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§ 3. REGULAR POLYGONS.

226. Formuloe for any polygon circumscribing a circle.

N

Х

B

8

Let AB be a side of a polygon circumscribing a circle whose centre is O. Let

= 1 sum of sides of polygon. r = radius of inscribed circle.

S= area of polygon.
Draw OX=r perpendicular to AB.
Then

A AOB = 10X. AB.
:. area of polygon = } (radius x perimeter)

S
i.e. S=s.r and r=

(1).

Х

=

8

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А A B sin } A sin B Also ge=0X=0A sin

.(2). 2 sin } (A + B) Thus the radius is determined by one side and the adjacent angles.

227. The perimeter and area of a regular polygon in terms of the radius of the inscribed circle.

Take the figure of last article.

Then the polygon being regular, and containing n sides, (say)

N LAOB=4 right-angles, :: LAOB=27/n.

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228. The perimeter and area of a regular polygon in terms of the radius of the circumscribing circle.

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Let AB be a side of a regular polygon of n sides. Let the centre of the circumscribing circle be 0. Draw 0X perpendicular to AB.

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EXAMPLES X.

=

=

[The following notation is employed :--area of ABC=A; inradius =r; circumradius=R; cosine-radius=p; eradii=r1, 12, 13; ex-cosine radii=

=P1, P2P3; half-sum of sides=8; and points are lettered as in the last chapter.]

Prove the following statements.
1. A8 = abc cos | A cos ? B cos įC.
2. A(8 - a)=abc cos } A sin ?B sin }C.
3. A (a + b) = abc cos C cos } (A B).
4. A (a - b) = abc sin C sin }(A - B).

(
5. 4A=a? sin 2B + b2 sin 2A.
6. 2A sin (A B) = (a62) sin A sin B.

16A’= (a + b + c)2 -2 (a* +64 + c). 8. abc = 4RA = 48rR=p (a? + b2 + c). 9. A = Rr (sin A + sin B + sin C). 10. pe = A tan. A tan} B tan 7 C. 11. 4R cos į A cos į B cos C. 12. abc r= 4R (: - a) (8 b) (8 c). 13. 2r + 2R = a cot A + b cot B + c cot C. 14. 4R sin A sin B sin c = a cos A + b cos B + c cos C. 15. A 2 R2 sin A sin B sin C. 16. 2A =p (a sin A + b sin B + c sin C). Р

sin A sin B sin C 17.

R 1 + cos A cos B cos C

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=

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tan w.

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18. сot w = cosec A cosec B cosec C + cot A cot B cot C.
19. cosec w= cosec? A + cosec? B + cosec C.

1 2 2 2 Δ sa
20.

+

+ a ū

Rp2 RA

+

с

le 1°

21.

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22.

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1 1 2a2 272 2c2

+ +
po R 6c ° ca? a>b2
a

5 с 1 23.

+
bc
ca ab

р 24. prarars = A.

25. ga + 7x2 + x = 8.
1 1 1 1
26.

+ +
11 72 13
++13

p=4R.
ri r2 73
28.

1 1
+ +
bc
ca ab r

2R'
rr'i

А 29.

= tan

27.

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31.

+

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=

+

+

+

abc 30. P1 = 12 + c2 - a

R tan A.
1 1 1 1

+
Pi P2 P3 р

1 1 1 1 32.

+ +
P2P3 P3P1 P1P2

R2
33.
P2P3 + P3P1 + P1P2

- 3R2 + R* (cosA + cos2 B + cos2 C) sec A sec B sec C.

1 1 1 1 bc + ca + ab 34. r273 rzli 7172

A? 35. 4r (rı +r, + r) = 2bc + 2ca + 2ab ao 12 c%. brr cr 1/2

T1r23 36.

r2 +r3 73 +11 Yu + 12 J(rar: + rzri + rir.) 37. J.(arz) + (br.) + (crz) 1(abc/r)

= 8 N (Rs) sin (45° - 4A) sin (45° - B) sin (45° - 1C). 38. 2rp (T. +rg + rz) = 2ps: - abc.

1
A2
13

1
=

+

72

+

arr3

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=

=

=P

cotw.

40. rcot} A = r, cot B=r; cot C =r cot 1 A cot B cot{C. 41.

Pi cot A = P2 cot B=Pz cot C = 42.

Px + P3 = {a sec B sec C. 43.

1*2 + 73 = a cot 1A.

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R 46. In no triangle is r greater than

2

nor p greater than R 13

47. Given the inradius r, the circumradius R, and the area A of a triangle, show that its sides are the roots of the equation

208 — 22-A/r + < (pu2 + 4rR+Ao/m2) = 4AR. 48. Given the half-sum of sides s, the half-sum of squares on sides of, and the area A of a triangle, show that the radii of its escribed circles are the roots of the equation

208 + 38 = 2* (8o – )/A+A. 49. If the sides of a triangle are the roots of the equation 208 + px = qace + v, its cosine-radius is v/(qo - 2p) and the rectangle contained by its inradius and circumradius is v/29.

50. If the squares on the sides of a triangle are the roots of the equation + px = qace + , its cosine-radius is v/q and its cir

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cumradius is

(4p - q?) 51. In an equilateral triangle, the incircle coincides with the nine-points circle, the cosine-circle with the Lemoine circle, and the centres of the escribed circles with those of the ex-cosine

13 circles. Also R=

, P
3
2 3?

Ag Pi = a, w = 30°.

a

a

a

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