Geometry with approximations

Given: Area of the shaded region (green) =14×Area of the whole circle (Diagram not to scale)\text{Area of the shaded region (green) } = \dfrac{1}{4} \times \text{Area of the whole circle} \text{ (Diagram not to scale)}

The vertical lines(each length a) are drawn parallel and equidistant from the axis of symmetry of the figure.\text{The vertical lines(each length a) are drawn parallel and equidistant from the axis of symmetry of the figure.}

The angle A is between two radii of the circle joining with the left line.\text{The angle A is between two radii of the circle joining with the left line.}

Let R = radius of the circle\text{Let R = radius of the circle}

12×(R)2×(A)12×(R)2×sin(A)=1234×π(R)2\dfrac{1}{2} \times (R)^2 \times (A) - \dfrac{1}{2} \times (R)^2 \times \sin(A) = \dfrac{1}{2} \cdot \dfrac{3}{4} \times \pi (R)^2 [Region of one unshaded part]\text{[Region of one unshaded part]}

(A)sin(A)=3π4 (A) - \sin(A) = \dfrac{3\pi}{4}

sin(A)=A3π4 \sin(A) = A - \dfrac{3\pi}{4}

a=2×sin(A2)×Ra = 2 \times \sin{ \left( \dfrac{A}{2} \right) } \times R

Perimeter of the unshaded region =2×(a+(R)(A)) = 2 \times ( a + (R) (A) )

Using (3rd order) Taylor Series approximation:\text{Using (3rd order) Taylor Series approximation:}

AA33!A3π4A - \dfrac{A^3}{3!} \approx A -\dfrac{3\pi}{4}

A39π2 A^3 \approx \dfrac{9\pi}{2}

A2.417988 radA \approx 2.417988 \text{ rad}

Perimeter of unshaded region2×[(2×sin(2.4179882)×R)+(2.417988)(R)]\text{Perimeter of unshaded region} \approx 2 \times \left[ \left( 2 \times \sin{ \left( \dfrac{2.417988}{2} \right) } \times R \right) + ( 2.417988)(R) \right]

Perimeter of unshaded region8.57702R\text{Perimeter of unshaded region} \approx 8.57702 R

Note : The answer is inaccurate because the approximation only works well near  \text{Note : The answer is inaccurate because the approximation only works well near }

x=0x = 0.  Nevertheless, Taylor Series is useful to approximate difficult calculations.\text{ Nevertheless, Taylor Series is useful to approximate difficult calculations.}

Note by Lin Shun Hao
1 month ago

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