# Circles and Arcs $$\text{Diagram shows a circle and } 4 \text{ congruent arcs. The } 4 \text{ arcs are:}$$

$1. \text{ Arc BGHD centered at A}$

$2. \text{ Arc AEHC centered at B }$

$3. \text{ Arc BFED centered at C }$

$4. \text{ Arc AFGC centered at D }$

$\text{ AB, BC, CD, and DA are each }90^\circ \text{arcs}$

$\text{Given:}$

$r = \text{Radius of the circle}$

$R = \text{Radius of each arc}$

$\text{Let EF } = 2q$

$\text{Let } \theta = \angle \text{ACF }$

$\text{Using Pythagorean Theorem}$

$R^2 = r^2 + r^2 \Leftrightarrow R = \sqrt2 r$

$R^2 = q^2 + (q + r)^2 \text{(Using Pythagorean Theorem)}$

$(\sqrt2 r)^2 = q^2 + q^2 + 2qr + r^2$

$2r^2 = 2q^2 + 2qr + r^2$

$2q^2 + 2qr - r^2 = 0$

$q = \dfrac{-2r \pm \sqrt{ (2r)^2 - 4(2)(-r^2) } }{ 2(2) }$

$q = \dfrac{-2r \pm \sqrt{ 12r^2 } }{4}$

$q = \dfrac{-r \pm \sqrt3 (r)}{2}$

$q = \dfrac{ \sqrt3 - 1 }{2} (r) \text{ , } q > 0$

$\sin( \theta ) = \dfrac{q}{R} = \dfrac{ \frac{ \sqrt3 - 1 }{2} (r)}{ \sqrt2 r}$

$\theta = \arcsin\left(\dfrac{ \sqrt3 -1}{2 \sqrt2} \right)$

$\theta = \dfrac{\pi}{12} \text{ rad} = 15^\circ$

$\text{Shaded area} = 4 \times( \text{Area of sector ECF} - \text{Area of} \triangle \text{ECF} ) + \text{Area of square EFGH}$

$\text{Shaded area} = 4 \times\left[ \dfrac{1}{2} \times (\sqrt2 r)^2 (\dfrac{\pi}{12} ) - \dfrac{1}{2} \times (\sqrt2 r)^2 \sin\left( \dfrac{\pi}{12} \right) \right] + \left[2 \times \dfrac{ \sqrt3 - 1 }{2} (r) \right]^2$

$\text{Shaded area} = \dfrac{\pi r^2}{3} - \sqrt2 (\sqrt3 - 1)r ^2 + (\sqrt3 - 1)^2 r^2$

$\text{Shaded area} = \dfrac{\pi r^2}{3} + (\sqrt3 - 1 - \sqrt2 )(\sqrt3 - 1)r ^2$

$\dfrac{ \text{Shaded area} }{ \text{ Total area } } = \dfrac{ \dfrac{\pi r^2}{3} + (\sqrt3 - \sqrt2 - 1 )(\sqrt3 - 1)r ^2 }{\pi r^2}$

$\dfrac{ \text{Shaded area} }{ \text{ Total area } } = \dfrac{ \pi + 3(\sqrt3 - \sqrt2 - 1 )(\sqrt3 - 1) }{3 \pi } = \dfrac{1}{3} + \dfrac{ (\sqrt3 - \sqrt2 - 1 )(\sqrt3 - 1) }{\pi }$ Note by Lin Shun Hao
1 year, 3 months ago

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