, while in the other two cases we obtain a 30 -60 -90 triangle. The formulas in Fig. 10-4 relate the lengths of the sides and radii of these regular polygons.
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Fig. 10-4
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CHAPTER 10 Regular Polygons and the Circle
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10.3 Applying line relationships in a regular hexagon In a regular hexagon, (a) find the lengths of the side and apothem if the radius is 12; (b) find the radius and length of the apothem if the side has length 8.
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(a) Since R (b) Since s 12, s 8, R s R 12 and r 8 and r
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6 23. 4 23.
10.4 Applying line relationships in a square In a square, (a) find the lengths of the side and apothem if the radius is 16; (b) find the radius and the length of the apothem if a side has length 10.
Solutions
(a) Since R (b) Since s 16, s 10, r R 22
1 2s
16 22 and r s 5 and R 22
1 2s 1 2s
8 22. 5 22.
10.5 Applying line relationships in an equilateral triangle In an equilateral triangle, (a) find the lengths of the radius, apothem, and side if the altitude has length 6; (b) find the lengths of the side, apothem, and altitude if the radius is 9.
Solutions
(a) Since h (b) Since R 6, we have r 9, s R 23
1 3h
2; R 9 23 ; r
2 3h 1 2R
4; and s
1 4 2;
R 23
4 23. 132.
and h
10.3 Area of a Regular Polygon
The area of a regular polygon equals one-half the product of its perimeter and the length of its apothem. As shown in Fig. 10-5, by drawing radii we can divide a regular polygon of n sides and perimeter p ns 1 1 into n triangles, each of area 1rs. Hence, the area of the regular polygon is n(1rs) 2 2 2 nsr 2 pr.
Fig. 10-5
CHAPTER 10 Regular Polygons and the Circle
SOLVED PROBLEMS
10.6 Finding the area of a regular polygon (a) Find the area of a regular hexagon if the length of the apothem is 5 23. (b) Find the area of a regular pentagon to the nearest integer if the length of the apothem is 20.
Solutions
(a) In a regular hexagon, r Then A
1 2 pr 1 2 (60) 1 2s
23. Since r 150 23.
5 23, s
1 2m
10 and p
6(10)
(5 23) 360 /5
1 2 (5)
(b) In Fig. 10-6, m/AOE or s 40 tan 36 . Now A
1 2 pr 1 2 nsr
72 and m/AOF 1453.
/AOE
36 . Then tan 36
1 2 s /20
s/40
(40 tan 36 ) (20)
Fig. 10-6
10.4 Ratios of Segments and Areas of Regular Polygons
PRINCIPLE 1: PRINCIPLE 2:
Regular polygons having the same number of sides are similar. Corresponding segments of regular polygons having the same number of sides are in proportion. Segments here includes sides, perimeters, radii or circumferences of circumscribed or inscribed circles, and such. Areas of regular polygons having the same number of sides are to each other as the squares of the lengths of any two corresponding segments.
PRINCIPLE 3:
SOLVED PROBLEMS
10.7 Ratios of lines and areas of regular polygons (a) In two regular polygons having the same number of sides, find the ratio of the lengths of the apothems if the perimeters are in the ratio 5:3. (b) In two regular polygons having the same number of sides, find the length of a side of the smaller if the lengths of the apothems are 20 and 50 and a side of the larger has length 32.5. (c) In two regular polygons having the same number of sides, find the ratio of the areas if the lengths of the sides are in the ratio 1:5. (d) In two regular polygons having the same number of sides, find the area of the smaller if the sides have lengths 4 and 12 and the area of the larger is 10,260.
CHAPTER 10 Regular Polygons and the Circle
Solutions
(a) By Principle 2, r : r (b) By Principle 2, s : s (c) By Principle 3, (d) By Principle 3, A Ar A Ar p: p 5 : 3. 20 : 50 and s 1 . 25 Q 4 2 R and A 12 1140. 13.
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