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Area of a Triangle
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The area K of any triangle equals one-half the product of its base and altitude. In general, if enough information about a triangle is known so that it can be solved, then its area can be found.
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12.2 Area Formulas
Cases I and II
Given two angles and a side of triangle ABC
The third angle is found using the fact that A B C 180 . The area of the triangle equals a side squared times the product of the sines of the angles including the side divided by twice the sine of the angle opposite the side; i.e., K a2 sin B sin C 2 sin A b2 sin A sin C 2 sin B c2 sin A sin B 2 sin C
For a derivation of these formulas, see Prob. 12.2. (See also Probs. 12.4 and 12.5.)
Case III
Given two sides and the angle opposite one of them in triangle ABC
A second angle is found by using the law of sines and the appropriate formula from Case I. Since there are sometimes two solutions for the second angle, there will be times when the area of two triangles must be found. (See Probs. 12.6 and 12.7.)
Case IV
Given two sides and the included angle of triangle ABC
The area of the triangle is equal to one-half the product of the two sides times the sine of the included angle; i.e., K
ab sin C
ac sin B
bc sin A (See also Probs. 12.8 and 12.9.)
For a derivation of these formulas, see Prob. 12.1.
CHAPTER 12 Area of a Triangle
Case V
Given the three sides of triangle ABC
The area of a triangle is equal to the square root of the product of the semiperimeter and the semiperimeter minus one side times the semiperimeter minus a second side times the semiperimeter minus a third side; i.e., K 2s(s a)(s b)(s c) where s
1 2 (a
[NOTE: The formula is known as Heron s (or Hero s) formula. For a derivation of the formula, see Prob. 12.3.]
(See Probs. 12.10 and 12.11.)
SOLVED PROBLEMS
12.1 Derive the formula K
1 2 bc
sin A. See Fig. 12.1.
c sin A. Thus,
Denoting the altitude drawn to side b of the triangle ABC by h, we have, from either figure, h 1 1 2 bh 2 bc sin A.
Fig. 12.1
12.2 Derive the formula K
From Prob. 12.1, K Then K
1 2 bc 1 2
c2 sin A sin B . 2 sin C
bc sin A; and by the law of sines, b sin A 1 c sin B c sin A 2 sin C c sin B . sin C
sin A
c2 sin A sin B . 2 sin C
12.3 Derive the formula K
2s(s
a)(s
a2 a2
b)(s
c), where s
1 2 (a
By the law of cosines, cos A
1 cos A cos A 1 1 b2 b2 c2 2bc c2 2bc
c2 2bc
2bc 2bc
so that
a2 a2 a2 (b (b 2bc c)2 2bc c)2 a2 (a (b b c c)(a 2bc a)(b 2bc b c c) a)
b2 c2 2bc b2 c2 2bc
and 1
a a and sin2 2 A
b b c
1 2 (1
c c a
2s, so a (a (a b b (a
b c) c) b
c 2c 2a
(a 2s 2s c)(a 4bc
b 2c 2a b
2b 2(s 2(s c) c)
a) 2(s b) # 2(s 4bc c) (s b)(s bc c)
cos A)
CHAPTER 12 Area of a Triangle
(b b)(s bc bc (s A c) c a)(b 4bc
1 cos2 2 A
1 2 (1
cos A) (s A
a) s(s A bc a) bc
2s # 2(s 4bc a) . Then 2s(s
s(s bc
Since 2 A < 90 , sin 2 A K
and cos 2 A c) A s(s
bc sin A
1 bc sin 2 A cos 2 A
b)(s bc
a)(s
b)(s
Case I 12.4 Find the area of triangle ABC, given c
23 cm, A
20 , and C
180 (A C) c2 sin A sin B 2 sin C 232 sin 20 sin 145 2 sin 15 200 cm2
Case II 12.5 Find the area of triangle ABC, given c
23 cm, A
20 , and B
180 (A B) 2 sin A sin B c 2 sin C 232 sin 20 sin 15 2 sin 145 41 cm2
Case III 12.6 Find the area of triangle ABC, given a
sin B b sin A a C K
112 m, b
219 m, and A
42 C Kr and Br 180 (A
138 . B) 22
219 sin 20 0.6688; B 112 180 (A B) 118 a2 sin B sin C 2 sin A 1122 sin 42 sin 118 2 sin 20 10,800 m2
a2 sin Br sin Cr 2 sin A 1122 sin 138 sin 22 2 sin 20 4600 m2
Case III 12.7 Find the area of triangle ABC, given A
sin B b sin A a C K
41 50 , a
123 ft, and b
31 30 .
