Circles and Arcs

\(\text{Diagram shows a circle and } 4 \text{ congruent arcs. The } 4 \text{ arcs are:}\)

1. Arc BGHD centered at A1. \text{ Arc BGHD centered at A}

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

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

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

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

Given:\text{Given:}

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

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

Let EF =2q\text{Let EF } = 2q

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

Using Pythagorean Theorem \text{Using Pythagorean Theorem}

R2=r2+r2R=2rR^2 = r^2 + r^2 \Leftrightarrow R = \sqrt2 r

R2=q2+(q+r)2(Using Pythagorean Theorem) R^2 = q^2 + (q + r)^2 \text{(Using Pythagorean Theorem)}

(2r)2=q2+q2+2qr+r2(\sqrt2 r)^2 = q^2 + q^2 + 2qr + r^2

2r2=2q2+2qr+r22r^2 = 2q^2 + 2qr + r^2

2q2+2qrr2=02q^2 + 2qr - r^2 = 0

q=2r±(2r)24(2)(r2)2(2)q = \dfrac{-2r \pm \sqrt{ (2r)^2 - 4(2)(-r^2) } }{ 2(2) }

q=2r±12r24q = \dfrac{-2r \pm \sqrt{ 12r^2 } }{4}

q=r±3(r)2q = \dfrac{-r \pm \sqrt3 (r)}{2}

q=312(r) , q>0q = \dfrac{ \sqrt3 - 1 }{2} (r) \text{ , } q > 0

sin(θ)=qR=312(r)2r \sin( \theta ) = \dfrac{q}{R} = \dfrac{ \frac{ \sqrt3 - 1 }{2} (r)}{ \sqrt2 r}

θ=arcsin(3122) \theta = \arcsin\left(\dfrac{ \sqrt3 -1}{2 \sqrt2} \right)

θ=π12 rad=15 \theta = \dfrac{\pi}{12} \text{ rad} = 15^\circ

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

Shaded area=4×[12×(2r)2(π12)12×(2r)2sin(π12)]+[2×312(r)]2\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

Shaded area=πr232(31)r2+(31)2r2\text{Shaded area} = \dfrac{\pi r^2}{3} - \sqrt2 (\sqrt3 - 1)r ^2 + (\sqrt3 - 1)^2 r^2

Shaded area=πr23+(312)(31)r2\text{Shaded area} = \dfrac{\pi r^2}{3} + (\sqrt3 - 1 - \sqrt2 )(\sqrt3 - 1)r ^2

Shaded area Total area =πr23+(321)(31)r2πr2\dfrac{ \text{Shaded area} }{ \text{ Total area } } = \dfrac{ \dfrac{\pi r^2}{3} + (\sqrt3 - \sqrt2 - 1 )(\sqrt3 - 1)r ^2 }{\pi r^2}

Shaded area Total area =π+3(321)(31)3π=13+(321)(31)π\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
2 months, 1 week ago

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