Sidebilder
PDF
ePub

§ 3. REGULAR POLYGONS.

226. Formulæ for any polygon circumscribing a circle.

N

A X

B

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

8 =

Let

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

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

A AOB = 10X. AB.
:. area of polygon = 1 (radius x perimeter)

S
i.e. S=8.r and r=-

(1).

[ocr errors]

A AB sin ] A sin B Also g=0X = OA 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.

[blocks in formation]

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

[blocks in formation]

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

[blocks in formation]

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, P2, P3; half-sum of sides=8; and points are lettered as in the last chapter.]

Prove the following statements.
1. As = abc cos į A cos 1B 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)=(a- 62) sin A sin B.
7. 16A?= (a + b2 +ca) - 2 (a* + 24 + c^).
8. abc = 4RA = 48rR=p (a+ b2 + c).
9. A= Rr (sin A + sin B + sin C).
10. me = A tan 1 A tan B tan 1 C.
11. 8 = 4R cosA cos } B cos } C.
12. abc r= 4R (8 a) (8 6) (8c).
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=2Ro 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

= tan w.
R 1 + cos A cos B cos C

+

17.

18. cot w = cosec A cosec B cosec C + cot A cot B cot C.
19. cosec w= coseco A + cosec2 B + cosec® C.
1 2 2 2

ga
7

Rp2 RA

20.

+

+

[ocr errors]

+

C

a

[merged small][merged small][merged small][merged small][merged small][ocr errors]

a

.

+

=

1 1 2a2 262 2c2 22.

+
p R2 6c ca?a 62
6

C 1 23.

+
bc
ca
ab

р
24.
prir273 = A?.

25.

r273 + rzr, + rir, = sa. 1 1 1 1 26.

+ 11 12 13 27. 11 +7, +13

-p= 4R. ri r2 13 1 1 28.

+ + bc

ca ab r 2R rri

A 29.

tan 7273

2

abc
30.
Pi=

R tan A.
72 + ca - a
1 1 1 1

+ +
Pi P2 P3 p.

1 1 1 1 32.

+ +
P2P3 P321 P1P2

R2
33.
P2P3 + P3Pi + P1P2

3R2 + R2 (cos? A + cos2 B + cos2 C) sec A sec B sec C. 1 1 1 1 bc + ca + ab 34. 7311 7172

2 35. 4r (ru + r+rs) = 2bc + 2ca + 2ab - a2 - 62 co. arr3 bryri cr,r2

1,7273 36.

r2 + 73 ra tri 7i+r, J(r.rs + rşrı +rır) 37. Jarı) + (brz) + J(crz) - J(abc/r)

= 8 / (Rs) sin (45° - 4A) sin (45° - 4B) sin (45° – 2C). 38. 2rp (ri+r2+ rz) = 2psa abc.

31.

[ocr errors]

+

+

+

[ocr errors]
[merged small][ocr errors][subsumed][subsumed]

41.

[ocr errors]

cot w.

[ocr errors]

40. } =}

ricot } A = r, cot B=r; cot C = r сot 1 A cot } B cotįC.

Picot A = P2 cot B=Pz cot C 42. P2+ P3= a sec B sec C. 43. 7, +rz = a cot A.

SI

+

+

44.

р
Pi + P2 + P3 =

1 -2RpA
2 cos A
2 cos B

2 cos C 45. р cot : B + cot 1C cot }C + cot A cot } A + cot B

a sin A + b sin B + csin C
a + b + c

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

203 — 22-A/1 + x (902 + 4rR + 4°/-2) = 4AR. 48. Given the half-sum of sides s, the half-sum of squares on sides o?, and the area A of a triangle, show that the radii of its escribed circles are the roots of the equation

2/8 + 38 = 2* (82 - 0°)/A + A. 49. If the sides of a triangle are the roots of the equation 208 + po = qacé + v, its cosine-radius is v/(qo 2p) and the rectangle contained by its inradius and circumradius is v/2q.

50. If the squares on the sides of a triangle are the roots of the equation 3* + px =qxo + v, its cosine-radius is vla and its circumradius is

(4p - 9%) 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=

3'
2 13'

A, P1 = a, w = 30°.

a

a

[ocr errors]
[ocr errors]

13

« ForrigeFortsett »