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101.6 213.9 214 m2
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1 2 (11.4)(19.8)
12.16 Prove that the area of a quadrilateral is equal to half the product of its diagonals and the sine of the included angle. See Fig. 12.6(a).
Let the diagonals of the quadrilateral ABCD intersect in O; let be an angle of intersection of the diagonals; and let O separate the diagonals into segments of length p and q, r and s, as in the figure. Area ABCD area AOB
1 2 rp
area AOD
1 2 qr
area BOC )
1 2 1 2
area DOC )
1 2 1 2 qs
sin qr
sin (180 qs) sin
ps sin (180r2 q)(r s) sin
1 2 (pr
(BD)(AC) sin .
CHAPTER 12 Area of a Triangle
Fig. 12.6
12.17 Prove that the area K of the smaller segment (shaded) of a circle of radius r and center O cut off by the chord AB of Fig. 12.6(b) is given by K 1r2( sin ), where radians is the central angle inter2 cepted by the chord.
The required area is the difference between the area of sector AOB and triangle AOB. is, The area S of the sector AOB is to the area of the circle as the arc AB is to the circumference of the circle; that S ru 1 and S 2r2 . 2pr pr 2 The area of triangle AOB Thus,
r # r sin u
1 2 2r u
1 2 2r 1 2 2r
sin u.
1 2 2 r (u
sin u
sin u)
12.18 Three circles with centers A, B, and C have respective radii 50, 30, and 20 in and are tangent to each other externally. Find the area of the curvilinear triangle formed by the three circles.
Let the points of tangency of the circles be R, S, and T as in Fig. 12.7. The required area is the difference between the area of triangle ABC and the sum of the areas of the three sectors ART, BRS, and SCT. Since the join of the centers of any two circles passes through their point of tangency, a BC 50, b CA 70, and c AB 80 in. Then s and K cos A cos B cos C Area ART (a b c) 100 a)(s (70)2 (50)2 (50)2 s a b)(s 50 c) s b 30 s c 20 1000 23 0.667 rad 1.047 rad 1.428 rad
1 2 2(20) (1.428)
area ABC b2 a2 a2 c2 2bc c2 2ac b2 2ab
2s(s a2 b2 c2
2100 (50) (30) (20) 0.7857 0.5 0.1429
1 2 2(30) (1.047)
(80)2 (50)2 2(70)(80) (80)2 (70)2 2(50)(80) (70)2 (80)2 2(50)(70)
A B C
38.2 60 81.8
1 2 2r
1 2 2(50) (0.667)
833.75, area BRS
471.15, area CST
285.60, and their sum is 1590.50. The required area is 1732 1590.50
141.50 or 142 in2.
Fig. 12.7
CHAPTER 12 Area of a Triangle
12.19 Find the area of the triangle ABC, given A 37 10 , C 62 30 , and b
34.9. See Fig. 12.8.
Fig. 12.8
B This is a Case II triangle, and K K K K
80 20 .
b2 sin C sin A . 2 sin B (34.9)2 sin 62 30r sin 37 10r 2 sin 80 20r 331.05 331 square units
12.20 Find the area of the triangle ABC, given b
28.6, c
44.3, and B
This is a Case III triangle in which there may be two solutions. See Fig. 12.9. sin C sin C sin C C 37.8 B C C c sin B b 44.3 sin 23.3 28.6 0.6127 180 23.3 37.8 142.2 142.2 165.5 < 180
Therefore, C is a solution. A Area of 180 (B C) 118.9 A 180 (B C) 14.5 c2 sin Ar sin B 2 sin Cr
ABC is K
c2 sin A sin B 2 sin C
Area of
A BC is K
K K K
(44.3)2 sin 1189.9 sin 23.3 2 sin 37.8 554.39 554 square units
Kr Kr Kr
(44.3)2 sin 14.5 sin 23.3 2 sin 142.2 158.55 159 square units
Two triangles are determined, their areas being 554 and 159 square units, respectively.
Fig. 12.9
CHAPTER 12 Area of a Triangle
16.4, b
sin 27.3
12.21 Find the area of the triangle ABC, given a
This is a Case IV triangle, and K K K K The area is 209 square units.
55.7, and C
ab sin C. See Fig. 12.10.
1 2 (16.4)(55.7)
209.48 209
Fig. 12.10
12.22 Find the area of the triangle ABC, given a
This is a Case V triangle, and K s a b c 2s s
1 2 (a
255, b
a)(s
290, and c
c). 2s(s K
419. See Fig. 12.11.
a)(s b)(s c)
2s(s b s s s s c) a b c 482
b)(s
255 290 419 964 482
227 192 63
K K K
2(482)(227)(192)(63) 36,379.551 36,400
The area is 36,400 square units.
Fig. 12.11
SUPPLEMENTARY PROBLEMS
Find the area of the triangle ABC, given:
12.23 b 12.24 a 12.25 a 12.26 A 12.27 b 13 ft, a 23.3 cm, c 4.1 m, b 65 , B 23.84, c 27 ft, C 85 121.0 Ans. 175 ft2 Ans. 215 cm2 Ans. 11 m2 Ans. 38 yd2 Ans. 324.5 square units
21.5 cm, B 5.2 m, c
6.7 m
35 , c
12 yd 50 32
35.26, A
CHAPTER 12 Area of a Triangle
12.28 a 12.29 a 12.30 b 12.31 a 12.32 a 456.3, b 512.3, B 444.8, A 384.2, b 28.16, b 586.8, C 52 15 , C 110 16 , B 492.8, c 60.15, c 28 17 63 46 30 10 677.8 51.17 Ans. 63,440 square units Ans. 103,600 square units Ans. 117,600 square units Ans. 93,080 square units Ans. 718.6 square units
12.33 To find the area of a triangular lot, the owner starts at one corner and walks due east 215 m to a second corner. After turning through an angle of 78.4 , the owner walks 314 m to the third corner. What is the area of the lot Ans. 33,100 m2
12.34 An artist wishes to make a sign in the shape of an isosceles triangle with a 42 vertex angle and a base of 18 m. What is the area of the sign Ans. 211 m2
12.35 Point C has a bearing of N28 E from point A and a bearing of N12 W from point B. What is the area of triangle ABC if B is due east of A and 23 km from A Ans. 355 km2
12.36 Three circles have radii of 7.72, 4.84, and 11.4 cm. If they are externally tangent, what is the area of the triangle formed by joining their centers Ans. 101 cm2
12.37 A woman hikes 503 m, turns and jogs 415 m, turns again, and runs 365 m returning to her starting point. What is the area of the triangle formed by her path Ans. 74,600 m2
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