96.2 ft.
96.2 sin 41 50r 0.5216; B 123 180 (A B) 106 40 2 sin A sin C b 2 sin B 96.22 sin 41 50r sin 106 40r 2 sin 31 31r 2 5660 ft
CHAPTER 12 Area of a Triangle
Case IV 12.8 Find the area of triangle ABC, given b
27 yd, c
1 2 bc
14 yd, and A
sin 43
sin A
1 2 (27)(14)
130 yd
Case IV 12.9 Find the area of triangle ABC, given a
14.27 cm, c
1 2 ac
17.23 cm, and B
sin86 14
86 14 .
sin B
1 2 (14.27)(17.23)
122.7 cm
Case V 12.10 Find the area of triangle ABC, given a
1 2 (a
5.00 m, b
b c) a)(s
7.00 m, and c
1 2 (5
10.0 m.
7 c) 7)(11
11 m.
2s(s 211(11 2264 16.2 m2
b)(s
5)(11
Case V 12.11 Find the area of triangle ABC, given a
1 2 (a
1.017 cm, b
1 2 (1.017
2.032 cm, and c
2.055)
2.055 cm.
c) a)(s
2.032 c)
2.552 cm
2s(s
b)(s
22.552(2.552 21.012392 1.006 cm2
1.017)(2.552
2.032)(2.552
2.055)
12.12 Find the area of an isosceles triangle with a base of 19.2 in and base angles of 23 10 each.
In Fig. 12.2, b 19.2 in, A B K 23 10 , and C 180 b2 sin A 2(23 10 ) 23 10 . Then 133 40
sin C 2 sin B
19.22 sin 23 10r sin 23 10r 2 sin 133 10r 39.4 in2
CHAPTER 12 Area of a Triangle
Fig. 12.2
12.13 A painter needs to find the area of the gable end of a house. What is the area of the gable if it is a triangle with two sides of 42.0 ft that meet at a 105 angle
In Fig. 12.3, a 42.0 ft, b 42.0 ft, and C K
1 2 1 2 (42)(42)
105 .
ab sin C sin 105
852 ft2
Fig. 12.3
12.14 Three circles with radii 3.0, 5.0, and 9.0 cm are externally tangent. What is the area of the triangle formed by joining their centers
In Fig. 12.4, a 8 cm, b 12 cm, and c s K
1 2 (a
14 cm. b c) a)(s 17 cm b)(s c) 12)(17 14)
2s(s 217(17 22295 48 cm2
8)(17
Fig. 12.4
CHAPTER 12 Area of a Triangle
12.15 In a quadrangular field ABCD, AB runs N62 10 E 11.4 m, BC runs N22 20 W 19.8 m, and CD runs S40 40 W 15.3 m. DA runs S32 10 E but cannot be measured. Find (a) the length of DA and (b) the area of the field.
Fig. 12.5
In Fig. 12.5, SN is the north-south line through D; the points E, F, and G are the feet of the perpendiculars to this line through A, B, and C, respectively; and the lines AH and CI are perpendicular to BF. (a) FB FI 9.97 FB EA FH FB 17.49 Since EA (b) DA sin 32 10 , DA Area ABCD IB GC IB 19.8 sin 22 20 17.49 EA HB; hence 17.49 10.08 7.41
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