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ss_7_4

# ss_7_4 - 1 2(4(6 sin 60 o = 12 √ 3 2 = 6 √ 3 Example...

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Section 7.4 The Area of a Triangle The area of a triangle is given by the following. A = 1 2 bh where b is the length of base and h is the length of the altitude. A = 1 2 ab sin γ ( h = a sin γ ) Example. Find the area of the triangle having a = 3, b = 4 and γ = 30 o . Solution. A = 1 2 ab sin γ = 1 2 (3)(4) sin 30 o = 1 2 (3)(4)( 1 2 ) = 3 Herron’s Formula Herron’s Formula is a convenient formula for finding the area of a triangle, when the lengths of all sides are know. A = p s ( s - a )( s - b )( s - c ) where s = 1 2 ( a + b + c ) Example. Find the area of the triangle if a = 3, b = 3 and c = 2. Solution. A = p s ( s - a )( s - b )( s - c ) where s = 1 2 ( a + b + c ) = 1 2 (3 + 3 + 2) = 4. Thus A = p 4(4 - 3)(4 - 3)(4 - 2) = 8 = 2 2 Alternate solution: By the law of cosines, c 2 = a 2 + b 2 - 2 ab cos γ . Hence, 4 = 9 + 9 - 2(3)(3) cos γ and cos γ = 14 18 = 7 9 . Thus sin γ = 32 9 = 4 2 9 . So A = 1 2 ab sin γ = 1 2 (3)(3) 4 2 9 = 2 2 Example. Find the area of the triangle having a = 6, b = 4 and γ = 60 o . Solution. A = 1 2 bh = 1

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Unformatted text preview: 1 2 (4)(6) sin 60 o = 12( √ 3 2 ) = 6 √ 3 Example. Find the area of the shaded segment shown below. 1 Solution. The pie-shaped sector, A s , has area equal to 70 360 times the area of the circle. Thus A s = 70 360 πr 2 = 7 36 π 8 2 = 112 9 π The area, A , of the shaded segment is A s minus the area of the triangle. Hence, A = A s-1 2 (8) 2 sin 70 o = 112 9 π-32 sin 70 o = 9 . 03 7.4 EXERCISES 1-2. In Figures 1, 7, ﬁnd the area of the triangle. Round answers to two decimal places. In Problems 3-5, ﬁnd the area of the triangle. Round answers to two decimal places. 3. a = 3 , b = 4 , γ = 40 o 4. a = 3 , c = 2 , β = 110 o 5. a = 5, b = 8, c = 9 2...
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