# Geometry with approximations Given: $\text{Area of the shaded region (green) } = \dfrac{1}{4} \times \text{Area of the whole circle} \text{ (Diagram not to scale)}$

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

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

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

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

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

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

$a = 2 \times \sin{ \left( \dfrac{A}{2} \right) } \times R$

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

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

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

$A^3 \approx \dfrac{9\pi}{2}$

$A \approx 2.417988 \text{ rad}$

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

$\text{Perimeter of unshaded region} \approx 8.57702 R$

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

$x = 0$. $\text{ Nevertheless, Taylor Series is useful to approximate difficult calculations.}$ Note by Lin Shun Hao
7 months, 2 weeks ago

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