Question

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?​

Answer 1

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