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A farmer has 36 individual lengths of fence. Each is I m long.What is the smallest are that can be enclosed using all 36 fences?What is the largest area that can be enclosed using all 36 fences? |
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Answer» There are 36 isosceles triangles each of base 1 m , r is the length of the TWO remaining sides Cosine rule : 1^2 = r^2 +r^2 -2r^2 COS (10) = 2r^2(1-cos (10)) r^2 = 1/[2(1-cos (10)] , Eq1 Now area of ∆ = (1/2) r^2 SIN (10). sub in Eq1 for r^2 A(∆) = sin(10)/ [(1/4)(1-cos(10)] area of 36 ∆ = 9 sin(10)/ (1-cos (10)) =102.87 m^2 The area of the disc of radius r ,would have been A = πr^2 = π/[2(1-cos (10)] = 103.4 m^2 Area of the 36 sectors = 103.4 - 102.87 = 0.523 m^2 |
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